Calculating Moment of Inertia for a Rotating Cylinder | Physics Tutorial

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The discussion focuses on calculating the moment of inertia for a rotating cylinder with a radius of 20 cm, subjected to a 50 g mass that drops 1 m in 12 seconds. The key equation used is the conservation of energy, represented as 0.5mv² + 0.5Iω² = mgh. The user correctly identifies that angular velocity (ω) relates to linear velocity (v) through the equation ω = v/r, but clarifies that the final speed is greater than the average speed due to the initial rest condition. This highlights the importance of accurately determining final velocities in rotational dynamics.

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dross
New user has been reminded to post using the Homework Help Template in future schoolwork threads.
Hi I am struggling with the following question.

A cylinder of radius 20cm is mounted on a horizontal axle coincident with its axis and is free to rotate. A light chord is wound onto it and a 50g mass hung from it. After release the mass drops 1m in 12seconds. What is the moment of inertia?

With energy conversation I know the gravitational potential energy lost must equal that of the kinetic energy gained my the mass + the rotational energy gained by the cylinder

0.5mv^2+0.5Iω^2 = mgh

I can rearrange this to find moment of Inertia, I. However is ω = v/r ? with v being 1/12 m/s ?
 
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v is not 1/12 m/s. The average speed is 1/12 m/s, but since it starts from rest, the final speed must be higher.
Otherwise you seem on the right track.
 
Yep got it. Thanks for your help!
 

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