# Find the angular speed of the cylinder

• PKay
In summary, a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R. The cylinder rotates without friction about a horizontal axle along the cylinder axis, starting with angular speed ω0. After one revolution of the cylinder, the rope has unwrapped and hangs vertically down, tangent to the cylinder. To find the angular speed of the cylinder at this time, you can use the equation U=Mgycm and express the answer in terms of the variables m, M, and R, along with appropriate constants. To eliminate one of the variables, you can use the relationship between v and ω. Additionally, in finding the center of mass, the circumference of the cylinder is divided by 2 because it is
PKay

## Homework Statement

Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R. The cylinder rotates without friction about a horizontal axle along the cylinder axis. One end of the rope is attached to the cylinder. The cylinder starts with angular speed ω0. After one revolution of the cylinder the rope has unwrapped and, at this instant, hangs vertically down, tangent to the cylinder.

Find the angular speed of the cylinder at this time. You can ignore the thickness of the rope. (Hint: Use Equation U=Mgycm.)
Express your answer in terms of the variables m, M, R, and appropriate constants.

## Homework Equations

$y = 1/2 M R^2$
ω=v/r

## The Attempt at a Solution

I know there's a post for the same problem. But I'm still confused. If I was supposed to ask that person on their thread or something, please tell me so I can delete this thread. I'm new to the forum. Anyways, this is what I tried:
mgh +1/2Iω02 = 1/2mv2 + 1/2Iwf2
mgR+1/2(1/2MR2ω02 = 1/2mv21/2(1/2MR2wf2
ωf = √{(mGR+1/4MR2ω02)/(1/2mv2+1/4MR2}

Apparently, according to masteringphysics, the answer does not depend on mv but now I'm a little stuck. I'm not sure what I should be substituting mv with. Thank you in advance!

PKay said:

## Homework Statement

Exactly one turn of a flexible rope with mass m is wrapped around a uniform cylinder with mass M and radius R. The cylinder rotates without friction about a horizontal axle along the cylinder axis. One end of the rope is attached to the cylinder. The cylinder starts with angular speed ω0. After one revolution of the cylinder the rope has unwrapped and, at this instant, hangs vertically down, tangent to the cylinder.

Find the angular speed of the cylinder at this time. You can ignore the thickness of the rope. (Hint: Use Equation U=Mgycm.)
Express your answer in terms of the variables m, M, R, and appropriate constants.

## Homework Equations

$y = 1/2 M R^2$
ω=v/r

## The Attempt at a Solution

I know there's a post for the same problem. But I'm still confused. If I was supposed to ask that person on their thread or something, please tell me so I can delete this thread. I'm new to the forum. Anyways, this is what I tried:
mgh +1/2Iω02 = 1/2mv2 + 1/2Iwf2
mgR+1/2(1/2MR2ω02 = 1/2mv21/2(1/2MR2wf2
ωf = √{(mGR+1/4MR2ω02)/(1/2mv2+1/4MR2}

Apparently, according to masteringphysics, the answer does not depend on mv but now I'm a little stuck. I'm not sure what I should be substituting mv with. Thank you in advance!
Hello PKay. Welcome to PF !

If you are referring to this thread as being the previous post, then you are correct in starting this new thread. For one thing, that thread is rather old. Also, it has been closed.

There is a relationship between v and ω . That should enable you to eliminate one of them.

SammyS said:
Hello PKay. Welcome to PF !

If you are referring to this thread as being the previous post, then you are correct in starting this new thread. For one thing, that thread is rather old. Also, it has been closed.There is a relationship between v and ω . That should enable you to eliminate one of them.

Yes! I was referring to that thread. I didn't know it was closed. Glad I did it correctly.

I actually got it now. I actually still have a related question. C=2piR but why do we have to divide by 2 for the center of mass?

Last edited:

## What is the definition of angular speed?

Angular speed is defined as the rate at which an object rotates or completes a full circle in a given amount of time.

## How is angular speed different from linear speed?

Angular speed measures how fast an object is rotating and is typically measured in radians per second. Linear speed, on the other hand, measures how fast an object is moving in a straight line and is typically measured in meters per second.

## What is the formula for calculating angular speed?

The formula for calculating angular speed is ω = θ/t, where ω is the angular speed, θ is the angular displacement in radians, and t is the time taken to complete the rotation.

## Can angular speed be negative?

Yes, angular speed can be negative if the object is rotating in the clockwise direction. Positive angular speed indicates counterclockwise rotation and negative angular speed indicates clockwise rotation.

## How does the radius of a cylinder affect its angular speed?

The radius of a cylinder does not directly affect its angular speed. However, a larger radius can result in a larger linear speed at the edge of the cylinder due to a larger distance traveled in one rotation. This means that the same angular speed can result in a faster linear speed for a larger cylinder compared to a smaller cylinder.

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