- #1

Adoniram

- 94

- 6

## Homework Statement

A uniform chain of length L = πR and mass M is placed on the upper half of a

uniform thin disc of radius R and mass M. The disc is placed vertically and can rotate freely

about its center that is fixed in space. With a small disturbance the chain starts to fall. Find out

the angular velocity of the disc when the chain leaves it.

(Assume the chain does not

*slide*off the disc, just falls off the side)

## Homework Equations

L = I w = F_ext R

I_disc = (MR^2)/2

F_ext = (d/dt) P = v (d/dt m) + m (d/dt) v

## The Attempt at a Solution

So, the chain is the cause of any change in rotation. Only the length of chain that is not counterbalanced across the disc causes any net acceleration, so mass (related to acceleration) is a function of time:

My guess: m(t) = M Sin[wt/2] = M Sin[theta/2]

When the disc has rotated 180 degrees, the chain is completely off and therefore there is no net acceleration on the disc (so that's why I picked the mass function above).

My naive feeling is that

w = (F_ext R)/ I

But before I even get that far, I need to account for I, which is more complicated than just the I of the disc, because the chain is there initially:

I = I_disc+ I_chain = (M^2 R/2) + (I_chain)

And I_chain is a function of rotation angle?

I_chain(theta) = ((MR^2)/Pi) * theta?

I got that by using I = Integral[r^2 dm] where dm= lambda dL, and dL = r dTheta, and lambda= M/(Pi * R)

So if that's true, how do I properly solve for this? I figure I can't just plug in theta=Pi and call it a day...

Thanks!