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Angle turned by disk

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Two men, each of mass 75 kg , stand on the rim of a horizontal large disc, diametrically opposite to each other. The disc has a mass 450 kg and is free to rotate about its axis. Each man simultaneously start along the rim clockwise with the same speed and reaches their original starting points on the disc. Find the angle turned through by the disc with respect to the ground.

In this we have to conserve angular momentum .
But I am unable to do it . As we have not given the radius and velocity (angular velocity).

asked Dec 20, 2016 in Physics Problems by koolman (4,286 points)
edited Dec 20, 2016 by koolman
You could assume a radius $R$ and velocity $v$. Please show your attempt.
What frame of reference are you using? Are you using the same frame for $u$ and $\omega$?

Also note that the angular momentum of the disk is $I\omega$.
Ground frame of reference
No, I think you are measuring $u$ wrt the disk and $\omega$ wrt the ground. Then your last eqn is ok, I think, except that on the left you have linear momentum instead of angular momentum. You have to conserve angular momentum.

When you have corrected the last eqn, try to continue the calculation. Insert an expression for $I$. Get $u$ on one side and $\omega$ on the other. Multiply by time, then $u$ becomes arc length $s=2\pi r$ through which the men walk, and $\omega$ becomes angle $\theta$ through which the disk turns.

You should be able to finish this yourself, I think.
Now what about this
Yes, that looks correct. Does it agree with the given answer? Why don't you post it as an answer, if you are able to post an answer your own question?
I would like you to answer that as you only helped me in this

1 Answer

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Best answer

My try m=mass of man M=mass of disks u=velocity of man w= angular velocity of disk r=radius of disc Now , conserving momentum 2m(u-wr) r= Iw
2m(2$\pi$r - $\theta$r)r= (1/2) Mr$^2$ $\theta$
4m$\pi$ = [(1/2)M+2m]$\theta$
$\theta$= 4$\pi$/5

answered Dec 20, 2016 by koolman (4,286 points)
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