What is the Angular Speed of a Freely Rotating Cylinder?

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Homework Help Overview

The problem involves a uniform cylinder rotating about a horizontal axis that is offset from its central longitudinal axis. The goal is to determine the angular speed of the cylinder as it passes through its lowest position, using principles of conservation of energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using conservation of energy and the parallel-axis theorem to find the moment of inertia. There are attempts to relate the angular speed to the height from which the cylinder is released. Some participants question the inclusion of translational motion in their calculations.

Discussion Status

There is ongoing exploration of the correct application of energy conservation principles and the moment of inertia. Some participants have provided guidance on focusing solely on rotational motion, while others are still trying to identify potential mistakes in their calculations.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share or the methods they can use. The problem setup includes specific measurements for the cylinder's radius and mass, as well as the distance from the axis of rotation.

tebes
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Homework Statement


A uniform cylinder of radius 12 cm and mass 25 kg is mounted so as to rotate freely about a horizontal axis that is parallel to and 6.6 cm from the central longitudinal axis of the cylinder.
If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position?

Homework Equations





The Attempt at a Solution


I'm using conservation of energy..
1/2 Iw^2= mgh
1/4 (r^2)(w^2) = gh
Then, solve for w
And i got 13.40 rad/s
But the answer is not correct.
Can someone point out my mistake. Thank you.
 
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You must apply parallel-axis theorem to find the moment of inertia (since the axis doesn't pass through the center of mass). Take in account the displacement of the center of mass and the rotational and translational kinetic energies when applying conservation of energy. Remember this equation: vCM = R*angular speed (vCM = speed of the center of mass). :)
 
BetoG93 said:
You must apply parallel-axis theorem to find the moment of inertia (since the axis doesn't pass through the center of mass). Take in account the displacement of the center of mass and the rotational and translational kinetic energies when applying conservation of energy. Remember this equation: vCM = R*angular speed (vCM = speed of the center of mass). :)

I found moment of inertia using parallel-axis theorem.
Then, I used the conservation of energy to solve for angular velocity.
But I still got it wrong.

I = Icom + MH^2
solve for I.
Then,
1/2 mv^2 + 1/2 Iw^2 = mgh
m(wr)^2 + Iw^2 = 2mgh
w^2 = ( 2mgh) / ( I + mr^2)
w = [( 2mgh) / ( I + mr^2)]^2

Maybe I missed something.
 
Try to ignore translation. It seems that there is just a rotation, since the axis doesn't move.
 
BetoG93 said:
Try to ignore translation. It seems that there is just a rotation, since the axis doesn't move.

ok . i ll try it.
 
tebes said:
ok . i ll try it.

You are right. We need to exclude the translation.
 
1/2mω^2=_{M}ΔP
 

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