M, a solid cylinder (M=2.39 kg, R=0.127 m) pivots on a thin, fixed, frictionless

Question

M, a solid cylinder (M=2.39 kg, R=0.127 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.730 kg mass, i.e., F = 7.161 N. Calculate the angular acceleration of the cylinder.

4.72×101 rad/s^2 (correct)

If instead of the force F an actual mass m = 0.730 kg is hung from the string, find the angular acceleration of the cylinder.

2.93×101 rad/s^2 (correct.)

How far does m travel downward between 0.550 s and 0.750 s after the motion begins?

4.84×10-1 m (correct).

The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.302 m in a time of 0.450 s. Find Icm of the new cylinder.

*You know how to write an equation for the torque from the second part. You are given quantities to calculate the acceleration (review motion in one dimension if this is not obvious), and in this case you need to solve for the moment of inertia.* (hint from professor)

Explanation / Answer

L = r*
= L/r = 0.302 m / 0.127 m = 2.38 rad

we know,
= 1/2**t^2
= 2*/ t^2
= (2*2.38)/(.45)^2
= 23.5 rad/s^2

I* = F*r
I = F*r/
I = (7.161*0.127)/(23.5)
I = 0.039 kg*m^2
Icm of the new cylinder, I = 0.039 kg*m^2

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