A solid disk, spinning counter-clockwise, has a mass of 8 kg and a radius of 5/2 m. If a point on the edge of the disk is moving at 5/8 m/s in the direction perpendicular to the disk's radius, what is the disk's angular momentum and velocity?

1 Answer
Apr 19, 2016

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Explanation:

r=5/2m, The circumference of the disk is C=2\pi r

Because a point on the outside of the disk moves along the circumference with v=5/8 m/s, one rotation is completed in

T=C/v={2\pi r}/v

The angular velocity, w, has units of radians per second. One rotation has 2\pi radians so

w={2\pi}/T=v/r= 0.25 radians per second

As a side note, you could obtain this immediately if you are familiar with the expression v_\text{perpendicular}=w\times r

The angular momentum of the disk is L=Iw, where I is the inertia of the disk. You could obtain the inertia by integration, or just look it up in a table.

For a uniform disk, I=1/2mR^2= 25 \quad kg\cdot m^2.

Therefore,

L=Iw=6.25 \quad kg\cdot m^2/s

Done, but if you wanted to solve for the inertia on your own without looking it up, you would have to integrate.

Assume the disk has uniform density \sigma={mass}/{area}=m/{\pi R^2}

I=\int r^2 dm=\int r^2 (\sigma dA)

where dm=\sigma dA. You just need to substitute (dA=r\quad d\phi\quad dr) and integrate to get the answer, I=1/2 mR^2.