Calculating Moment of Inertia for a Rotating Cylinder | Physics Tutorial

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To calculate the moment of inertia for a rotating cylinder, the gravitational potential energy lost by a hanging mass equals the sum of the kinetic energy gained by the mass and the rotational energy gained by the cylinder. The equation 0.5mv^2 + 0.5Iω^2 = mgh is used, where m is the mass, v is the linear velocity, I is the moment of inertia, and ω is the angular velocity. The relationship between linear and angular velocity is given by ω = v/r, but the final speed must be higher than the average speed since the cylinder starts from rest. Clarifications on the correct value of v are essential for accurate calculations. Understanding these principles is crucial for solving the problem effectively.
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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