A uniform disk rotates at a fixed angular velocity on an axis through its center

Question

A uniform disk rotates at a fixed angular velocity on an axis through its center normal to the plane of the disk, and has kinetic energy E. If the same disk rotates at the same angular velocity about an axis on the edge of the disk (still normal to the plane of the disk), what is its kinetic energy?

1/2E

3/2E

2E

3E

4E

A sphere (I = (2/5) MR2) rolls without slipping with a center of mass speed v. What is its kinetic energy?

(7/10) Mv^2

(5/4) Mv^2

(1/2) Mv^2

(1/5) Mv^2

(3/10) Mv^2

A hoop (I = MR2) of mass M and radius R rolls without slipping down an incline of height h. The speed of its center of mass at the bottom of the ramp is v. What is vin terms of the height h?

(gh/2)^1/2

(2gh)^1/2

(4gh)^1/2

(3gh/4)^1/2

(gh)^1/2

Explanation / Answer

The Kinetic Energy of a rotating body with Moment of Ineria "I" and angular velocity "w" is;

E = (1/2)Iw^2

For a disk spinning about an axis thru its center, I = (1/2)MR^2
So;
E = (1/4)MR^2w^2

For a disk spinning about an axis on the rim, use the parallel axis theorem to get the moment of inertia to be;

I = (1/2)MR^2 + MR^2 = (3/2)MR^2

So now the kinetic energy is;

E = (3/4)MR^2w^2

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