Physics 2210 Homework 18 Spring 2015

Physics 2210 Homework 18

Spring 2015

Charles Jui

April 12, 2015

IE Sphere Incline

Wording

A solid sphere of uniform density starts from rest and rolls without slipping down an inclined plane with angle = 30.The sphere has mass M = 8 kg and radius R = 0.19 m . The coefficient of static friction between the sphere and the plane is ? = 0.64. What is the magnitude of the frictional force on the sphere? See Figure 1.

Figure 1: Sphere Incline

Solution

II.Newton law gives

Fnet = Ma

1

The net force is composed of the three forces

Fnet = N + Fg + f

Since the sphere is moving along the incline, only its parallel component will be non-zero, i.e.

Fnet = M g sin - f On the other hand, for rotation we have

net = I Taking the center as origin, we get1

net = Rf No slipping condition requires

a = R

Combining the above results2

M g sin - f M

=

Fnet M

=R?

net I

=

R2f (2/5)M R2

Solving for f

2 f = M g sin

7

f = 11.211 N

Gymnast

Wording

A gymnast with mass m1 = 41 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 108 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam. See Figure 2.

Figure 2: Gymnast 1

1Only the frictional force gives non-zero contribution.

2Moment

of

inertia

for

solid

sphere

is

I

=

2 5

M

R2

.

2

Figure 3: Gymnast 2

1. What is the force the left support exerts on the beam?

2. What is the force the right support exerts on the beam?

3. How much extra mass could the gymnast hold before the beam begins to tip?

4. Now the gymnast (not holding any additional mass) walks directly above the right support. What is the force the left support exerts on the beam? See Figure 3.

5. What is the force the right support exerts on the beam?

6. At what location does the gymnast need to stand to maximize the force on the right support?

Solution

Forces supports push up are FL and FR. Since the system is at rest, we got

0 = Fnet = FL + FR - m1g - m2g Since it's not rotating, it's true that3

L

L

L

0 = net = 6 FR - 6 FL + 2 m1g

1) Solving for the left force

1 FL = 2 (4m1 + m2)g

FL = 1334.16 N

2)

Solving for the right force

1 FR = 2 (m2 - 2m1)g

FR = 127.53 N

3Taking the center of the beam as the origin.

3

3)

The beam starts to tip when FR = 0, thus

0

=

1 2 (m2

- 2m1)g

m1 = m2/2

And so

m1

=

m1

-

m1

=

m2 2

-

m1

m1 = 13 kg

4)

Again we have

0 = Fn et = FR + FL - m1g - m2g

Site of the force has changed, thus4

0

=

n et

=

L 6

FR

-

L 6

FL

-

L 6 m1g

Thus the new left force

FL

=

1 2 (2m1

+

m2)g

FL = 931.95 N

5) Solving for the new right force

FR

=

1 2 m2g

FR = 529.74 N

6) At the right edge of the beam

Hanging Beam

Wording

A purple beam is hinged to a wall to hold up a blue sign. The beam has a mass of mb = 6.1 kg and the sign has a mass of ms = 16.8 kg. The length of the beam is L = 2.44 m. The sign is attached at the very end of the beam, but the horizontal wire holding up the beam is attached 2/3 of the way to the end of the beam. The angle the wire makes with the beam is = 32.6. See Figure 4.

4Origin at the beam's center.

4

Figure 4: Hanging Beam

1. What is the tension in the wire? 2. What is the net force the hinge exerts on the beam? 3. The maximum tension the wire can have without breaking is T = 977 N .

What is the maximum mass sign that can be hung from the beam? 4. What else could be done in order to be able to hold a heavier sign?

Solution

Since the beam is not moving, we have

0 = Fnet = Fb + T + Fs + Fh 0 = net = b + t + s + h

In coordinates5

0 = T - Fhx

(1)

0 = Fhy - msg - mbg

(2)

L

2L

0 = 2 mbg cos - 3 T sin + Lmsg cos

(3)

1) Solving for tension(from the Eq. (3))

T = 3(mb + 2ms)g 4 tan

T = 456.733 N

5Taking the position of the hinge as the origin.

5

2) Components of the hinge force are

Fhx = T Fhy = (ms + mb)g

And thus its magnitude

Fh = (Fhx)2 + (Fhy)2 = T 2 + (ms + mb)2g2

or

Fh = g ?

9(mb + 2ms)2 16 tan2

+

(ms

+

mb)2

Fh = 508.991 N

3)

Turning around the formula for the tension

ms,max

=

2Tmax tan 3g

-

mb 2

ms,max = 39.411 kg

4) ? while still keeping it horizontal, attach the wire to the end of the beam ? keeping the wire attached at the same location on the beam, make the wire perpendicular to the beam

? attach the sign on the beam closer to the wall

Meterstick

Wording

A meterstick (L = 1 m) has a mass of m = 0.145 kg. Initially it hangs from two short strings: one at the 25 cm mark and one at the 75 cm mark. See Figure 5.

Figure 5: Meterstick 1 1. What is the tension in the left string?

6

Figure 6: Meterstick 2

2. Now the right string is cut! What is the magnitude of the initial angular acceleration of the meterst about its pivot point? (You may assume the rod pivots about the left string, and the string remains vertical)

3. What is the tension in the left string right after the right string is cut?

4. After the right string is cut, the meterstick swings down to where it is vertical for an instant before it swings back up in the other direction. What is the angular speed when the meterstick is vertical? See Figure 6.

5. What is the magnitude of the acceleration of the center of mass of the meterstick when it is vertical?

6. What is the tension in the string when the meterstick is vertical?

7. Where is the angular acceleration of the meterstick a maximum?

Solution

1) Since the stick is stable so6

0 = Fnet = TL - mg + TR 0 = net = (L/4)TR - (L/4)TL

Solving for the tension in the left string

1

TL

=

mg 2

TL = 0.711 N

2) The torque about the pivot is

= F sin = (L/4)mg

6Origin is in the middle of the stick.

7

The second Newton law for rotation

= I

Moment of inertia of the rod about this pivot can be obtained by parallel

axis theorem

I

=

I0

+

md2

=

1 mL2 12

+

m(L/4)2

=

7 mL2 12

So solving for the angular acceleration

12g = =

I 7L

= 16.817 rad/s2

3)

At this moment, the acceleration of the center of mass is a = (L/4). So the second Newton law gives

mg - TL = Fnet = ma = m(L/4)

Solving for the tension

TL

=

4 mg

7

TL = 0.813 N

4)

Using conservation of energy

1 I2 = mg L

2

4

Solving for the angular speed7

mgL

6g

=

=

2I

7L

= 5.800 rad/s

5)

Acceleration of the center of mass is completely centripetal8

and hence

ac

=

2

=

6g 7L

?

L 4

3

ac

=

g 14

ac = 8.409 m/s2

7Using the moment of inertia about the pivot - see above. 8The radius of rotation being = L/4.

8

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download