What is the Angular Speed of a Freely Rotating Cylinder?

AI Thread Summary
The discussion centers on calculating the angular speed of a freely rotating cylinder using conservation of energy principles. The initial attempt incorrectly applied the conservation of energy without considering the parallel-axis theorem for the moment of inertia, which is necessary since the rotation axis is offset from the center of mass. Participants emphasize the importance of accounting for both rotational and translational kinetic energies in the calculations. The correct approach involves using the formula that incorporates the moment of inertia derived from the parallel-axis theorem. Ultimately, the focus is on refining the calculations to accurately determine the angular speed as the cylinder passes through its lowest position.
tebes
Messages
38
Reaction score
0

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.
 
Physics news on Phys.org
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
 
Back
Top