A spindle is wrapped with a very fine silk thread and is sitting initially at rest on a horizontal table. The outer radius of the spindle is 10cm and the total mass of the spindle with the thread is 100 grams. You begin to pull on the thread with a constant force at an angle of 90 degrees to radius of the spindle. The thread is applying this force a distance of 5 cm from the center of the spindle. After the spindle has rolled 1m without slipping the total kinetic energy is 2 joules and the angular velocity is 2 rads/sec.A. What is the moment of inertia of the spindle? Assuming that the distance from the center of the spindle the thread is pulling is constant and the thread's mass can be ignored. B. Assuming that the coefficient of static friction is 0.5, what is the force of tension on the thread?

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Asked Nov 27, 2019
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A spindle is wrapped with a very fine silk thread and is sitting initially at rest on a horizontal table. The outer radius of the spindle is 10cm and the total mass of the spindle with the thread is 100 grams. You begin to pull on the thread with a constant force at an angle of 90 degrees to radius of the spindle. The thread is applying this force a distance of 5 cm from the center of the spindle. After the spindle has rolled 1m without slipping the total kinetic energy is 2 joules and the angular velocity is 2 rads/sec.

A. What is the moment of inertia of the spindle? Assuming that the distance from the center of the spindle the thread is pulling is constant and the thread's mass can be ignored. 
B. Assuming that the coefficient of static friction is 0.5, what is the force of tension on the thread? 

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Expert Answer

Step 1

(A)

Consider the spindle as a solid cylinder, so the moment of inertia of the spindle is,

Substitute the values,

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0.001kg 1(100g) '0.01m (10 cm cm 0.0005 kg m

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Step 2

(B)

The angular displacement of the spindle is,

According to the rotational kinematics of equation.

Substitute the values,

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r 5 cm 10 cm -0.5 rad 2 o-a2a0 20 2rad/s 0rad/s 2(0.5 rad) 2rad/s

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Step 3

The torque exerted on the spindle is,

Substitute the values ...

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rT sin 0= Ia Ia Т- rsin (0.0005kg m2)(2rad/s?) (0.05m)sin (90°) Т- 0.02N

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Angular Motion

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