M, a solid cylinder (M=2.35kg, R=0.113 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 0.870 kg mass, i.e., F=8.535N. 1. Calculate the angular acceleration of the cylinder. 2. If instead of the force F an actual mass (m=0.870kg) is hung from the string, find the angular acceleration of the cylinder. 3. How far does m travel between 0.690 s and .890s after the motion begins? 4. 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 0.477m in a time of 0.530s. Find Icm of the new cylinder.

College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter8: Rotational Equilibrium And Dynamics
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M, a solid cylinder (M=2.35kg, R=0.113 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 0.870 kg mass, i.e., F=8.535N. 1. Calculate the angular acceleration of the cylinder. 2. If instead of the force F an actual mass (m=0.870kg) is hung from the string, find the angular acceleration of the cylinder. 3. How far does m travel between 0.690 s and .890s after the motion begins? 4. 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 0.477m in a time of 0.530s. Find Icm of the new cylinder.
Expert Solution
Step 1

"As per the guidelines i solved the first three subpart".

1.

The torque due to motion of the solid cylinder is,

τs=τsFR=IαFR=MR22αF=MR2α

Solve for angular acceleration of the cylinder.

α=2FMR=28.535 N2.35 kg0.113 m=64.3 rad/s2

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