Physics
AP Physics 5: Circular and Rotational Motion Name __________________________
A. Circular Motion
1. constant perimeter (tangential) speed: vt = 2πr/T
a. distance = circumference of the circle: 2πr
b. time = time for one revolution: T (period)
2. constant inward (centripetal) acceleration: ac = v2/r
3. centripetal force, Fc = mac = mv2/r
a. turning on a road problems
| v = 2πr/T |
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|ac |
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|when the road is horizontal: Fc = Ff = μsmg |
|roads are banked in order to reduce the amount of friction (component of|
|the Fg is || to Fc) |
b. horizontal loop problem (mass on a string)
| Ft-x = Fc = mv2/r |
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|θ |
|Ft Ft-y = Fg = mg |
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|v = 2πr/T |
|Ft = (Fc2 + Fg2)½ |
|tanθ = Fg/Fc (θ is measured from horizontal) |
c. vertical loop problem (mass on a string)
|top: Fnet = Fc = Ft + Fg ∴ Ft = Fc – Fg |
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|Fg Ft |
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|Fg Ft |
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|bottom: Fnet = Fc = Ft – Fg ∴ Ft = Fc + Fg |
|if on a roller coaster: Fn = Ft |
4. Newton's law of universal gravity, Fg = GMm/r2
a. G = 6.67 x 10-11 N•m2/kg2
b. M = mplanet and m = msatellite
c. r is the distance, measured from center to center
d. g = GM/r2
e. Fg = Fc: GMm/r2 = mv2/r ∴ v = (GM/r)½
v = 2πr/T
m Fg = Fg M
r
| |Mass (kg) |Radius (m) |r from Earth (m) |
|Earth |5.98 x 1024 |6.38 x 106 | |
|Moon |7.35 x 1022 |1.74 x 106 |3.84 x 108 |
|Sun |1.99 x 1030 |6.96 x 108 |1.50 x 1011 |
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B. Newton's Laws—Rotation
1. torque, τ = r⊥Fr (tau—Greek letter for t)
rotation
r 90o Fr
a. r = perpendicular distance from axis of rotation to rotating force Fr
b. when r is not perpendicular to Fr, then τ = rFrsinθ
c. torque units are m•N (not N•m—work)
2. First Law: Object remains at rest or uniform rotation as long as no net torque (τnet) acts on it
a. measured as the moment of inertia, I = βmr2
b. β corrects for mass distribution (β = 1 for a hoop)
c. equilibrium (τnet = 0)
1. center of mass for a complex system
a. balance point where an upward force equal to the total weight equals the sum of all the downward weight torques
|rcm Fcm = (m1 + m2 + m3)g |
|m1 m2 |
|m3 |
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|r1 F1 |
|r2 F2 |
|r3 |
|F3 |
|not rotating ∴ τCM = τ1 + τ2 + τ3 |
|rcm(m1 + m2 + m3)g = r1m1g + r2m2g + r3m3g |
|rcm = (r1m1 + r2m2 + r3m3)/(m1 + m2 + m3) |
|rcm = Σ(rimi)/Σmi |
2. cantilever problems—how far from the edge can m2 be placed without tipping?
| mass of plank at its center |
|m1 m2 |
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|r1 r2 |
|Fg1 Fg2 |
|not rotating ∴ τ1 = τ2 |
|r1m1g = r2m2g |
|r1m1 = r2m2 |
3. two supports problems—what are the tensions?
| FL mass of plank at its center FR |
|rR |
|m1 m2 |
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|r1 |
|r2 |
|assume left side is point of rotation |
|not rotating ∴ τR = τ1 + τ2 |
|rRFR = r1m1g + r2m2g |
|FR = (r1m1g + r2m2g)/rR |
|solve for FL: FL + FR = m1g + m2g |
3. Second Law: τnet = Iα (τ = rFr, Ι = βmr2, α = a/r)
a. rFr = (βmr2)(a/r) ∴ Fr = βma (acceleration at rim)
b. pulley problems—what is the acceleration
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|m1 |
|rough surface (μ) m2 |
|Fp – Ff = m1a + βm2a β |
|Fp – μm1g = (m1 + βm2)a |
|solve for d, v or t using kinematics Fp|
c. rolling problems—what is the acceleration
| β Frolling = ma + βma |
|m Frolling = (1 + β)ma |
|Fgsinθ = (1 + β)ma |
|solve for d, v or t using kinematics |
|Fgsinθ |
|θ |
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C. Conservation Laws—Rotation
1. rotational momentum, L = rβmv (kg•m2/s)
a. when τnet = 0, then ΔL = 0
b. change r and/or β will change v
orbiting planet spinning diver
Kepler's Law
(A1 → 2 = A3 → 4)
r1v1 = r2v2 r1β1v1 = r2β2v2
2. rotational kinetic energy, Kr = ½βmv2 (J)
a. v is the velocity at the rim
b. rolling kinetic energy: Krolling = ½(1 + β)mv2
3. mixed linear and rotation motion problems
a. Summary of translational and rotational formulas
|Variable |Translational |Rotational |Rolling |
|force |F = ma |Fr = βma |F = (1 + β)ma |
|momentum |p = mv |L = rβmv |p + L = (1 + rβ)mv |
|kinetic energy |K = ½mv2 |Kr = ½βmv2 |K = ½(1 + β)mv2 |
b. collision problems (conservation of momentum)—what is the velocity after the boy jumps on the merry-go-round?
| m v |
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|convert linear into rotational (r = rM, β = 1) |
|Lm + LM = L' M, rM, βM |
|rMβmmvm + rMβMMvM = (rMβmm + rMβMM)v' |
|mvm = (mm + βMM)v' |
|v' = mvm/(mm + βMM) |
c. pulley problems (conservation of energy)—what is the velocity after descending h meters?
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|m1 |
|rough surface (μ) m3 |
|Ug3 – Wf1 = K1 + K2 + K3 β |
|m3gx – μkm1gx = ½(m1 + m2 + βm3)v'2 |
|v' = [2(m3 – μkm1)gx/(m1 + m2 + βm3)]½ |
|m2 |
d. pulley problems (conservation of energy)—what is the velocity after descending h meters?
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|Ug1 + Ug2 = Ug1' + Ug2' + K1 + K2 + K3 m3 |
|0 + m2gh = m1gh + 0 + ½(m1 + m2 + βm3)v'2 β |
|(m2 – m1)gh = ½(m1 + m2 + βm3)v'2 |
|v = [2(m2 – m1)gh/(m1 + m2 + βm3)]½ |
|m2 |
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|m1 h |
e. rolling problems (conservation of energy)—what is the velocity after descending h meters?
| m Ug = Krolling |
|β mgh = ½(1 + β)mv2 |
|h gh = ½(1 + β)v2 |
|v = [2gh/(1 + β)]½ |
f. cart problems (conservation of energy)—what is the velocity after descending h meters?
| m1 Ug = K + Krolling |
|? ? β, m2 (m1 + 4m2)gh = ½m1v2 + ½(1 + β)4m2v2 |
|(m1 + 4m2)gh = ½(m1 + (1 + β)4m2)v2 |
|h |
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D. Simple Harmonic Motion (SHM)
1. oscillating mass on a spring
a. acceleration is NOT constant ∴ kinematic formulas are invalid
b. displacement, velocity and acceleration oscillate between +A and –A, where A = amplitude
1. x = +A, when t = 0 (pictured above)
|time |t = 0 |t = ¼T |t = ½T |t = ¾T |t = T |
|displacement |+A |0 |-A |0 |+A |
|velocity |0 |-vmax |0 |+vmax |0 |
|acceleration |-amax |0 |+amax |0 |-amax |
2. x = 0, when t = 0 (heading downward)
|time |t = 0 |t = ¼T |t = ½T |t = ¾T |t = T |
|displacement |0 |-A |0 |+A |0 |
|velocity |-vmax |0 |+vmax |0 |-vmax |
|acceleration |0 |+amax |0 |-amax |0 |
c. maximum acceleration, amax = A(k/m) = vmax2/A
|Steps |Algebra |
|start with |Fs = ma |
|substitute kA for Fs |kA = ma |
|solve for a |amax = A(k/m) |
d. maximum velocity, vmax = A(k/m)½ = 2πA/T
|Steps |Algebra |
|start with |Us = K |
|substitute ½kA2 for Us and ½mv2 for K |½kA2 = ½mv2 |
|solve for v |vmax = A(k/m)½ |
e. velocity at x, in terms of vmax: vx = vmax[1 – (x2/A2)]½
|Steps |Algebra |
|start with |Kx + USx = Umax |
|substitute for Us and K |½mvx2 + ½kx2 = ½kA2 |
|solve for vx2 |vx2 = (k/m)(A2 – x2) |
|multiply-divide by A2 |vx2 = A2(k/m)[(A2/A2) – (x2/A2)] |
|square root both sides |vx = A(k/m)½[(1 – (x2/A2)]½ |
|substitute vmax for A(k/m)½ |vx = vmax[1 – (x2/A2)]½ |
f. time for one cycle, period, T = 2π(m/k)½
|Steps |Algebra |
|start with |vmax = A(k/m)½ |
|substitute 2πA/T for vmax |2πA/T = A(k/m)½ |
|simplify |2π/T = (k/m)½ |
|solve for T |T = 2π/(k/m)½ |
|substitute (m/k)½ for 1/(k/m)½ |T = 2π(m/k)½ |
g. formulas at midpoint, 0, and extremes, A
| |midpoint |extreme |
|x |0 |xmax = A |
|v |vmax = A(k/m)½ = 2πA/T |0 |
|a |0 |amax = -A(k/m) = vmax2/A |
|F |0 |F = -ma = -kA |
|U |0 |Umax = ½kA2 |
|K |Kmax = ½mv2 |0 |
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2. pendulum
[pic]
a. period of the simple pendulum, T = 2π(L/g)½
|Steps |Algebra |
|start with |F = kx |
|substitute mgsinθrad for F |mgsinθrad = kx |
|substitute Lθrad for x |mgsinθrad = kLθrad |
|for small angles sinθrad = θrad |mg = kL |
|solve for k |k = mg/L |
|start with |T = 2π(m/k)½ |
|substitute mg/L for k |T = 2π(m/mg/L)½ |
|simplify |T = 2π(L/g)½ |
b. notice that m cancels out of the equation, so the period only depends on the L and g
3. damped harmonic motion
[pic]
a. amplitude of oscillating spring or swinging pendulum will decrease until it stops—damping
b. damping is due to friction and air resistance
1. forces always oppose direction of velocity
2. damping is enhanced if oscillator is placed in viscous fluid (car shock absorbers)
c. forced damping is accomplished with motors that are programmed to oppose velocity (earthquake protected buildings)
4. resonance
a. object can be set to oscillate by an external force—forced vibration
b. when forced vibration matches natural vibration, then amplitude builds with each vibration—resonance
c. examples
1. child swinging
2. building during an earthquake
3. air inside a musical instrument
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A. Circular Motion
1. Centripetal Force Lab
Measure the period of a whirling mass using two techniques, and then vary the tension and radius to see their effects on the period.
a. Collect the following data.
|Control |
|string length, L |0.5 m |
|hanging weight, m1 |200 g |
|stopper mass, m2 | |
|time (10 orbits), t | |
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|Double L |Half m1 |
|m1 |200 g |m1 |100 g |
|string length, L |1.0 m |string length, L |0.5 m |
|time (10 orbits), t | |time (10 orbits), t | |
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| |m1g |m2g |Ftx |Fty |
|What is Fty equal to? | | | | |
|Which force equals Fc? | | | | |
|What is Ft equal to? | | | | |
c. Calculate the following from the data.
| |Control |Double L |Half m1 |
|T1 based on t | | | |
|Ft based on m1 | | | |
|Fty based on m2 | | | |
|θ based on Ft and Fty | | | |
|Fc based on Ft and Fty | | | |
|r based on L and θ | | | |
|v based on Fc | | | |
|T2 based on v | | | |
|%Δ T1 and T2 (use average | | | |
|as standard) | | | |
c. How was the period affected by the following?
|Doubling L | |
|Halving m1 | |
Questions 2-10 Briefly explain your answer.
2. When a tetherball is whirling around the pole, the net force is directed
(A) toward the top of the pole
(B) toward the ground
(C) horizontally away from the pole
(D) horizontally toward the pole
|D—net force equals Fc, which is toward the center of the circle |
3. You are standing in a bus that makes a sharp left turn. Which of the following is true?
(A) you lean to the left because of centripetal force
(B) you lean to the right because of inertia
(C) you lean forward because of the net force is forward
(D) you lean to the right because of centrifugal force
|B—the turn to the left results in the right side of the bus being in |
|"front" of you as you continue forward due to inertia |
4. You drive your car too fast around a curve and the car starts to skid. What is the correct description of this situation?
(A) car's engine is not strong enough to keep the car from being pushed out
(B) friction between the tires and the road is not strong enough to keep the car in a circle
(C) car is too heavy to make the turn
(D) none of the above
|B—the centripetal force needed to make the turn is generated by friction|
|between the road and tires, which is not sufficient at the high speed |
5. A steel ball is whirling around in a circle on the end of a string when the string breaks. Which path will it follow?
A B C
|B—without the tension in the string to supply the centripetal force, the|
|ball continues in a straight line |
6. Two stones A and B have the same mass. They are tied to strings and whirled in horizontal circles. The radius of the circular path for stones A is twice the radius of stone B's path. If the period of motion is the same for both stones, what is the tension in cord A (FA) compared to cord B (FB)?
(A) FA = FB (B) FA = 2FB (C) FA = ½FB
|B—Fc = mv2/r = m(2πr/T)2/r = 4π2mr/T2, since 4π2m/T2 is constant ∴ Fc ∝ |
|r |
7. You driving along a rural road. Which is true when you are at the lowest point along a dip in the road?
(A) Fn < Fg (B) Fn = Fg (C) Fn > Fg
|C—at the bottom of the vertical circle: |
|Fnet = Fc = Fn – Fg > 0 (upward) ∴ Fn > Fg |
8. You swing a ball on the end of a string in a vertical circle. Which is true of the centripetal force at the top of the circle?
(A) Fc = Ft + Fg (B) Fc = Ft – Fg (C) Fc = Fg – Ft
|A—at the top of the vertical circle, both Ft and Fg are downward, which |
|is the direction of Fc ∴ Fc = Ft + Fg |
9. Which is stronger the Earth's pull on the Moon or the Moon's pull on the Earth?
(A) Earth's pull (B) Moon's pull (C) they are equal
|C—Newton's third law states that for every force there is an equal but |
|opposite force ∴ the pulls are equal |
10. Is there a net force acting on an astronaut floating in orbit around the Earth while on a space walk?
(A) yes (B) no
|A—The net force is Fg, which provides the Fc necessary to keep the |
|astronaut in orbit |
11. A car is traveling clockwise on the north side of a circular track. (r = 50 m) takes 16 s to make one lap. Determine
|v |v = 2πr/T = 2π(50 m)/16 s = 20 m/s |
|direction of v |east |
|ac |ac = v2/r = (20 m/s)2/(50 m) = 8 m/s2 |
|direction of ac |south |
12. The earth is 1.5 x 1011 m from the sun and makes one complete circular orbit in 1 year.
a. What is the period of orbit in seconds?
|T = 365 days x 24 hr/day x 3600 s/hr = 3.2 x 107 s |
b. What is the earth’s orbital velocity?
|v = 2πr/T = 2π(1.5 x 1011 m)/3.2 x 107 s = 3.0 x 104 m/s |
c. What is the centripetal acceleration of the earth toward the sun?
|aC = v2/r = (3.0 x 104 m/s)2/1.5 x 1011 m = 6 x 10-3 m/s2 |
13. A driver of a 1000-kg sports car attempts a turn whose radius of curvature is 50 m on a road where μ = 0.8.
a. What is the fastest that the driver can make the turn?
|Fc = Ff → mv2/r = μmg |
|v2/50 m = (0.8)(10 m/s2) ∴ v = 20 m/s |
b. Could the driver make the turn at this speed
(1) with a 2,000-kg SUV? Explain
|Yes, because mass cancels out of the equation. |
(2) when the road is wet? Explain
|No, because the coefficient of friction is less. |
14. A 2-kg mass is moving at 5 m/s in a horizontal circle of radius 1 m at the end of a cord.
a. What is the horizontal component of tension?
|Ft-x = Fc = mv2/r |
|Ft-x = (2 kg)(5 m/s)2/(1 m) = 50 N |
b. What is the vertical component of tension?
|Ft-y = Fg = mg |
|Ft-y = (2 kg)(10 m/s2) = 20 N |
c. What is the overall tension in the cord?
|Ft = (Ft-x2 + Ft-y2)½ |
|Ft = (502 + 202)½ = 54 N |
d. What angle does the cord make with the horizontal?
|tanθ = Ft-y/Ft-x = Fg/Fc |
|tanθ = 20/50 ∴ θ = 22o |
15. A 2-kg mass is moving at 5 m/s in a vertical circle of radius 1 m at the end of a cord.
a. What is the tension in the cord at the top of the circle?
|Ft = Fc – Fg = mv2/r – mg |
|Ft = (2 kg)(5 m/s)2/(1 m) – (2 kg)(10 m/s2) = 30 N |
b. What is the tension in the cord at the bottom?
|Ft = Fc + Fg = mv2/r + mg |
|Ft = (2 kg)(5 m/s)2/(1 m) + (2 kg)(10 m/s2) = 70 N |
16. A 1-kg pendulum bob swings back and forth from a 2-m string that can support 15 N of tension without breaking.
a. What is the maximum speed that the bob can reach at the bottom of the swing without breaking the string?
|Ft = Fg + Fc = mg + mv2/r |
|15 N = (1kg)(10 m/s2) +(1 kg)v2/(2 m) ∴ v = 3.16 m/s |
b. What is the maximum height measured from vertical that the bob can reach?
|Ug + K = Ug' + K' → mgh = ½mv2 |
|(10 m/s2)h = ½(3.16 m/s)2 ∴ h = 0.50 m |
17. How would the force of gravity be affected if the Earth
a. had the same mass but a smaller radius?
|Between Earth and Moon |On the Earth's surface |
|the same |greater |
b. had the same radius but a smaller mass?
|Between Earth and Moon |On the Earth's surface |
|less |Less |
18. Determine the acceleration due to gravity on the planet compared to Earth.
|Mass |Radius (x Earth) |Acceleration (x gEarth) |
|m = mEarth |r = rEarth |g |
|m = mEarth |r = 2rEarth |¼g |
|m = mEarth |r = ½rEarth |4g |
|m = 2mEarth |r = rEarth |2g |
|m = ½mEarth |r = rEarth |½g |
19. What is the acceleration due to gravity (g) on Mars?
( m = 6.4 x 1023 kg, r = 3.4 x 106 m)
|g = GM/r2 |
|g = (6.67x10-11N•m2/kg2)(6.4x1023kg)/(3.4x106m)2=3.7m/s2 |
20. When the Apollo Missions went to the moon they passed a point where the gravitational attractions from the moon and the earth are equal. What is the ratio rEarth/rMoon where this happened if mEarth/mMoon = 100?
|GmMm/rM2 = GmEm/rE2 |
|rE2/rM2 = mE/mM = 100 ∴ rM/rE = 10 |
21. Consider the following changes to earth.
I Increase earth's mass
II Decrease earth's mass
III Increase earth's radius
IV Decrease earth's radius
|Which changes would decrease the acceleration due to gravity|II and III |
|on the earth's surface? | |
|Which changes would increase the acceleration due to gravity|I and IV |
|on the earth's surface? | |
|Which changes would decrease the acceleration due to gravity|II only |
|on the moon? | |
|Which changes would increase the acceleration due to gravity|I only |
|on the moon? | |
22. The Earth's mass is 81 times the mass of the moon and the Earth's radius is 4 times the radius of the moon.
a. What is gMoon in terms of gEarth?
|g = GM/r2 ∴ gMoonrMoon2/mMoon = gEarthrEarth2/mEarth |
|gMoon(1)2/1 = g(42)/81 |
|gMoon = 16g/81 = 0.2 g |
b. What is the mass of a 50 kg person on the Moon?
|Mass doesn't change: 50 kg |
c. What is the weight of a 50 kg person on the Moon?
|Fg = mg = (50 kg)(0.2) = 10 kg |
B. Newton's Laws—Rotation
23. Equilibrium Lab
a. Extend from the table edge a ½-m stick with a 50-g mass at 0 cm and measure the balance point (CM).
50 g
½-m stick |← rr → |← r50 →
table | | |
50 cm 25 cm 0 cm
(1) Collect the following data.
|rr |r50 |ruler mass, mr |
| | | |
(2) Use the cantilever technique to determine the mass of the ruler.
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(3) Determine the percent difference between the measured mr (true) and the calculated mr from (2).
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b. Extend from the table edge a ½-m stick with 50-g at 40 cm, 10-g at 15 cm and 20-g at 5 cm and measure the balance point (rcm).
50 g 10 g 20 g
mr
½-m stick| | | | |
table 50 40 25 15 5 0 cm
|(1) Record the balance point rcm. | |
(2) Use the center of mass formula to determine rcm.
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(3) Determine the percent difference between the measured rcm (true) and the calculated rcm from (2).
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c. Explore the relationship between center-of-mass and balance by performing the following.
(1) Stand with your heels and back against a wall and try to bend over and touch your toes. Explain
|You fall over because your hips can't move back to maintain a center of |
|mass over your feet. |
(2) Stand facing the wall with your toes against the wall and try to stand on your toes. Explain
|You fall back because you can't shift your weight forward to keep you |
|weight over the ball's of your feet. |
(3) Rest a meter stick on two fingers. Slowly bring your fingers together. Explain
|The finger closest to the CM supports more weight ∴ greater Fn and Ff, |
|which locks that finger in place. |
Questions 24-36 Briefly explain your answer.
24. You are using a wrench to loosen a rusty nut. Which will produce the greatest torque?
A B
C
D
|B—torque is the greatest when rsinθ is maximum, which is when the |
|distance is the greatest and the angle is 90o |
Questions 25-26 Four objects have the same mass and radius.
F
axis of rotation → ?
(A) hollow cylinder, β = 1 (B) solid cylinder, β = 1/2
(C) hollow ball, β = 2/3 (D) solid ball, β = 2/5
25. Which object would have the greatest moment of inertia?
|A—most of the mass is along its rim ∴ β in I = βmr2 is the greatest |
26. Which object would have the greatest rotational acceleration?
|D—Fr = βma: the same force would generate a greater acceleration for the|
|object with the smallest β |
27. 3 identical balls descend 3 identical ramps (except for μs). Ball A slides down ramp A (μs = 0), ball B rolls down ramp B (μs = .3) and ball C rolls down ramp C (μs = .6). Which is true of their velocities when the reach the end of their ramp?
(A) vA > vB = vC (B) vA > vB > vC (C) vA = vB = vC
|A—sliding is greater (Fgsinθ = ma) than rolling |
|(Fgsinθ = (1 + β)ma), and μs doesn't effect acceleration |
28. A 1-kg block is hung at the end of a rod 1-m long. The balance point is 0.25 m from the end holding the block, what is the mass of the rod?
|← 0.25 m →|← 0.25 m →|
center of rod
1 kg
(A) 0.25 kg (B) 0.5 kg (C) 1 kg (D) 2 kg
|C—placing all the rod's mass at its center |
|rbmbg = rrmrg → (0.25 m)(1 kg) = (0.25 m)(mrod) ∴ mr = 1 kg |
29. What is the total mass of the mobile? (rods are massless)
1 m 2 m A
B 1 m 3 m 1 kg
(A) 5 kg (B) 6 kg (C) 7 kg (D) 8 kg
|B—(1 m)B = (3 m)(1 kg) ∴ B = 3 kg |
|(1 m)(1 kg + 3 kg) = (2 m)(A) ∴ A = 2 kg |
|1 kg + 3 kg + 2 kg = 6 kg |
30. Consider the two configurations of interlocking blocks on the edge of a table. Which of the following is true?
A B
(A) A tips (B) B tips (C) both tip (D) neither tip
|A—the blocks to right of the table's edge are not balanced by an equal |
|number of blocks to the left |
31. Consider the door as viewed from above.
Determine
a. The torque when F1 = 45 N and r1 = 1 m.
|τ1 = r⊥F1 = (1 m)(45 N) = 45 m•N |
b. The force, F2, where r2 = 0.4 m, that will generate the same torque as part a.
|τ1 = r⊥F2 |
|45 m•N = (0.4 m)F2 ∴ F2 = 113 N |
32. A 5-kg disk (β = ½) rolls down a 30o incline. Determine
a. The parallel component of Fg.
|Fg-|| = Fgsinθ = (5 kg)(10 m/s2)sin30 = 25 N |
b. The disk's acceleration at the rim.
|Fg-|| = Frolling = (1 + β)ma |
|25 N = (1 + ½)(5 kg)a ∴ a = 3.3 m/s2 |
33. A 25-kg box rests on the edge of a merry-go-round (r = 2 m).
a. What is the maximum force of friction between the box and merry-go-round (μs = 0.80)?
|Ff = μFn = (0.80)(250 N) = 200 N |
b. What is the maximum velocity before the box slips off?
|Ff = Fc = mv2/r |
|200 N = (25 kg)v2/(2 m) ∴ v = 4 m/s |
c. What is the acceleration of the 200-kg merry-go-round (β = ½) exerting by a 50-N force along the outer rim?
|Fr = βma |
|50 N = ½(200 kg)a ∴ a = 0.5 m/s2 |
d. How much time will it take to reach the maximum velocity before the box slips off of the merry-go-round?
|vt = vo + at |
|4 m/s = 0 + (0.5 m/s2)t ∴ t = 8 s |
e. Would this time increase or decrease if μ = 1.0?
|Increase |
34. A 5-m, 75-kg plank is extended 2 m over the edge of a building. What is the maximum distance that a 25-kg child walks out from the building's
edge without tipping the plank?
|τchild = τplank |
|(x m)(250 N) = (0.5 m)(750 N) |
|x = 1.5 m |
35. Consider the diagram of the printing press
on a table. Determine
a. F1.
|τF1 ’ τ1500 + τ15,000 |
|(20 m)F1 = (10 m)(15,000 N) + (5 m)(150,000 N) |
|F1 = 45,000 N |
b. F2.
|F1 + F2 = 1500g + 15,000g |
|45,000 N + F2 = 15,000 N + 150,000 N ∴ F2 = 120,000 N |
36. A plank is placed on two scales, which are then zeroed. A 172-cm-tall student lies on the plank resulting in the reading shown.
[pic]
a. What is the student's mass?
|mstudent = 35.1 kg + 31.6 kg |
|mstudent = 66.7 kg |
b. What is the distance from her feet to her center-of-mass?
|rcm = (r1m1 + r2m2)/(m1 + m2) |
|rcm = [(172 cm)(35.1 kg) + (0 cm)(31.6 kg)](66.7 kg) |
|rcm = 90.5 cm |
37. A 2200-kg trailer is attached to a stationary truck.
[pic]
Determine the
a. normal force on the trailer tires at A.
|τA = τcm |
|(8 m)FA = (5.5 m)(22000 N) ∴ FA = 15,000 N |
b. normal force on the support B.
|FA + FB = Fg |
|15,000 N + FB = 22,000 N ∴ FB = 7,000 N |
38. A 200-N sign hangs from the end of a 5-m pole, which is held at a 37o angle by a horizontal guy wire.
guy wire
pole
37o
Determine the tension in the guy wire.
|τcc = τc |
|(5 m)Fguy wiresin37 = (5 m)(200 N)sin53 ∴ F = 265 N |
C. Conservation Laws—Rotation
39. 1 + β lab
Roll different objects down an incline and calculate the final velocity and (1 + β) for each and compare the calculated values with the ideal values.
a. Collect the following data.
|Ring |Disk |
|height, h | |height, h | |
|distance, d | |distance, d | |
|time, t | |time, t | |
| | |
|height, h | |height, h | |
|distance, d | |distance, d | |
|time, t | |time, t | |
| | | | |
| | | | |
c. Calculate (1 + β) for using conservation of energy.
|Ring |Disk |Ball |Cart |
| | | | |
d. Calculate the percent difference with the ideal values.
|Ring |Disk |Ball |Cart |
|2 |3/2 |7/5 |1 |
| | | | |
40. A hoop, cylinder and sphere roll down a 1-m ramp inclined 30o at the same time that a box slides down a frictionless ramp that is also 1 m long and inclined 30o.
a. Derive a formula for determining the velocity of each object when it reaches the bottom of the ramp.
|Ug = Krolling and h = 1.0 m(sinθ) = 0.5 m |
|mgh = ½(1 + β)mv2 → v = [g/(1 + β)]½ |
b. What are the velocities of each when they reach the bottom of the ramp?
|Hoop |v = [g/(1 + β)]½ = [10/(1 + 1)]½ = 2.24 m/s |
|(β = 1) | |
|Cylinder |v = [g/(1 + β)]½ = [10/(1 + 1/2)]½ = 2.58 m/s |
|(β = 1/2) | |
|Sphere |v = [g/(1 + β)]½ = [10/(1 + 2/5)]½ = 2.67 m/s |
|(β = 2/5) | |
|Box |v = [g/(1 + β)]½ = [10/(1 + 0)]½ = 3.16 m/s |
|(β = 0) | |
c. What is the order in which they reach the bottom?
|Box-sphere-Cylinder-Hoop |
41. Determine the velocity of a Yo-Yo (β = ½) that "rolls" straight down its string a distance of 0.50 m.
|Ug = K'rolling → mgh = ½(1 + ½)mv2 |
|(10 m/s2)(0.50 m) = ¾v2 ∴ v = 2.6 m/s |
42. A marble (β = 2/5) rolls from rest down a ramp and around a loop (radius = 10 m). Determine
A
B
H 10 m
a. the minimum velocity at B.
|Fg = Fc |
|mg = mv2/r ∴ v = (rg)½ = 10 m/s |
b. the minimum height H at A.
|Ug-A = U'g-B + K'rolling-B → mgH + 0 = mgh + ½(1 + β)mv2 |
|(10 m/s2)H = (10 m/s2)(20 m) + ½( 1 +2/5)(10 m/s)2 |
|H = 27 m |
43. A string is attached to a 1.0-kg block and is wrapped round a pulley (β = ½, m = 2.0 kg). The block is released from rest and accelerates downward while the pulley rotates.
What is the block's velocity after descending 1 m?
|Ub = K'b + K'r-p |
|mbgh = ½mbv2 + ½βmpv2 = ½(mb + βmp)v2 |
|(1 kg)(10 m/s2)(1 m) = ½(1.0 kg + 1.0 kg)v2 ∴ v = 3.2 m/s |
44. Two weights (m1 = 0.60 kg, m2 = 0.40 kg) are connected by a cord that hangs from a pulley (β = ½, M = 0.50 kg).
M
m1
1 m
m2
What is the velocity of m1 after descending 1 m?
|Um2 + Um1 = U'm2 + U'm1 + K'r-M + K'm1 + K'm2 |
|m2gh + 0 = 0 + m1gh + ½βMv2 + ½m1v2 + ½m2v2 |
|gh(m2 – m1) = ½(βM + m1 + m2)v2 |
|(10 m/s2)(1.0 m)(0.20 kg) = (0.625 kg)v2 ∴ v = 1.8 m/s |
45. A string attached to a 20-kg block resting on a table passes over a pulley (β = ½, m = 4 kg) and attaches to a 14-kg mass hanging over the edge of the table. The 20-kg box slide along the table (μ = 0.25) while the 14-kg mass descends 1 m.
20 kg
1 m
14 kg
What is the hanging mass' velocity after descending 1 m?
|Um – Wf = K'b + K'r-p + K'm |
|mmgh – μmbgd = ½mbv2 + ½βmpv2 + ½mmv2 |
|(14)(10)(1) – (.25)(20)(10)(1) = (10 + 1 + 7)v2 ∴ v = 2.2 m/s |
46. What is the angular momentum of a 0.2-kg ball traveling at 9 m/s on the end of a string in a circle of radius 1 m?
|L = rβmv |
|L = (1 m)(1)(0.2 kg) (9 m/s) = 1.8 kg•m2/s |
47. What is the angular momentum of Earth, m = 6.0 x 1024 kg?
a. about its axis of rotation (β = 2/5, rplanet = 6.4 x 106 m)
|L = rβmv = r(2/5)m(2πr/T) = 4πmr2/5T |
|L = 4π(6.0 x 1024 kg)(6.4 x 106 m)2/(5)(60 x 60 x 24 s) |
|L = 7.1 x 1033 kg•m2/s |
b. in its orbit around the Sun (β = 1, rorbit = 1.5 x 1011 m)
|L = rβmv = r(1)m(2πr/T) = 2πmr2/T |
|L = 2π(6.0 x 1024 kg)(1.5 x 1011 m)2/(60 x 60 x 24 x 365 s) |
|L = 2.7 x 1040 kg•m2/s |
48. Halley's comet follows an elliptical orbit, where its closest approach to the sun is 8.9 x 1010 m and its farthest distance is 5.3 x 1012 m. How many times faster does the comet travel at its fastest compared to its slowest?
|r1v1 = r2v2 |
|v1/v2 = r2/r1 = (5.3 x 1012 m)/(8.9 x 1010 m) = 60 |
49. A child (m = 42 kg) runs toward a stationary merry-go-round (β = ½, m = 180 kg, r = 1.2 m) along a tangent at 3 m/s. The child jumps on the merry-go-round and sets it rotating.
3 m/s
42 kg β = ½
180 kg
1.2 m
What is the speed of the merry-go-round after the child jumps on?
|rCβCmCvC + rMβMmMvM = (rCβCmC + rMβMmM)v' |
|(1)(42 kg)(3 m/s) + 0 = [(1)(42 kg) + (½)(180 kg)]v' |
|v' = 0.95 m/s |
50. The rim of a disk (β = ½, m = M, r = R) rotates at a velocity, V. A ring (β = 1, m = M, r = R) is dropped on top of the disk.
a. Calculate Ltotal before the ring is dropped on the disk.
|Ltotal = rdβdmdvd + rrβrmrvr |
|Ltotal = (R)(½)(M)(V) + (R)(1)(M)(0) = ½RMV |
b. Calculate the velocity after the ring is dropped.
|Ltotal = Ltotal' = (rdβdmd + rrβrmr)v' |
|½RMV = (½RM + RM)v' ∴ v' = ⅓V |
51. Tarzan (100 kg) is on a ledge that is 20 m above Jane (45 kg), who is trapped on a lower ledge. Tarzan grabs a long vine and swings down from the ledge and grabs Jane, who is stationary. The two swing over to a rock ledge on the other side of the river gorge that is 10 m higher than the rock ledge where Jane is trapped. Assuming the vine is long enough, can Tarzan and Jane reach the other side?
T
J
a. Calculate Tarzan's velocity when he grabs Jane.
|mgh = ½mv2 |
|v = (2gh)½ = [(2)(10 m/s2)(20 m)]½ = 20 m/s |
b. Calculate the velocity after Tarzan grabs Jane.
|rβmTvT + rβmJvJ = (rβmT + rβmJ)v' |
|(100 kg)(20 m/s) + 0 = (145 kg)v' ∴ v' = 13.8 m/s |
c. Calculate how high Tarzan swings to the other side.
|mgh = ½mv2 |
|h = v'2/2g = (13.8 m/s)2/(2)(10 m/s2) = 9.5 m |
|d. Did Tarzan and Jane make it? |No |
e. What could Tarzan have done to save Jane?
|Start from a higher ledge or with a running start. |
f. How high would Tarzan have to start to save Jane?
|v' = (2gh)½ = [(2)(10 m/s2)(10 m)]½ = 14 m/s |
|rβmTvT = rβ(mT + mJ)v' |
|vT = (145 kg)(14 m/s)/100 kg = 20.3 m/s |
|h = v2/2g = (20.3 m/s)2/(2)(10 m/s2) = 20.6 m |
g. What minimum initial velocity would Tarzan need to save Jane starting from the original ledge?
|½mv2 + mgh = ½mv'2 |
|½v2 + (10 m/s2)(20 m) = ½(20.3)2 ∴ v = 3.5 m/s |
52. A 1-kg, disk (β = ½) is placed on a 2-m ramp where the top is 1 m above the base of the ramp. The disk is placed at the top and rolls down to the base of the ramp.
a. What is the disk's velocity when reaches the base?
|Ug = Krolling → mgh = ½(1 + ½)mv2 |
|(10 m/s2)(1 m) = ¾v2 ∴ v = 3.65 m/s |
b. How much time does it take the disk to travel the 2 m?
|d = ½(v + vo)t |
|2 m = ½(3.65 m/s + 0)t ∴ t = 1.1 s |
c. Predict how the following alterations would change the disk's velocity at and time to reach the base of ramp?
|Alteration | Final Velocity |Time |
|A 2.0-kg disk is used |same |same |
|A 1.0-kg ring (β = 1) is used |less |more |
|A 3-m ramp is used, but h = 1 m |same |more |
53. A string attached to a 10-kg box resting on a table passes over a pulley (β = ½, m = 1 kg) and attaches to a 5-kg mass hanging over the edge of the table. The 10-kg box slide 1 m along the table (μ = 0.3) while the 5-kg mass descends.
1 m
a. How much kinetic energy does the system have at the point where the 5-kg mass has descended 1 m?
|K' = Ug – Wf = mmgh – μmbgd |
|K' = (5 kg)(10 m/s2)(1 m) – (0.3)(10 kg)(10 m/s2)(1 m) |
|K' = 20 J |
b. What is the maximum velocity of the system?
|K' = K'b + K'r-p + K'm = ½mbv2 + ½βmpv2 + ½mmv2 |
|20 J = (5 kg + 0.25 kg + 2.5 kg)v2 ∴ v = 1.6 m/s |
54. Halley's Comet has a velocity of 3.88 x 104 m/s when it is 8.9 x 1010 m from the sun. How fast is it traveling when it is 5.3 x 1012 m from the sun?
|(8.9 x 1010 m)(3.88 x 104 m/s) = (5.3 x 1012 m)v2 |
|v2 = 652 m/s |
55. What is the angular momentum of the Moon?
(m = 7.35 x 1022 kg, rmoon = 1.74 x 106 m, rorbit = 3.84 x 108 m, Torbit = Trotation = 2.42 x 106 s)
a. about its axis of rotation (β = 2/5)
|L = rβmv = r(2/5)m(2πr/T) = 4πmr2/5T |
|L = 4π(7.35 x 1022 kg)(1.74 x 106 m)2/(5)(2.42 x 106 s) |
|L = 2.3 x 1029 kg•m2/s |
b. in its orbit around the Earth (β = 1)
|L = rβmv = r(1)m(2πr/T) = 2πmr2/T |
|L = 2π(7.35 x 1022 kg)(3.84 x 108 m)2/(2.42 x 106 s) |
|L = 2.8 x 1034 kg•m2/s |
56. A student (m = 75 kg) runs at 5 m/s tangentially toward a merry-go-round (β = ½, m = 150 kg, r = 2 m) rotating at 2 m/s, jumps on the merry-go-round and sets it rotating. What is the velocity of the student after he jumps on to the merry-go-round?
|rβsmsvs + rβMmMvM = (rβsms + rβmmm)v' |
|(1)(75)(5) + (½)(150)(2)= [(1)(75) + (½)(150)]v' |
|v' = 3.5 m/s |
57. A 2-kg block and a 1-kg sphere hang from 2-m strings. The sphere is raised to a horizontal position and swings toward the block and collides with it.
1 kg
2 kg
a. What is the sphere's velocity before the collision?
|Ug = K → msgh = ½msvs2 |
|vs = (2gh)½ = [2(10 m/s2)(2.0 m)]½ = 6.3 m/s |
Assume that the collision is inelastic.
b. What is the sphere-block's velocity after the collision?
|rβmSvS + rβmBvB = rβ (mS + mB)v' |
|(1 kg)(6.3 m/s) + 0 = (1 kg +2 kg)v' ∴ v' = 2.1 m/s |
c. What is the maximum height reached after the collision?
|Ug = K → (mS + mB)gh = ½(mS + mB)v'2 |
|h = v'2/2g = (2.1 m/s)2/(2)(10 m/s2) = 0.22 m |
d. What is the maximum height reached after the collision if the block and sphere exchange positions initially?
|v' = mBvB/(mB + mS) = 12.6 kg•m/s/3 kg = 4.2 m/s |
|h = v'2/2g = (4.2 m/s)2/(2)(10 m/s2) = 0.88 m |
The sphere is raised to a horizontal position initially and then collides elastically with the block.
e. What are the velocities of the block and sphere after the collision?
|vS + vS' = vB + vB' |
|6.3 m/s + vS' = 0 + vB' ∴ vS' = vB' – 6.3 |
|rβmSvS + rβmBvB = rβmSvS' + rβmBvB' |
|(1 kg)(6.3 m/s) = (1 kg)(vB' – 6.3) + (2 kg)vB' vB' = 4.2 m/s |
|vS' = vB' – 6.3 = 4.2 m/s – 6.3 m/s = -2.1 m/s |
f. What are the maximum heights reached by the block and sphere?
|hB = v'2/2g = (4.2 m/s)2/(2)(10 m/s2) = 0.88 m |
|hS = v'2/2g = (-2.1 m/s)2/(2)(10 m/s2) = 0.22 m |
g. Was potential energy conserved after the collision?
|mSghS + mBghB = (1 kg)(2 m) + 0 = 2 J |
|mBgh'B + mSgh'S = (2)(0.88) + (1)(0.22) = 1.98 J ∴ yes |
D. Simple Harmonic Motion (SHM)
58. Simple Harmonic Motion Lab
a. Measure the length L and time t for 10 oscillations of a spring with different hanging masses m, determine k using two methods and compare the results.
(1) Collect the following data.
|m (kg) |0 |0.10 |0.20 |0.30 |0.40 |0.50 |
|L (m) | | | | | | |
|t (s) | | | | | |
|F using the added| | | | | |
|weight | | | | | |
|x using ΔL | | | | | |
|k using F = kx | | | | | |
|Average k | |
(3) Determine the following using time data.
|added mass |0.10 |0.20 |0.30 |0.40 |0.50 |
|T using t | | | | | |
|k using | | | | | |
|T = 2π(m/k)½ | | | | | |
|Average k | |
(4) Calculate the percent difference between the two values of k using the average as true.
| |
b. Measure the pendulum period for different releasing angles θ and see which angle gives the most ideal values for T.
(1) Collect the following data.
|L (m) | |
|θ (o) |5o |10o |15o |20o |25o |
|T1 (s) | | | | | |
| | | | | | |
| | | | | | |
| | | | | | |
(2) Calculate the following from the data.
|angle |5o |10o |15o |20o |25o |
|x = L(2πθ/360) | | | | | |
|F = mgsinθ | | | | | |
|k = F/x | | | | | |
|T2 = 2π(m/k)½ | | | | | |
|%Δ = |T1 – T2|/T1 | | | | | |
(3) Calculate T3 = 2π(L/g)½.
| |
(4) Which angle produced T closest to the one based on the pendulum's length?
| |
Questions 59-85 Briefly explain your answer.
Questions 59-62 A spring bob in SHM has amplitude A and period T.
59. What is the total distance traveled by the bob after time T?
(A) 0 (B) ½A (C) 2A (D) 4A
|D—starting at 0: down A, up A, up A, down A |
60. What is the total displacement after time T?
(A) 0 (B) ½A (C) 2A (D) 4A
|A—you end where you started |
61. How long does it take the bob to travel a distance of 6A?
(A) ½T (B) ¾T (C) 5/4T (D) 3/2T
|D—Each A is ¼T ∴ 6 x ¼T = 3/2T |
62. At what point in the motion is v = 0 and a = 0 simultaneously?
(A) x = 0 (B) 0 < x < A(C) x = A (D) no point
|D—when x = 0, a = 0, but v = max; when x = A, v = 0, but a = max; at |
|intermediate points a ≠ 0 and v ≠ 0 |
63. A mass on the end of a spring oscillates in simple harmonic motion with amplitude A. If the mass doubles but the amplitude is not changed, what happen to the total energy?
(A) decrease (B) no change (C) increase
|B—E = ½kA2 ∴ mass will not change the total energy |
64. If the amplitude of a simple harmonic oscillator is doubled, which quantity will change the most?
(A) T (B) v (C) a (D) K + U
|D—E = ½kA2, whereas v = A(k/m)½ and a = A(k/m) change the same, and T = |
|2π(m/k)½ doesn't change at all |
65. A spring with mass m has period T. If m is doubled, what is the new T?
(A) T/√2 (B) T (C) √2T (D) 2T
|C—T = 2π(m/k)½: when m doubles T increases by √2 |
Questions 66-67 Consider the periods of pendulums A and B,
66. Which period is greater when LA = LB, but mA > mB?
(A) A (B) B (C) the same
|C—T = 2π(L/g)½: only changing L or g can change T |
67. Which period is greater when mA = mB, but LA > LB?
(A) A (B) B (C) the same
|A—T = 2π(L/g)½: greater L = greater T |
68. A grandfather clock has a weight at the bottom of the pendulum that can be moved up or down. If the clock is running slow, should the weight be moved up or down?
(A) up (B) down (C) neither will work
|A—T = 2π(L/g)½: to speed up the clock (reduce T), you need to decrease L|
|by moving the weight up |
69-70 Consider the following options.
(A) add mass to the oscillator
(B) move the oscillator to an elevator accelerating down
(C) move the oscillator to an elevator accelerating up
(D) move the oscillator to the Moon
69. Which will decrease the period of a pendulum?
|C—T = 2π(L/g)½: "g" increases when accelerating up ∴ T decreases |
70. Which will change the period of oscillation on a spring?
|A—T = 2π(m/k)½: adding mass or changing the spring will change the |
|period |
71. After a pendulum starts swinging, its amplitude gradually decreases with time because of friction. What happens to the period of the pendulum during this time
(A) decreases (B) no change (C) increases
|B—T = 2π(L/g)½: amplitude has no effect on T |
72. When you sit on a swing, the period of oscillation is T1. When you stand on the same swing, the period of oscillation is T2. Which is true?
(A) T1 < T2 (B) T1 = T2 (C) T1 > T2
|C—T = 2π(L/g)½: by standing on the swing, you have reduced the distance |
|L (distance from fulcrum to center of mass), which decreases T |
73. When a 50 kg person sits on a swing, the period of oscillation is T1, when a 100 kg person sits on the same swing, the period of oscillation is T2. Which is true?
(A) T1 < T2 (B) T1 = T2 (C) T1 > T2
|B—T = 2π(L/g)½: mass has no effect on T |
74. Consider the graph
of one cycle of SHM.
a. Determine the time in terms of T for each situation.
| |Maximum up |Zero |Maximum down |
|Acceleration |½ T |¼ T, ¾ T |0 T, 1 T |
|Velocity |¾ T |0 T, ½ T, 1 T |¼ T |
b. Determine the following when m = 1 kg, k = 100 N/m and A = 0.1 m.
(1) maximum acceleration
|aA = A(k/m) = (0.1 m)(100 N/m)/(1 kg) = 10 m/s2 |
(2) maximum velocity
|vo = A(k/m)½ = (0.1 m)[(100 N/m)/(1 kg)]½ = 1 m/s |
(3) period
|T = 2π(m/k)½ = 2π(1/100)½ = 0.63 s |
(4) maximum kinetic energy
|Ko = ½mv2 = ½(1 kg)(1 m/s)2 = 0.5 J |
(5) maximum potential energy
|UA = ½kA2 = (100 N/m)(0.1 m)2 = 0.5 J |
(6) velocity when x = 0.05 m
|v = vo[1 – (x2/A2)]½ = (1 m/s)[1 – 0.52/0.12)]½ = 0.866 m/s |
c. Graph the potential energy (----), kinetic energy (•••) and total energy (––) for one complete oscillation.
|0.5 J | | | | |
| | | | | |
|0 J | | | | |
| |¼T |¾T |
d. complete the following chart (x = +A at t = 0 s)
|t |¼T |½T |¾T |1T |
|x |0 m |-0.1 m |0 m |0.1 m |
|v |-1 m/s |0 m/s |1 m/s |0 m/s |
|a |0 m/s2 |10 m/s2 |0 m/s2 |-10 m/s2 |
|F |0 N |10 N |0 N |- 10 N |
e. How do the following change if the amplitude is 0.2 m?
|Max acceleration |Max velocity |Period |
|doubles |doubles |remains the same |
75. A 1-kg ball on the end of a 1-m string is set in motion by pulling the ball out so that it is raised 0.015 m. Determine
a. the maximum speed
|UG = K → mgh = ½mv2 |
|v = (2gh)½ = [2(10 m/s2)(0.015 m)]½ = 0.55 m/s |
b. the period of oscillation.
|T = 2π(L/g)½ = 2π(1.0 m/10 m/s2)½ = 2.0 s |
c. What would the period be with the following changes?
|m = 4 kg |L = 4 m |g = 40 m/s2 |
|2.0 s |4.0 s |1.0 s |
76. Consider the diagram of one cycle of SHM.
[pic]
a. Determine the time (in terms of T) for each of the following.
| |Maximum up |Zero |Maximum down |
|Acceleration |¾ T |0 T, ½ T, 1 T |¼ T |
|Velocity |0 T, 1 T |¼ T, ¾ T |½ T |
b. Determine the following when m = 1 kg, k = 100 N/m and A = 0.25 m.
(1) maximum acceleration
|aA = A(k/m) = (0.25 m)(100 N/m)/(1 kg) = 25 m/s2 |
(2) maximum velocity
|vo = A(k/m)½ = (0.25 m)[(100 N/m)/(1 kg)]½ = 2.5 m/s |
(3) period
|T = 2π(m/k)½ = 2π(1/100)½ = 0.63 s |
(4) maximum kinetic energy
|Ko = ½mv2 = ½(1 kg)(2.5 m/s)2 = 3.125 J |
(5) maximum potential energy
|UA = ½kA2 = ½(100 N/m)(0.25 m)2 = 3.125 J |
(6) velocity when x = 0.20 m
|vx = vo[1 – (x2/A2)]½ = 2.5 m/s[1 – (.22/.252)]½ = 1.5 m/s |
c. Graph the potential energy, kinetic energy and total energy for one complete oscillation.
|3 J | | | | |
| | | | | |
|0 J | | | | |
| |¼T |¾T |
d. complete the following chart (x = 0 at t = 0 s)
|t |¼T |½T |¾T |1T |
|x |+0.25 m |0 m |-0.25 m |0 m |
|v |0 m/s |-2.5 m/s |0 m/s |2.5 m/s |
|a |-25 m/s2 |0 m/s2 |25 m/s2 |0 m/s2 |
|F |-25 N |0 N |25 N |0 N |
e. Determine the following values when A = 0.50 m.
(1) maximum acceleration
|aA = |-A(k/m)| = (0.50 m)(100 N/m)/(1.0 kg) = 50 m/s2 |
(2) maximum velocity
|vo = A(k/m)½ = (0.50 m)[(100 N/m)/(1.0 kg)]½ = 5 m/s |
(3) period
|T = 2π(m/k)½ = 2π(1/100)½ = 0.63 s |
77. A 1-kg ball swings from the ceiling on the end of a 2-m string. The ball starts its swing from a position that is
0.2 m above its lowest point.
a. What is the maximum speed of the ball?
|Ug = K ∴ mgh = ½mv2 |
|v = (2gh)½ = [2(10 m/s2)(0.20 m)]½ = 2 m/s |
b. What is the period of oscillation for the pendulum?
|T = 2π(L/g)½ = 2π(2.0 m/10 m/s2)½ = 2.8 s |
Practice Multiple Choice (No calculator)
1. In the diagram, a car travels clockwise at constant speed.
[pic]
Which letters represent the directions of the car's velocity, v, and acceleration, a?
v a v a
(A) A C (B) C B
(C) C A (D) D A
|C—In uniform circular motion velocity is tangent to the circle and |
|acceleration is toward the center |
2. A racing car is moving around the circular track of radius 300 m. At the instant when the car's velocity is directed due east, its acceleration is 3 m/s2 directed due south. When viewed from above, the car is moving
(A) clockwise at 30 m/s (B) counterclockwise at 30 m/s
(C) clockwise at 10 m/s d. counterclockwise at 10 m/s
|A—Traveling east and turning south ∴ clockwise. |
|ac = v2/r ∴ v = (acr)½ = [(3 m/s2)(300 m)]½ = 30 m/s |
3. The disk is rotating counterclockwise when the ball is projected outward at the instant the disk is in the position shown.
Which of the following best shows the subsequent direction of the ball relative to the ground?
(A) ∧ (B) ⇐ (C) ⋄ (D) ⇒
|D—Ball simultaneously is directed to the right (due to the spring) and |
|toward the top (due to the rotation) |
4. A person weighing 800 N on earth travels to another planet with the same mass as earth, but twice the radius. The person's weight on this other planet is most nearly
(A) 200 N (B) 400 N (C) 800 N (D) 1600 N
|A—Fg = GMm/r2, where r is doubled |
|∴ Fg should be (½)2 = ¼ as much → ¼(800 N) = 200 N |
5. A ball is released from rest at position P swings through position Q then to position R where the string is again horizontal.
What are the directions of the ball's acceleration at positions, Q and R?
Q R Q R
(A) ⇔ ⇐ (B) ⇔ ⋄
(C) ⇐ ⇔ (D) ⇐ ⇓
|A—At Q, acceleration is ⇔, at R acceleration is ⇐ |
6. A 5-kg sphere is connected to a 10-kg sphere by a rod.
[pic]
The center of mass is closest to
(A) A (B) B (C) C (D) D
|B—The center of mass is the balance point, which is closer to the more |
|massive side |
7. A ball attached to a string is moved at constant speed in a horizontal circular path. A target is located near the path of the ball as shown in the diagram.
[pic]
At which point along the ball's path should the string be released, if the ball is to hit the target?
(A) A (B) B (C) C (D) D
|B—The released ball will move tangent to the circle at that spot |
8. The diagram shows a 5.0-kg bucket of water being swung in a horizontal circle of 0.70-m radius at a constant speed of 2.0 m/s.
The centripetal force on the bucket of water is
(A) 5.7 N (B) 29 N (C) 14 N (D) 200 N
|B—Fc = mv2/r = (5.0 kg)(2.0 m/s)2/0.70 m = 29 N |
Questions 9-10 refer to a ball that is tossed straight up from the surface of a small asteroid with no atmosphere. The ball rises to a height equal to the asteroid's radius and then falls straight down toward the surface of the asteroid.
9. What forces act on the ball while it is on the way up?
(A) a decreasing gravitational force that acts downward
(B) an increasing gravitational force that acts downward
(C) a constant gravitational force that acts downward
(D) a constant gravitational force that acts downward and a decreasing force that acts upward
|A—The force of gravity is the only force acting on the ball and Fg |
|decreases as r increases (Fg = GMm/r2) |
10. The acceleration of the ball at the top of its path is
(A) at its maximum value for the ball's flight
(B) equal to the acceleration at the surface
(C) equal to one-half the acceleration at the surface
(D) equal to one-fourth the acceleration at the surface
|D—g = GM/r2, r is doubled ∴ g is (½)2 = ¼ as much |
Questions 11-12 A 125-N board is 4 m long and is supported by vertical chains at each end. A person weighing 500 N is sitting on the board. The tension in the right chain is 250 N.
11. What is the tension in the left chain?
(A) 250 N (B) 375 N (C) 500 N (D) 625 N
|B—F↑ = F↓ |
|250 N + FL = 125 N + 500 N ∴ FL = 375 N |
12. How far from the left end of the board is the person sitting?
(A) 0.4 m (B) 1.5 m (C) 2 m (D) 2.5 m
|B—rFperson + rFboard = rFright chain |
|r(500 N) + (2 m)(125 N) = (4 m)(250 N) ∴ r = 1.5 |
13. A square piece of plywood on a horizontal tabletop is subjected to the two horizontal forces shown above.
[pic]
Where should a third force of magnitude 5 N be applied to put the piece of plywood into equilibrium?
| A B |
|C |
| | | | |
| D | | | |
| |
|A—C and D would cancel torque, B would cancel up and down, but only A |
|will cancel both. |
14. The diagram represents two satellites of equal mass, A and B, in circular orbits around a planet.
[pic]
Comparing the gravitational force between satellite and planet, B's gravitational force compared to A's is
(A) half as great (B) twice as great
(C) one-fourth as great (D) four times as great
|C—Fg = GMm/r2, where r is doubled |
|∴ Fg is (½)2 = ¼ as much. |
15. The radius of the earth is approximately 6,000 km. The acceleration of an astronaut in a perfectly circular orbit 6,000 km above the earth would be most nearly
(A) 0 m/s2 (B) 2.5 m/s2 (C) 5 m/s2 (D) 10 m/s2
|B—g = GM/r2, where r is doubled |
|∴ g is (½)2 = ¼ as much. ¼(10 m/s2) = 2.5 m/s2 |
16. A 5-m, 100-kg plank rests on a ledge with 2 m extended out.
[pic]
How far can a 50-kg person walk out on the plank past the edge of the building before the plank just begins to tip?
(A) ½ m (B) 1 m (C) 3/2 m (D) 2 m
|B—rFperson = rFplank (rplank is 2.5 m – 2 m = 0.5 m) |
|r(50 kg)(10 m/s2) = (0.5 m)(100 kg)(10 m/s2) ∴ r = 1 m |
17. The system is balanced when hanging by the rope.
[pic]
What is the mass of the fish?
(A) 1.5 kg (B) 2 kg (C) 3 kg (D) 6 kg
|C—CM = Σ(rimi)/Σ(mi)—assume the rod is 4 m |
|2 = [(0)(3.5) + (1)(X) + (4)(5)]/(8.5 + X) ∴ X = 3 |
18. A ball attached to a string is whirled around in a horizontal circle with radius r, speed v and tension T. If the radius is increased to 4r and the tension remains the same, then the speed of the ball is
(A) ¼v (B) ½v (C) v (D) 2v
|D—T = Fc = mv2/r = mv'2/r': r increases to 4r without changing T ∴ v'2 |
|also increase by 4. v;2 = 4 ∴ v' = 2v |
19. A 0.4-kg object swings on the end of a string. At the bottom of the swing, the tension in the string is 6 N. What is the centripetal force acting on the object at the bottom of the swing?
(A) 2 N (B) 4 N (C) 6 N (D) 10 N
|A—At the bottom of the swing, Fnet = Fc = Ft – Fg |
|Fc = 6 N – (0.4 kg)(10 m/s2) = 2 N |
20. Two wheels, fixed to each other, are free to rotate about a frictionless axis perpendicular to the page. Four forces are exerted tangentially to the rims of the wheels.
[pic]
The net torque on the system about the axis is
(A) zero (B) FR (C) 2FR (D) 5FR
|C—τnet = τcc – τc |
|τnet = (3R)F + (3R)F + (2R)F – (3R)(2F) = 8RF – 6RF = 2RF |
21. Mars has a mass 1/10 that of Earth and a diameter 1/2 that of Earth. The acceleration of a falling body near the surface of Mars is most nearly
(A) g/5 (B) 2g/5 (C) g/2 (D) g
|B—g = GM/r2 ∴ gE = GME/rE2 and GM = GMM/rM2 |
|gM/gE = (MM/ME)(rE/rM)2 = (1/10)(1/½)2 = 4/10 ∴ gM = 2g/5 |
22. A satellite of mass m and speed v moves in a stable, circular orbit around a planet of mass M. What is the radius of the satellite’s orbit?
(A) GM/mv (B) Gv/mM (C) GM/v2 (D) GmM/v
|C—Fc = Fg |
|mv2/r = GMm/r2 ∴ r = GM/v2 |
23. A wheel of radius R is mounted on an axle so that the wheel is in a vertical plane. Three small objects having masses m, M, and 2M, respectively, are mounted on the rim.
[pic]
What is m in terms of M when the wheel is stationary?
(A) 1/2 M (B) M (C) 3/2 M (D) 2 M
|C—Rm + (rcos60)M = R(2M) |
|m + ½M = 2M ∴ m = 3/2M |
24. In each case the unknown mass m is balanced by a known mass M1 or M2.
[pic]
What is the value of m in terms of the known masses?
(A) M1 + M2 (B) (M1 + M2)/2
(C) M1M2 (D) (M1M2)½
|D—I1m = l2M1 and l1M2 = l2m → l1/l2 = M1/m = m/M2 → |
|M1M2 = m2 ∴ m = (M1M2)½ |
25. An asteroid moves in an elliptic orbit with the Sun at one focus.
Which of the following increases as the asteroid moves from point P in its orbit to point Q?
(A) Speed (B) Angular momentum
(C) Total energy (D) Potential energy
|D—Energy & angular momentum are conserved, |
|r ↑ v ↓ (r1v1 = r2v2), Ug increases (Ug = -GMm/r) |
26. A satellite S is in an elliptical orbit around a planet P with r1 and r2 being its closest and farthest distances, respectively, from the center of the planet. If the satellite has a speed v1 at its closest distance, what is its speed at its farthest distance?
(A) (r1/r2)v1 (B) (r2/r1)v1
(C) (r2 – r1)v1 (D) ½(r1 + r2)v1
|A—r1v1 = r2v2 ∴ v2 = (r1/r2)v1 |
27. A satellite of mass M moves in a circular orbit of radius R at a constant speed v. Which must be true?
I. The net force on the satellite is equal to Mv2/R and is directed toward the center of the orbit.
II. The net work done on the satellite by gravity in one revolution is zero.
III. The angular momentum of the satellite is a constant.
(A) I only (B) III only (C) I and II (D) I, II, and III
|D—I is true (Fg = Fc = Mv2/R), II is true (F⊥r ∴ no work done), and III |
|is true (angular momentum is constant) |
Questions 28-29 A sphere of mass M, radius R, and β = 2/5, is released from rest at the top of an inclined plane of height h.
[pic]
28. If the plane is frictionless, what is the speed of the center of mass of the sphere at the bottom of the incline?
(A) (2gh)½ (B) 2Mgh (C) 2MghR2 (D) 5gh
|A—Ug = K (no rolling because of zero friction) |
|mgh = ½mv2 ∴ v = 2gh = (2gh)½ |
29. If the plane has friction so that the sphere rolls without slipping, what is the speed at the bottom of the incline?
(A) (2gh)½ (B) 2Mgh (C) 2MghR2 (D) (10gh/7)½
|D—mgh = K + Kr = ½mv2 + ½βmv2 = ½(1 + β)mv2 |
|v2 = 2gh/(1 + 2/5) ∴ v = (10gh/7)½ |
30. For which motions is there a variable force involved?
(A) Constant speed in a straight line
(B) Simple harmonic motion
(C) Constant speed in a circle
(D) Constant acceleration in a straight line
|B—A has no force, C and D have a constant force, only SHM has a variable|
|force ( Fs = kx) |
31. A particle of mass, m, moves with a constant speed v along the dashed line y = a.
When the x-coordinate of the particle is xo, the magnitude of the angular momentum of the particle with respect to the origin of the system is
(A) zero (B) amv (C) xomv (D) (vo2 + a2)½mv
|B—L = rβmv = (a)(1)(m)(v) = amv |
Questions 32-36 A block oscillates without friction on the end of a spring. The minimum and maximum lengths of the spring as it oscillates are, respectively, xmin and xmax.
[pic]
The graphs below can represent quantities associated with the oscillation as functions of the length x of the spring.
(A) (B)
(C) (D)
32. Which graph can best represent the total mechanical energy of the block-spring system as a function of x?
|C—Total mechanical energy is constant when a conservative force is |
|involved |
33. Which graph can best represent the kinetic energy of the block as a function of x?
|D—The kinetic energy is greatest when the velocity is greatest (K = |
|½mv2), which occurs at the midpoint |
34. Which graph can best represent the potential energy of the block as a function of x?
|B—Potential energy is greatest when x is greatest |
|(Ug = ½kx2), which occurs at xmin and xmax, |
35. Which graph can best represent the acceleration of the block as a function of x?
|B—Acceleration is greatest when force is greatest (Fs = kx), which |
|occurs at xmin and xmax |
36. Which graph can best represent the velocity of the block as a function of x?
|D—Velocity is greatest at the midpoint and is zero at xmin and xmax |
|(same as K) |
37. A block attached to the lower end of a vertical spring oscillates up and down. The period of oscillation depends on which of the following?
I. Mass of the block
II. Amplitude of the oscillation
III. Spring constant
(A) I only (B) II only (C) III only (D) I and III only
|D—T = 2π(m/k)½ ∴ T depends on both m and k, but not on amplitude, A. |
38. When a 1-kg bob is attached to a spring, the period of oscillation is 2 s. What is the period of oscillation when a 2-kg bob is attached to the same spring?
(A) 0.5 s (B) 1.0 s (C) 1.4 s (D) 2.8 s
|D—T = 2π(m/k)½ ∴ T ∝ √m. |
|The mass is doubled ∴ T = 2√2 = 2.8 s |
39. A pendulum and a mass hanging on a spring both have a period of 1 s on Earth. They are taken to planet X, which has twice the gravitational acceleration g as Earth. Which is true about the periods of the two objects on planet X compared to their periods on Earth?
(A) Both are shorter.
(B) Both are longer.
(C) The pendulum is longer and the spring is the same.
(D) The pendulum is shorter and the spring is the same.
|D—The period of a pendulum, T = 2π(L/g)½ ∴ T ∝ (1/g)½ and it is shorter,|
|but the period of a spring is the same |
40. The graph is of the displacement x versus time t for a particle in simple harmonic motion with a period of 4 s.
[pic]
Which graph shows the potential energy of the particle as a function of time t for one cycle of motion?
(A) (B) (C) (D)
|C—Potential energy is greatest when x is greatest |
|(Ug = ½kx2), which occurs at 1 s and 3 s. |
41. Two identical springs are hung from a horizontal support. When a 1.2-kg block is suspended from the pair of springs, each spring is stretched an additional 0.15 m.
The spring constant of each spring is most nearly
(A) 40 N/m (B) 48 N/m (C) 60 N/m (D) 80 N/m
|A—Each spring supports half of the weight (12 N). |
|Fs = kx → 6 N = k(0.15 m) ∴ k = 40 N/m |
42. A ball is dropped from a height of 10 m onto a hard surface so that the collision at the surface may be assumed elastic. Under such conditions the motion of the ball is
(A) simple harmonic with a period of about 1.4 s
(B) simple harmonic with a period of about 2.8 s
(C) simple harmonic with an amplitude of 5 m
(D) periodic but not simple harmonic
|The force on ball is constant, except when it is in contact with the |
|floor, ∴ not SHM, where force ∝ x.. |
43. An object swings on the end of a cord as a simple pendulum with period T. Another object oscillates up and down on the end of a vertical spring, also with period T. If the masses of both objects are doubled, what are the new values for the periods?
Pendulum Spring Pendulum Spring
(A) T/√2 √2T (B) T √2T
(C) T T (D) √2T T
|B—Pendulum: T = 2π(L/g), ∴T = T |
|Spring: T = 2π(m/k), where T ∝ √m ∴ T = √2T |
44. When a mass m is hung on a spring, the spring stretches a distance d. If the mass is then set oscillating on the spring, the period of oscillation is proportional to
(A) (d/g)½ (B) (g/d)½ (C) (d/mg)½ (D) (m2g/d)½
|A—Fs = kx → mg = kd ∴ k = gm/d |
|T ∝ (m/k)½ = (md/gm)½ = (d/g)½ |
45. A 3-kg block is hung from a spring, causing it to stretch 12 cm at equilibrium. The 3-kg block is then replaced by a 4-kg block, and the new block is released from the spring when it is unstretched. How far will the 4-kg block fall before its direction is reversed?
(A) 9 cm (B) 18 cm (C) 24 cm (D) 32 cm
|D—The 4-kg block stretches the spring 16 cm (Fs ∝ x), the spring falls |
|another 16 cm before turning around |
46. A spring is fixed to the wall at one end. A block of mass M attached to the other end of the spring oscillates with amplitude A on a frictionless, horizontal surface. The maximum speed of the block is v.
[pic]
The spring constant is
(A) Mg/A (B) Mgv/2A (C) Mv2/2A (D) M(v/A)2
|D—v = A(k/m)½ |
|v2 = A2k/M ∴ k = Mv2/A2 = M(v/A)2 |
47. A sphere of mass m1 is attached to a spring. A second sphere of mass m2 is suspended from a string of length L, If both spheres have the same period of oscillation, which of the following is an expression for the spring constant?
(A) L/m1g (B) g/m2L (C) m1L/g (D) m1g/L
|D—Tp = 2π(L/g)½ = Ts = 2π(m/k)½ |
|L/g = m1/k ∴ k = m1g/L |
Practice Free Response
1. A roller coaster ride at an amusement park lifts a car of mass 700 kg to point A at a height of 90 m above the lowest point on the track, as shown above. The car starts from rest at point A, rolls with negligible friction down the incline and follows the track around a loop of radius 20 m. Point B, the highest point on the loop, is at a height of 50 m above the lowest point on the track.
P
a. (1) Indicate on the figure the point P at which the maximum speed of the car is attained.
(2) Calculate the value vmax of this maximum speed.
|Ug = K → mgh = ½mv2 |
|v = (2gh)½ = [2(10 m/s2)(90 m)]½ = 42 m/s |
b. Calculate the speed vB of the car at point B.
|v = (2gh)½ = [2(10 m/s2)(40 m)]½ = 28 m/s |
c. (1) Draw and label vectors to represent the forces acting on the car when it is upside down at point B.
| |
| |
|Fg Fn |
(2) Calculate all the forces identified in (c1).
|Fg = mg = (700 kg)(10 m/s2) = 7,000 N |
|Fc = mv2/r = (700 kg)(28 m/s)2/(20 m) = 27,000 N |
|Fc = Fg + Fn ∴ Fn = Fc – Fg = 27,000 – 7,000 = 20,000 N |
2. A string attached to a 20-kg block resting on a table passes over a pulley (β = ½, m = 10 kg) and attaches to a 10-kg mass hanging over the edge of the table. The 20-kg box slide along the table (μ = 0.30) while the 10-kg mass descends 2 m.
2 m
Determine the
a. force of friction on the 20-kg block as it slides.
|F = μmbg = (0.30)(20 kg)(10 m/s2) = 60 N |
b. force of gravity on the 10 kg mass.
|Fg = mmg = (10 kg)(10 m/s2) = 100 N |
c. net force rotating the pulley.
|Fnet = 100 N – 60 N = 40 N |
d. acceleration of the pulley at the rim.
|Fnet = (mb + βmp + mm)a |
|40 N = (20 kg + 5 kg + 10 kg)a ∴ a = 1.1 m/s2 |
e. velocity when the system has moved 2 m.
|v2 = vo2 + 2ad |
|v2 = 02 + 2(1.1 m/s2)(2 m) ∴ v = 2.1 m/s |
f. loss of potential energy as the 10-kg mass falls 2 m.
|Ug = mmgh = (10 kg)(10 m/s2)(2 m) = 200 J |
g. work done by friction as the 20-kg block slides 2 m.
|Wf = μmbgd = (.30)(20 kg)(10 m/s2)(2 m) = 120 J |
h. velocity of the system
|Um – Wf = K'b + K'r-p + K'm = ½mbv2 + ½βmpv2 + ½mmv2 |
|200 J – 120 J = ½(20 kg + 5 kg + 10 kg)v2 ∴ v = 2.1 m/s |
3. A 0.5-kg hoop (β = 1) rolls from rest at the top of the ramp of length L = 2 m and angle θ = 30o. The table height H = 1 m.
a. Determine the potential energy of the hoop at the top of the ramp, where Ug = 0 at the floor.
|Ug = mg(H + Lsinθ) |
|Ug = (0.5 kg)(10 m/s2)[1 m + (2 m)(sin30)] = 10 J |
b. The hoop rolls down the ramp and then onto the floor. Determine the hoop's
(1) speed at the bottom of the ramp.
|Ug = Krolling → mgLsinθ = ½(1 + β)mv2 |
|(10 m/s2)(2 m)sin30 = ½(1 + 1)v2 ∴ v = 3.2 m/s |
(2) speed just before it hits the floor.
|Kfloor = Ktable + Utable → ½mvf2 = ½mvt2 + mgh |
|vf2 = (3.2 m/s)2 + 2(10 m/s2)(1 m)∴ vf = 5.5 m/s |
(3) translational kinetic energy before it hits the floor.
|K = ½mv2 = ½(0.5 kg)(5.5 m/s)2 = 7.5 J |
(4) percentage of total energy that is rotational kinetic energy just before it hits the floor.
|(10 J – 7.5 J)/10 J x 100 = 25 % |
c. The hoop is replaced by a 0.5 kg solid sphere (β = 2/5), which rolls down the ramp and then onto the floor. Determine the sphere's
(1) speed at the bottom of the ramp.
|Ug = Krolling → mgLsinθ = ½(1 + β)mv2 |
|(10 m/s2)(2 m)sin30 = ½(1 + 2/5)v2 = 7/10 v2 ∴ v = 3.8 m/s |
(2) speed just before it hits the floor.
|Kfloor = Ktable + Utable → ½mvf2 = ½mvt2 + mgh |
|vf2 = (3.8 m/s)2 + 2(10 m/s2)(1 m) = 34 m2/s2∴vf = 5.8 m/s |
(3) translational kinetic energy before it hits the floor.
|K = ½mv2 = ½(0.5 kg)(5.8 m/s)2 = 8.4 J |
(4) percentage of total energy that is rotational kinetic energy just before it hits the floor.
|(10 J – 8.4 J)/10 J x 100 = 16 % |
d. Comparing a hoop (β = 1), disk (β = ½) and a sphere (β = 2/5) just before it lands on the floor, which would
(1) have the greatest % rotational kinetic energy?
|The hoop (largest β = largest % rotational energy) |
(2) land furthest from the base of the table?
|The sphere (smallest β = most translational energy) |
(3) have the most kinetic energy just before it landed?
|All three would have the same amount of kinetic energy |
4. The graph shows a system in simple harmonic motion.
[pic]
Complete the chart with either +, 0, or –.
|t |0 s |1 s |2 s |3 s |
|x (m) |0 |– |0 |+ |
|v (m/s) |– |0 |+ |0 |
|a (m/s2) |0 |+ |0 |– |
|F (N) |0 |+ |0 |– |
|U (J) |0 |+ |0 |+ |
|K (J) |+ |0 |+ |0 |
5. A 3.0 kg bob swings on the end of a 1.0 m string. The potential energy U of the object as a function of distance x from its equilibrium position is shown. This particular object has a total energy E of 0.4 J.
[pic]
a. What is the bob's potential energy when its displacement is +4 cm from its equilibrium position?
|U = 0.05 J |
b. What is the greatest distance x for the pendulum bob? Explain your reasoning.
|10 cm, the mass stops at this point because all of the energy is in the |
|form of potential energy. |
c. How much time does it take the pendulum to go from the greatest +x to the greatest –x?
|T = 2π(L/g)½ = 2π(1.0 m/10 m/s2)½ = 2.0 s ∴ 1.0 s |
d. Determine the bob's kinetic energy when x = -7 cm.
|K = 0.4 J – U = 0.4 J – 0.2 J = 0.2 J |
e. What is the object's speed at x = 0?
|K = ½mv2 → 0.4 J = ½(3.0 kg)v2 ∴ v = 0.5 m/s |
6. The cart of mass m with four wheels of mass m/4 and β = ½ is released from rest and rolls from height h. After rolling down the ramp and across the horizontal surface, the cart collides and sticks with a bumper of mass 3m attached to a spring, which has a spring constant k.
[pic]Given: m = 1 kg, h = 0.50 m, k = 250 N/m, determine
a. the potential energy of the cart at the top of the ramp.
|Ug = 2mgh |
|Ug = 2(1 kg)(10 m/s2)(0.50 m) = 10 J |
b. the speed of the cart at the bottom of the ramp.
|Ug = K = ½mv2 + ½(1 + β)mv2 = 5/4 mv2 |
|10 J = 5/4(2 kg)v2 ∴ v = 2.8 m/s |
c. the carts translational kinetic energy before the collision.
|Kt = ½mv2 |
|Kt = ½(2 kg)(2.8 m/s)2 = 8 J |
d. the speed of the cart just after the collision.
|mcvc + mbvb = (mc + mb)v' |
|2mvc + 0 = 5mv' ∴ v' = 2/5vc = 1.1 m/s |
e. the translational kinetic energy of the cart and bumper just after the collision.
|Kt' = ½mv2 |
|Kt'= ½(5 kg)(1.1 m/s)2 = 3.2 J |
f. the amount that the spring is compressed.
|K = Us = ½kx2 |
|3.2 J = ½(250 N/m)x2 ∴ x = 0.16 m |
Unit 5 Answers (Don't look until after you have tried the problem)
|2 |D—net force equals Fc, which is toward the center of the circle |
|3 |B—the turn to the left results in the right side of the bus being in |
| |"front" of you as you continue forward due to inertia |
|4 |B—the centripetal force needed to make the turn is generated by friction|
| |between the road and tires, which is not sufficient at the high speed |
|5 |B—without the tension in the string to supply the centripetal force, the|
| |ball continues in a straight line |
|6 |B—Fc = mv2/r = m(2πr/T)2/r = 4π2mr/T2, since 4π2m/T2 is constant ∴ Fc ∝ |
| |r |
|7 |C—at the bottom of the vertical circle: |
| |Fnet = Fc = Fn – Fg > 0 (upward) ∴ Fn > Fg |
|8 |A—at the top of the vertical circle, both Ft and Fg are downward, which |
| |is the direction of Fc ∴ Fc = Ft + Fg |
|9 |C—Newton's third law states that for every force there is an equal but |
| |opposite force ∴ the pulls are equal |
|10 |A—The net force is Fg, which provides the Fc necessary to keep the |
| |astronaut in orbit |
|11 |v = 2πr/T = 2π(50 m)/16 s = 20 m/s |
| |east |
| |ac = v2/r = (20 m/s)2/(50 m) = 8 m/s2 |
| |south |
|12 |T = 365 days x 24 hr/day x 3600 s/hr = 3.2 x 107 s |
| |v = 2πr/T = 2π(1.5 x 1011 m)/3.2 x 107 s = 3.0 x 104 m/s |
| |aC = v2/r = (3.0 x 104 m/s)2/1.5 x 1011 m = 6 x 10-3 m/s2 |
|13 |Fc = Ff → mv2/r = μmg → v2/50 m = (0.8)(10 m/s2) ∴ v = 20 m/s |
| |Yes, because mass cancels out of the equation. |
| |No, because the coefficient of friction is less. |
|14 |Ft-x = Fc = mv2/r = (2 kg)(5 m/s)2/(1 m) = 50 N |
| |Ft-y = Fg = mg = (2 kg)(10 m/s2) = 20 N |
| |Ft = (Ft-x2 + Ft-y2)½ = (502 + 202)½ = 54 N |
| |tanθ = Ft-y/Ft-x = Fg/Fc = 20/50 ∴ θ = 22o |
|15 |Ft = Fc – Fg = mv2/r – mg |
| |Ft = (2 kg)(5 m/s)2/(1 m) – (2 kg)(10 m/s2) = 30 N |
| |Ft = Fc + Fg = mv2/r + mg |
| |Ft = (2 kg)(5 m/s)2/(1 m) + (2 kg)(10 m/s2) = 70 N |
|16 |Ft = Fg + Fc = mg + mv2/r |
| |15 N = (1kg)(10 m/s2) +(1 kg)v2/(2 m) ∴ v = 3.16 m/s |
| |Ug + K = Ug' + K' → mgh = ½mv2 |
| |(10 m/s2)h = ½(3.16 m/s)2 ∴ h = 0.50 m |
|17 |the same |greater |
| |less |Less |
|18 |g |¼g |4g |2g |½g |
|19 |g = GM/r2= (6.67x10-11N•m2/kg2)(6.4x1023kg)/(3.4x106m)2=3.7m/s2 |
|20 |GmMm/rM2 = GmEm/rE2 → rE2/rM2 = mE/mM = 100 ∴ rM/rE = 10 |
|21 |II and III |I and IV |II only |I only |
|22 |g = GM/r2 ∴ gMoonrMoon2/mMoon = gEarthrEarth2/mEarth |
| |gMoon(1)2/1 = g(42)/81= 16g/81 = 0.2 g |
| |Mass doesn't change: 50 kg |
| |Fg = mg = (50 kg)(0.2) = 10 kg |
|24 |B—torque is the greatest when rsinθ is maximum, which is when the |
| |distance is the greatest and the angle is 90o |
|25 |A—most of the mass is along its rim ∴ β in I = βmr2 is the greatest |
|26 |D—Fr = βma: the same force would generate a greater acceleration for the|
| |object with the smallest β |
|27 |A—sliding is greater (Fgsinθ = ma) than rolling |
| |(Fgsinθ = (1 + β)ma), and μs doesn't effect acceleration |
|28 |C—placing all the rod's mass at its center |
| |rbmbg = rrmrg → (0.25 m)(1 kg) = (0.25 m)(mrod) ∴ mr = 1 kg |
|29 |B—(1 m)B = (3 m)(1 kg) ∴ B = 3 kg |
| |(1 m)(1 kg + 3 kg) = (2 m)(A) ∴ A = 2 kg |
| |1 kg + 3 kg + 2 kg = 6 kg |
|30 |A—the blocks to right of the table's edge are not balanced by an equal |
| |number of blocks to the left |
|31 |τ1 = r⊥F1 = (1 m)(45 N) = 45 m•N |
| |τ1 = r⊥F2 → 45 m•N = (0.4 m)F2 ∴ F2 = 113 N |
|32 |Fg-|| = Fgsinθ = (5 kg)(10 m/s2)sin30 = 25 N |
| |Fg-|| = Frolling = (1 + β)ma → 25 N = (1 + ½)(5 kg)a ∴ a = 3.3 m/s2 |
|33 |Ff = μFn = (0.80)(250 N) = 200 N |
| |Ff = Fc = mv2/r → 200 N = (25 kg)v2/(2 m) ∴ v = 4 m/s |
| |Fr = βma → 50 N = ½(200 kg)a ∴ a = 0.5 m/s2 |
| |vt = vo + at → 4 m/s = 0 + (0.5 m/s2)t ∴ t = 8 s |
| |Increase |
|34 |τchild = τplank → (x m)(250 N) = (0.5 m)(750 N) ∴ x = 1.5 m |
|35 |τF1 ’ τ1500 + τ15,000 |
| |(20 m)F1 = (10 m)(15,000 N) + (5 m)(150,000 N) ∴ F1 = 45,000 N |
| |F1 + F2 = 1500g + 15,000g |
| |45,000 N + F2 = 15,000 N + 150,000 N ∴ F2 = 120,000 N |
|36 |mstudent = 35.1 kg + 31.6 kg = 66.7 kg |
| |rcm = (r1m1 + r2m2)/(m1 + m2) |
| |rcm = [(172 cm)(35.1 kg) + (0 cm)(31.6 kg)](66.7 kg) = 90.5 cm |
|37 |τA = τcm → (8 m)FA = (5.5 m)(22000 N) ∴ FA = 15,000 N |
| |FA + FB = Fg → 15,000 N + FB = 22,000 N ∴ FB = 7,000 N |
|38 |τcc = τc → (5 m)Fguy wiresin37 = (5 m)(200 N)sin53 ∴ F = 265 N |
|40 |Ug = Krolling and h = 1.0 m(sinθ) = 0.5 m |
| |mgh = ½(1 + β)mv2 → v = [g/(1 + β)]½ |
| |v = [g/(1 + β)]½ = [10/(1 + 1)]½ = 2.24 m/s |
| |v = [g/(1 + β)]½ = [10/(1 + 1/2)]½ = 2.58 m/s |
| |v = [g/(1 + β)]½ = [10/(1 + 2/5)]½ = 2.67 m/s |
| |v = [g/(1 + β)]½ = [10/(1 + 0)]½ = 3.16 m/s |
| |Box-sphere-Cylinder-Hoop |
|41 |Ug = K'rolling → mgh = ½(1 + ½)mv2 |
| |(10 m/s2)(0.50 m) = ¾v2 ∴ v = 2.6 m/s |
|42 |Fg = Fc → mg = mv2/r ∴ v = (rg)½ = 10 m/s |
| |Ug-A = U'g-B + K'rolling-B → mgH + 0 = mgh + ½(1 + β)mv2 |
| |(10 m/s2)H = (10 m/s2)(20 m) + ½( 1 +2/5)(10 m/s)2 ∴ H = 27 m |
|43 |Ub = K'b + K'r-p |
| |mbgh = ½mbv2 + ½βmpv2 = ½(mb + βmp)v2 |
| |(1 kg)(10 m/s2)(1 m) = ½(1.0 kg + 1.0 kg)v2 ∴ v = 3.2 m/s |
|44 |Um2 + Um1 = U'm2 + U'm1 + K'r-M + K'm1 + K'm2 |
| |m2gh + 0 = 0 + m1gh + ½βMv2 + ½m1v2 + ½m2v2 |
| |gh(m2 – m1) = ½(βM + m1 + m2)v2 |
| |(10 m/s2)(1.0 m)(0.20 kg) = (0.625 kg)v2 ∴ v = 1.8 m/s |
|45 |Um – Wf = K'b + K'r-p + K'm |
| |mmgh – μmbgd = ½mbv2 + ½βmpv2 + ½mmv2 |
| |(14)(10)(1) – (.25)(20)(10)(1) = (10 + 1 + 7)v2 ∴ v = 2.2 m/s |
|46 |L = rβmv = (1 m)(1)(0.2 kg) (9 m/s) = 1.8 kg•m2/s |
|47 |L = rβmv = r(2/5)m(2πr/T) = 4πmr2/5T |
| |L = 4π(6.0 x 1024 kg)(6.4 x 106 m)2/(5)(60 x 60 x 24 s) |
| |L = 7.1 x 1033 kg•m2/s |
| |L = rβmv = r(1)m(2πr/T) = 2πmr2/T |
| |L = 2π(6.0 x 1024 kg)(1.5 x 1011 m)2/(60 x 60 x 24 x 365 s) |
| |L = 2.7 x 1040 kg•m2/s |
|48 |r1v1 = r2v2 → v1/v2 = r2/r1 = (5.3 x 1012 m)/(8.9 x 1010 m) = 60 |
|49 |rCβCmCvC + rMβMmMvM = (rCβCmC + rMβMmM)v' |
| |(1)(42 kg)(3 m/s) + 0 = [(1)(42 kg) + (½)(180 kg)]v' |
| |v' = 0.95 m/s |
|50 |Ltotal = rdβdmdvd + rrβrmrvr = (R)(½)(M)(V) + (R)(1)(M)(0) = ½RMV |
| |Ltotal = Ltotal' = (rdβdmd + rrβrmr)v' |
| |½RMV = (½RM + RM)v' ∴ v' = ⅓V |
|51 |mgh = ½mv2 → v = (2gh)½ = [(2)(10 m/s2)(20 m)]½ = 20 m/s |
| |rβmTvT + rβmJvJ = (rβmT + rβmJ)v' |
| |(100 kg)(20 m/s) + 0 = (145 kg)v' ∴ v' = 13.8 m/s |
| |mgh = ½mv2 ∴ h = v'2/2g = (13.8 m/s)2/(2)(10 m/s2) = 9.5 m |
| |No |
| |Start from a higher ledge or with a running start. |
| |v' = (2gh)½ = [(2)(10 m/s2)(10 m)]½ = 14 m/s |
| |rβmTvT = rβ(mT + mJ)v' |
| |vT = (145 kg)(14 m/s)/100 kg = 20.3 m/s |
| |h = v2/2g = (20.3 m/s)2/(2)(10 m/s2) = 20.6 m |
| |½mv2 + mgh = ½mv'2 |
| |½v2 + (10 m/s2)(20 m) = ½(20.3)2 ∴ v = 3.5 m/s |
|52 |Ug = Krolling → mgh = ½(1 + ½)mv2 |
| |(10 m/s2)(1 m) = ¾v2 ∴ v = 3.65 m/s |
| |d = ½(v + vo)t → 2 m = ½(3.65 m/s + 0)t ∴ t = 1.1 s |
| |same |same |
| |less |more |
| |same |more |
|53 |K' = Ug – Wf = mmgh – μmbgd |
| |K' = (5 kg)(10 m/s2)(1 m) – (0.3)(10 kg)(10 m/s2)(1 m) = 20 J |
| |K' = K'b + K'r-p + K'm = ½mbv2 + ½βmpv2 + ½mmv2 |
| |20 J = (5 kg + 0.25 kg + 2.5 kg)v2 ∴ v = 1.6 m/s |
|54 | (8.9 x 1010 m)(3.88 x 104 m/s) = (5.3 x 1012 m)v2 ∴ v2 = 652 m/s |
|55 |L = rβmv = r(2/5)m(2πr/T) = 4πmr2/5T |
| |L = 4π(7.35 x 1022 kg)(1.74 x 106 m)2/(5)(2.42 x 106 s) |
| |L = 2.3 x 1029 kg•m2/s |
| |L = rβmv = r(1)m(2πr/T) = 2πmr2/T |
| |L = 2π(7.35 x 1022 kg)(3.84 x 108 m)2/(2.42 x 106 s) |
| |L = 2.8 x 1034 kg•m2/s |
|56 |rβsmsvs + rβMmMvM = (rβsms + rβmmm)v' |
| |(1)(75)(5) + (½)(150)(2)= [(1)(75) + (½)(150)]v' ∴ v' = 3.5 m/s |
|57 |Ug = K → msgh = ½msvs2 |
| |vs = (2gh)½ = [2(10 m/s2)(2.0 m)]½ = 6.3 m/s |
| |rβmSvS + rβmBvB = rβ (mS + mB)v' |
| |(1 kg)(6.3 m/s) + 0 = (1 kg +2 kg)v' ∴ v' = 2.1 m/s |
| |Ug = K → (mS + mB)gh = ½(mS + mB)v'2 |
| |h = v'2/2g = (2.1 m/s)2/(2)(10 m/s2) = 0.22 m |
| |v' = mBvB/(mB + mS) = 12.6 kg•m/s/3 kg = 4.2 m/s |
| |h = v'2/2g = (4.2 m/s)2/(2)(10 m/s2) = 0.88 m |
| |vS + vS' = vB + vB' |
| |6.3 m/s + vS' = 0 + vB' ∴ vS' = vB' – 6.3 |
| |rβmSvS + rβmBvB = rβmSvS' + rβmBvB' |
| |(1 kg)(6.3 m/s) = (1 kg)(vB' – 6.3) + (2 kg)vB' vB' = 4.2 m/s |
| |vS' = vB' – 6.3 = 4.2 m/s – 6.3 m/s = -2.1 m/s |
| |hB = v'2/2g = (4.2 m/s)2/(2)(10 m/s2) = 0.88 m |
| |hS = v'2/2g = (-2.1 m/s)2/(2)(10 m/s2) = 0.22 m |
| |mSghS + mBghB = (1 kg)(2 m) + 0 = 2 J |
| |mBgh'B + mSgh'S = (2)(0.88) + (1)(0.22) = 1.98 J ∴ yes |
|59 |D—starting at 0: down A, up A, up A, down A |
|60 |A—you end where you started |
|61 |D—Each A is ¼T ∴ 6 x ¼T = 3/2T |
|62 |D—when x = 0, a = 0, but v = max; when x = A, v = 0, but a = max; at |
| |intermediate points a ≠ 0 and v ≠ 0 |
|63 |B—E = ½kA2 ∴ mass will not change the total energy |
|64 |D—E = ½kA2, whereas v = A(k/m)½ and a = A(k/m) change the same, and T = |
| |2π(m/k)½ doesn't change at all |
|65 |C—T = 2π(m/k)½: when m doubles T increases by √2 |
|66 |C—T = 2π(L/g)½: only changing L or g can change T |
|67 |A—T = 2π(L/g)½: greater L = greater T |
|68 |A—T = 2π(L/g)½: to speed up the clock (reduce T), you need to decrease L |
| |by moving the weight up |
|69 |C—T = 2π(L/g)½: "g" increases when accelerating up ∴ T decreases |
|70 |A—T = 2π(m/k)½: adding mass or changing the spring will change the period|
|71 |B—T = 2π(L/g)½: amplitude has no effect on T |
|72 |C—T = 2π(L/g)½: by standing on the swing, you have reduced the distance L|
| |(distance from fulcrum to center of mass), which decreases T |
|73 |B—T = 2π(L/g)½: mass has no effect on T |
|74 |½ T |¼ T, ¾ T |0 T, 1 T |
| |¾ T |0 T, ½ T, 1 T |¼ T |
| |aA = A(k/m) = (0.1 m)(100 N/m)/(1 kg) = 10 m/s2 |
| |vo = A(k/m)½ = (0.1 m)[(100 N/m)/(1 kg)]½ = 1 m/s |
| |T = 2π(m/k)½ = 2π(1/100)½ = 0.63 s |
| |Ko = ½mv2 = ½(1 kg)(1 m/s)2 = 0.5 J |
| |UA = ½kA2 = (100 N/m)(0.1 m)2 = 0.5 J |
| |v = vo[1 – (x2/A2)]½ = (1 m/s)[1 – 0.52/0.12)]½ = 0.866 m/s |
| |0 m |-0.1 m |0 m |0.1 m |
| |-1 m/s |0 m/s |1 m/s |0 m/s |
| |0 m/s2 |10 m/s2 |0 m/s2 |-10 m/s2 |
| |0 N |10 N |0 N |- 10 N |
| |doubles |doubles |remains the same |
|75 |UG = K → mgh = ½mv2 |
| |v = (2gh)½ = [2(10 m/s2)(0.015 m)]½ = 0.55 m/s |
| |T = 2π(L/g)½ = 2π(1.0 m/10 m/s2)½ = 2.0 s |
| |2.0 s |4.0 s |1.0 s |
|76 |¾ T |0 T, ½ T, 1 T |¼ T |
| |0 T, 1 T |¼ T, ¾ T |½ T |
| |aA = A(k/m) = (0.25 m)(100 N/m)/(1 kg) = 25 m/s2 |
| |vo = A(k/m)½ = (0.25 m)[(100 N/m)/(1 kg)]½ = 2.5 m/s |
| |T = 2π(m/k)½ = 2π(1/100)½ = 0.63 s |
| |Ko = ½mv2 = ½(1 kg)(2.5 m/s)2 = 3.125 J |
| |UA = ½kA2 = ½(100 N/m)(0.25 m)2 = 3.125 J |
| |vx = vo[1 – (x2/A2)]½ = 2.5 m/s[1 – (.22/.252)]½ = 1.5 m/s |
| |+0.25 m |0 m |-0.25 m |0 m |
| |0 m/s |-2.5 m/s |0 m/s |2.5 m/s |
| |-25 m/s2 |0 m/s2 |25 m/s2 |0 m/s2 |
| |-25 N |0 N |25 N |0 N |
| |aA = |-A(k/m)| = (0.50 m)(100 N/m)/(1.0 kg) = 50 m/s2 |
| |vo = A(k/m)½ = (0.50 m)[(100 N/m)/(1.0 kg)]½ = 5 m/s |
| |T = 2π(m/k)½ = 2π(1/100)½ = 0.63 s |
|77 |Ug = K ∴ mgh = ½mv2 ∴ |
| |v = (2gh)½ = [2(10 m/s2)(0.20 m)]½ = 2 m/s |
| |T = 2π(L/g)½ = 2π(2.0 m/10 m/s2)½ = 2.8 s |
Practice Multiple Choice (No calculator)
|1 |C—In uniform circular motion velocity is tangent to the circle and |
| |acceleration is toward the center |
|2 |A—Traveling east and turning south ∴ clockwise. |
| |ac = v2/r ∴ v = (acr)½ = [(3 m/s2)(300 m)]½ = 30 m/s |
|3 |D—Ball simultaneously is directed to the right (due to the spring) and |
| |toward the top (due to the rotation) |
|4 |A—Fg = GMm/r2, where r is doubled |
| |∴ Fg should be (½)2 = ¼ as much → ¼(800 N) = 200 N |
|5 |A—At Q, acceleration is ⇔, at R acceleration is ⇐ |
|6 |B—The center of mass is the balance point, which is closer to the more |
| |massive side |
|7 |B—Ball will move tangent to the circle at that spot |
|8 |B—Fc = mv2/r = (5.0 kg)(2.0 m/s)2/0.70 m = 29 N |
|9 |A—The force of gravity is the only force acting on the ball and Fg |
| |decreases as r increases (Fg = GMm/r2) |
|10 |D—g = GM/r2, r is doubled ∴ g is (½)2 = ¼ as much |
|11 |B—F↑ = F↓: 250 N + FL = 125 N + 500 N ∴ FL = 375 N |
|12 |B—rFperson + rFboard = rFright chain |
| |r(500 N) + (2 m)(125 N) = (4 m)(250 N) ∴ r = 1.5 |
|13 |A—C and D would cancel torque, B would cancel up and down, but only A |
| |will cancel both. |
|14 |C—Fg = GMm/r2, where r is doubled ∴ Fg is (½)2 = ¼ as much. |
|15 |B—g = GM/r2, where r is doubled |
| |∴ g is (½)2 = ¼ as much. ¼(10 m/s2) = 2.5 m/s2 |
|16 |B—rFperson = rFplank (rplank is 2.5 m – 2 m = 0.5 m) |
| |r(50 kg)(10 m/s2) = (0.5 m)(100 kg)(10 m/s2) ∴ r = 1 m |
|17 |C—CM = Σ(rimi)/Σ(mi)—assume the rod is 4 m |
| |2 = [(0)(3.5) + (1)(X) + (4)(5)]/(8.5 + X) ∴ X = 3 |
|18 |D—T = Fc = mv2/r = mv'2/r': r increases to 4r without changing T ∴ v'2 |
| |also increase by 4. v;2 = 4 ∴ v' = 2v |
|19 |A—At the bottom of the swing, Fnet = Fc = Ft – Fg |
| |Fc = 6 N – (0.4 kg)(10 m/s2) = 2 N |
|20 |C—τnet = τcc – τc |
| |τnet = (3R)F + (3R)F + (2R)F – (3R)(2F) = 8RF – 6RF = 2RF |
|21 |B—g = GM/r2 ∴ gE = GME/rE2 and GM = GMM/rM2 |
| |gM/gE = (MM/ME)(rE/rM)2 = (1/10)(1/½)2 = 4/10 ∴ gM = 2g/5 |
|22 |C—Fc = Fg → mv2/r = GMm/r2 ∴ r = GM/v2 |
|23 |C—Rm + (rcos60)M = R(2M) → m + ½M = 2M ∴ m = 3/2M |
|24 |D—I1m = l2M1 and l1M2 = l2m → l1/l2 = M1/m = m/M2 → |
| |M1M2 = m2 ∴ m = (M1M2)½ |
|25 |D—Energy & angular momentum are conserved, |
| |r ↑ v ↓ (r1v1 = r2v2), Ug increases (Ug = -GMm/r) |
|26 |A—r1v1 = r2v2 ∴ v2 = (r1/r2)v1 |
|27 |D—I is true (Fg = Fc = Mv2/R), II is true (F⊥r ∴ no work done), and III |
| |is true (angular momentum is constant) |
|28 |A—Ug = K (no rolling because of zero friction) |
| |mgh = ½mv2 ∴ v = 2gh = (2gh)½ |
|29 |D—mgh = K + Kr = ½mv2 + ½βmv2 = ½(1 + β)mv2 |
| |v2 = 2gh/(1 + 2/5) ∴ v = (10gh/7)½ |
|30 |B—A has no force, C and D have a constant force, only SHM has a variable |
| |force ( Fs = kx) |
|31 |B—L = rβmv = (a)(1)(m)(v) = amv |
|32 |C—Total mechanical energy is constant when a conservative force is |
| |involved |
|33 |D—The kinetic energy is greatest when the velocity is greatest (K = |
| |½mv2), which occurs at the midpoint |
|34 |B—Potential energy is greatest when x is greatest |
| |(Ug = ½kx2), which occurs at xmin and xmax, |
|35 |B—Acceleration is greatest when force is greatest (Fs = kx), which occurs|
| |at xmin and xmax |
|36 |D—Velocity is greatest at the midpoint and is zero at xmin and xmax (same|
| |as K) |
|37 |D—T = 2π(m/k)½ ∴ T depends on both m and k, but not on amplitude, A. |
|38 |D—T = 2π(m/k)½ ∴ T ∝ √m. |
| |The mass is doubled ∴ T = 2√2 = 2.8 s |
|39 |D—The period of a pendulum, T = 2π(L/g)½ ∴ T ∝ (1/g)½ and it is shorter, |
| |but the period of a spring is the same |
|40 |C—Potential energy is greatest when x is greatest |
| |(Ug = ½kx2), which occurs at 1 s and 3 s. |
|41 |A—Each spring supports half of the weight (12 N). |
| |Fs = kx → 6 N = k(0.15 m) ∴ k = 40 N/m |
|42 |The force on ball is constant, except when it is in contact with the |
| |floor, ∴ not SHM, where force ∝ x.. |
|43 |B—Pendulum: T = 2π(L/g), ∴T = T |
| |Spring: T = 2π(m/k), where T ∝ √m ∴ T = √2T |
|44 |A—Fs = kx → mg = kd ∴ k = gm/d |
| |T ∝ (m/k)½ = (md/gm)½ = (d/g)½ |
|45 |D—The 4-kg block stretches the spring 16 cm (Fs ∝ x), the spring falls |
| |another 16 cm before turning around |
|46 |D—v = A(k/m)½ |
| |v2 = A2k/M ∴ k = Mv2/A2 = M(v/A)2 |
|47 |D—Tp = 2π(L/g)½ = Ts = 2π(m/k)½ |
| |L/g = m1/k ∴ k = m1g/L |
Practice Free Response
|1 |Ug = K → mgh = ½mv2 |
| |v = (2gh)½ = [2(10 m/s2)(90 m)]½ = 42 m/s |
| |v = (2gh)½ = [2(10 m/s2)(40 m)]½ = 28 m/s |
| |Fg = mg = (700 kg)(10 m/s2) = 7,000 N |
| |Fc = mv2/r = (700 kg)(28 m/s)2/(20 m) = 27,000 N |
| |Fc = Fg + Fn ∴ Fn = Fc – Fg = 27,000 – 7,000 = 20,000 N |
|2 |F = μmbg = (0.30)(20 kg)(10 m/s2) = 60 N |
| |Fg = mmg = (10 kg)(10 m/s2) = 100 N |
| |Fnet = 100 N – 60 N = 40 N |
| |Fnet = (mb + βmp + mm)a |
| |40 N = (20 kg + 5 kg + 10 kg)a ∴ a = 1.1 m/s2 |
| |v2 = vo2 + 2ad = 02 + 2(1.1 m/s2)(2 m) ∴ v = 2.1 m/s |
| |Ug = mmgh = (10 kg)(10 m/s2)(2 m) = 200 J |
| |Wf = μmbgd = (.30)(20 kg)(10 m/s2)(2 m) = 120 J |
| |Um – Wf = K'b + K'r-p + K'm = ½mbv2 + ½βmpv2 + ½mmv2 |
| |200 J – 120 J = ½(20 kg + 5 kg + 10 kg)v2 ∴ v = 2.1 m/s |
|3 |Ug = mg(H + Lsinθ) |
| |Ug = (0.5 kg)(10 m/s2)[1 m + (2 m)(sin30)] = 10 J |
| |Ug = Krolling → mgLsinθ = ½(1 + β)mv2 |
| |(10 m/s2)(2 m)sin30 = ½(1 + 1)v2 ∴ v = 3.2 m/s |
| |Kfloor = Ktable + Utable → ½mvf2 = ½mvt2 + mgh |
| |vf2 = (3.2 m/s)2 + 2(10 m/s2)(1 m)∴ vf = 5.5 m/s |
| |K = ½mv2 = ½(0.5 kg)(5.5 m/s)2 = 7.5 J |
| |(10 J – 7.5 J)/10 J x 100 = 25 % |
| |Ug = Krolling → mgLsinθ = ½(1 + β)mv2 |
| |(10 m/s2)(2 m)sin30 = ½(1 + 2/5)v2 = 7/10 v2 ∴ v = 3.8 m/s |
| |Kfloor = Ktable + Utable → ½mvf2 = ½mvt2 + mgh |
| |vf2 = (3.8 m/s)2 + 2(10 m/s2)(1 m) = 34 m2/s2∴vf = 5.8 m/s |
| |K = ½mv2 = ½(0.5 kg)(5.8 m/s)2 = 8.4 J |
| | (10 J – 8.4 J)/10 J x 100 = 16 % |
| |The hoop (largest β = largest % rotational energy) |
| |The sphere (smallest β = most translational energy) |
| |All three would have the same amount of kinetic energy |
|4 |0 |– |0 |+ |
| |– |0 |+ |0 |
| |0 |+ |0 |– |
| |0 |+ |0 |– |
| |0 |+ |0 |+ |
| |+ |0 |+ |0 |
|5 |U = 0.05 J |
| |10 cm, the mass stops at this point because all of the energy is in the |
| |form of potential energy. |
| |T = 2π(L/g)½ = 2π(1.0 m/10 m/s2)½ = 2.0 s ∴ 1.0 s |
| |K = 0.4 J – U = 0.4 J – 0.2 J = 0.2 J |
| |K = ½mv2 → 0.4 J = ½(3.0 kg)v2 ∴ v = 0.5 m/s |
|6 |Ug = 2mgh = 2(1 kg)(10 m/s2)(0.50 m) = 10 J |
| |Ug = K = ½mv2 + ½(1 + β)mv2 = 5/4 mv2 |
| |10 J = 5/4(2 kg)v2 ∴ v = 2.8 m/s |
| |Kt = ½mv2 = ½(2 kg)(2.8 m/s)2 = 8 J |
| |mcvc + mbvb = (mc + mb)v' |
| |2mvc + 0 = 5mv' ∴ v' = 2/5vc = 1.1 m/s |
| |Kt' = ½mv2 = ½(5 kg)(1.1 m/s)2 = 3.2 J |
| |K = Us = ½kx2 → 3.2 J = ½(250 N/m)x2 ∴ x = 0.16 m |
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Physics
is Phun
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