Boddeker's Mechanics Notes
|Chapter 10 Outline: Angular |
|10.1: Angular Position, Velocity, and Acceleration: radian, average angular speed, instantaneous|10.5: Calculation of Moments of Inertia: moment of inertia of a rigid object, parallel-axis theorem |
|angular speed, average angular acceleration, instantaneous angular acceleration | |
| |10.6: Torque: moment arm |
|10.2: Rotational Kinematics: Rotational Motion with Constant Angular Acceleration | |
| |10.7: Relationship Between Torque and Angular Acceleration |
|10.3: Angular and Linear Quantities | |
| |10.8: Work, Power, and Energy in Rotational Motion Rolling Motion of a Rigid Object |
| | |
|10.4: Rotational Kinetic Energy: moment of inertia, rotational kinetic energy |10.9: Rolling Motion of a Rigid Object |
|10.1: Angular Position, Velocity, and Acceleration |
|sin (½dθ) = ½dx / r |As dθ approaches zero ½ds approaches ½dx which results in |[pic] |
|sin (½dθ) = ½ds / r | | |
|½dθ = ½ds / r |If θ is in radians, sin(θ) = θ if less than 10° (10°π / 180°) | |
|dθ = ds / r | | |
|s = rθ | | |
|vave = Δx / Δt |a = Δv / Δt | |
|ωave = Δθ / Δt |α = Δω / Δt | |
|v = dx / dt |a = dv / dt | |
|ω = dθ / dt |α = dω / dt | |
|Example |α = Δω / Δt |
|If the Merry-Go-Round accelerates for |0.1 = Δω / 5 |
|5 seconds at a rate of 0.1 rad/s2, what is |Δω = ½ rad / sec |
|the final angular velocity of the Unicorn. |(initial is zero…so the change is equal to the final) |
|The Unicorn is at the outside edge 6 meters | |
|from the center? |Same answer for both…angular velocity doesn’t depend on radius. |
|At this time what is the angular velocity of | |
|the Satyr on the inside edge which is 3 meters | |
|from the center? | |
| |
|10.2: Rotational Kinematics |
|d = ½at2 + v0t + d0 |
|θ = ½αt2 + ω0t + θ0 |
|Example |ω = dθ / dt |
|If the angular position is described by the following equation, θ = 6t2 - 5t - 20, what is |α = dω / dt α = d2θ / dt2 |
|the angular acceleration at 4 seconds? | |
| |α = d2(6t2 - 5t – 20) / dt2 |
| |α = 12 radians/s2 |
|Example |[pic] | τ = I α |Key Point |
|If a frictionless pulley has an inertia of 1 kgm2 (which can| |FNet r = I α |This is a frictionless pulley, |
|be modeled as a solid disk), where m1 = 8 kg, m2 = 12 kg, | |[m2(g-a) – m1(g+a)] r = I (a / r) |but NOT massless, so the |
|and R = 40 cm; what is the acceleration of the system? | |a = (m2 - m1)g / (I/r2 + m2 + m1) |tension on each side is NOT |
| | |a = 1.53 m/s2 |equal. |
| |
|10.3: Angular and Linear Quantities |
|v = dx / dt |at = dv / dt |ac = v2 / r |a2 = at2 + ac2 |
|s = r θ |at = r dω/dt |ac = (ωr)2/ r |a2 = (αr)2 + (ω²r)2 |
|dx = r dθ |at = r α |ac = ω2r | |
|v = r (dθ) / dt |at = α r | |a = r √(α2 + ω4) |
|ω = dθ / dt | | | |
|dθ = ω dt |at ( tangential |ac ( centripetal |[pic] |
|v = r (ω dt) / dt | | | |
|v = ω r | | | |
|Example | |v = ω r |
|If the Merry-Go-Round accelerates for | |v = ½ rad/sec (6 meters) |
|5 seconds at a rate of 0.1 rad/s2, what is |α = Δω / Δt |v = 3.0 m/s |
|the final (tangential) velocity of the Unicorn. |0.1 = Δω / 5 | |
|The Unicorn is at the outside edge 6 meters |Δω = ½ rad / sec |v = ω r |
|from the center? | |v = ½ rad/sec (3 meters) |
|At this time, what is the (tangential) velocity of | |v = 1.5 m/s |
|the Satyr on the inside edge which is 3 meters | | |
|from the center? | | |
|Demo: Rotational Inertia: ME-Q-RI |
|10.4: Rotational Kinetic Energy |
|Ki = ½ mivi² |All objects can be broken up into smaller components, thus rotational |KR = ½Σ mivi² |where v = ωr so |
| |kinetic energy is the sum of the kinetic energies of all these smaller |KR = ½Σ miri²ωi² |At this point we define a new term, Inertia. |
| |components, or |KR = ½I ω² |I = ½Σ miri² |
|K = Kt + Kr t ( translational r ( rotational |
|K = ½mv² + ½Iω² |
|Demo: Rotational Inertia Rods: ME-Q-RR |
|10.5: Calculation of Moments of Inertia |
| | |[pic] |
|Parallel Axis Theorem | | |
| | | |
|The moment of inertia about any axis parallel to and a distance D away from this axis | | |
|is | | |
|I = ICM + MD2 | | |
|Figure (a) shows a disk that can rotate about an axis at a radial distance h from the center of the disk. Figure (b) gives the rotational inertia I of the disk about the axis as a function of that distance h, from |
|the center out to the edge of the disk. The scale on the I axis is set by IA = 0.150 kg·m2 and IB = 0.450 kg·m2. What is the mass of the disk? |
|Calculate I at h = 0.1 m (or 0.2 m) |Then apply |
| |Parallel Axis Theorem with the following given |
|(IB - IA = .3) | |
|Ih=0.1 = 0.15 + 2.5div (0.3kgm2 / 10div) | |
|Ih=0.1 = 0.225 kgm2 | |
|Then write two Parallel Axis Theorem Equations. |ICM = Ih=0 = 0.15 kgm2 |ICM + md2 |
| | |0.15 + md2 |
|h = 0.1 |h = 0.2 |Two Unknowns…two equations. |
|½ mr2 + m h2 |½ mr2 + m h2 |To solve…subtract the two equations |
|½ mr2 + m 0.12 |½ mr2 + m 0.22 | |
|0.15 + 0.225 |0.15 + 0.45 |½ mr2 + m 0.22 = 0.6 |
| | |½ mr2 + m 0.12 = 0.375 |
|½ mr2 + m 0.12 = 0.375 |½ mr2 + m 0.22 = 0.6 |0 + m (0.03) = 0.225 |
| | |solve for m |
| | |m = 7.5 kg |
| | | |
|Another example this time using the basic definition ( r2dm | |I = (as Δmi approaches 0) |
| | |I = Σ ri2Δmi |
|Example |I = ∫ r2dm |I = ∫r2dm |
|What's the moment of inertia through the central axis of a symmetrical |I = ∫ r2 (2πρL r dr) | |
|homogenous cylinder? |I = 2πρ L ∫0 to R r3 dr |dm = ρ dV ρ = m / V |
| |I = ½πρL R4 | |
| |I = ½ MR2 |& by substituting ρ = M/V where V = πR2L back into the equation we get |
| |
|What role do center of mass of an object and inertia play? |
|Below are three different pendulums |
|The one to the right is obvious, |[pic] |
|CM = L/2 (the middle of the meter stick), thus, h is zero to L/2 | |
|PEmax or | |
|Ug = mg(L/2) and KR = ½Iω² | |
| | |
|How do we calculate Inertia (r2dm)? | |
|Linear density is λ = m / L | |
|(but L is the radius, so λ = m / r) | |
|I = r2 dm dm = λ dr |Ug = mg(L/2) KR = ½Iω2 |And if we want the linear speed at the bottom or the pendulum |
|I = r2 λ dr | |v = ωr (r = L) |
|I = λ ∫r2dr |mg(L/2) = ½Iω2 |v = ωL |
|I = λ r3/3 from 0 to L |mg(L/2) = ½(m L2/3)ω2 |v = L √(3g/L) |
|I = λ L3/3 λ = m / L |ω = √(3g/L) |v = √(3gL2/L) |
|I = (m/L) L3/3 | |v = √(3gL) |
|I = m L2/3 | | |
|Once again CM is very simple (it's the center of the ball since the rod is massless), and the Inertia doesn't even have to be calculated |[pic] |
|PEmax or Ug = mg(L) and KR = ½Iω2 | |
|mgL = ½Iω2 | |
|mgL = ½(m L2)ω2 | |
|ω = (2g/L)½ | |
|This time we simply add together the Inertia of each portion of the rod-ball pendulum. |[pic] |
|I = Irod + Iball |Ug = mg(L) and KR = ½Iω2 | |
|I = m L2/3 + m L2 |mgL = ½ I ω2 | |
|I = 4m L2/3 |mgL = ½(4m L2/3)ω2 | |
| |ω = (3g/2L)½ | |
|As you can see the center of mass, thus the inertia, for all of three of these changes. |
| |[pic] |
|Table 10.2 | |
|[pic] | |
| | |
| |[pic] |
| |[pic] |
|Demo: Rolling Objects: ME-Q-RO |
|10.6: Torque |
|Force times the lever arm (distance) is Torque, τ |[pic] |
|τ = Fd | |
|Where d = sinφ r | |
|d is the perpendicular distance from the pivot point to the force vector | |
|τ = Fd where d = sinφ r | |
|τ = r F sinφ | |
| | |
|Our Lab class…shows Torque is | |
|τ = F(d, which is a little easier for beginning students to understand. | |
|τ = F( d | |
|τ = sin60°(100N) (0.1m) | |
|τ = 8.66 Nm | |
| | |
|Proper METHOD | |
|τ = F d( | |
|τ = 100 N sin60°(0.1 m) | |
|τ = 8.66 Nm | |
| | |
|Doesn’t look too different…right? | |
|[pic] | |
| |[pic] |
| | |
| |[pic] |
| |
|10.7: Relationship Between Torque and Angular Acceleration |
|F = m a | |
|τ = I ( | |
|What is the angular acceleration of a 100 gram meter stick (a hole is drilled at the 0 cm and the stick | inertia of a solid rod is 1/3 mr2 |
|pivots about the 0 cm position) if a force of 0.2 Newtons is applied to the 100 cm end of the stick? |τ = I ( |
| |F d = m r2/3 ( |
| |0.2N*1m = 0.1kg(1)2/3 ( |
| |( = 6 rad/s2 |
|10.8: Work, Power, and Energy in Rotational Motion Rolling Motion of a Rigid Object |
|dW = F ds |[pic] |
|dW = F sinφ r dθ | |
|dW = τ dθ | |
|So the rate at which work is being done | |
| | |
|(just like the clock in our room where second hand requires work to be done all the time to keep it moving) | |
|dW = τ dθ |Σ τ = I α |
|dW/dt = τ dθ/dt P = dW / dt |Σ τ = I dω / dt |
|Power = τ dω dω = dθ/dt |Σ τ = I dω/dθ (dθ/dt) (apply chain rule) |
| |Στdθ = I dω ( ω ) dW = Σ τ dθ |
| |dW = I ∫ ω dω |
| |Σ W = ΔKR |
| |Σ W = ½I ωf² - ½I ωi² |
| |
|10.9: Rolling Motion of a Rigid Object |
|vCM = ds / dt |aCM = dvCM / dt |Parallel Axis Theorem |
|vCM = R dθ / dt |aCM = R dω / dt |The moment of inertia about any axis parallel |
|vCM = R ω |aCM = R α |to and a distance D away from this axis is |
| | |I = ICM + MD² (see above 10.5) |
| | | |
| | |[pic] |
| | |
| | |
| | |
| | |
| | |
|Applying the parallel axis theorem | |
|the Total Kinetic Energy (K = ½Ipω²) can now be expressed as |
| K = ½ICMω² + ½MR²ω² |
| K = ½ICMω² + ½M vCM² |
|Demo: Double Cone and Plane: ME-J-DC |
|Actually belongs in Ch 9…but the center of mass is actually lowering…but it looks like the cone is going up…so could also be included in conservation of energy…so a good review for Total Kinetic Energy. |
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related searches
- mechanics flat rate guide free
- mechanics flat rate time chart
- mechanics flat rate manual
- notes to make someone s day
- laws of mechanics physics
- physics mechanics problems
- journal of the mechanics and
- physics mechanics topics
- examples of mechanics in physics
- physics mechanics practice problems
- european journal of mechanics a
- international journal of mechanics science