U.S. Physics Olympiad F=ma 2008: Problem 12

[Ritted]

12. 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?
(a) 1/2E
(b) 3/2E
(c) 2E
(d) 3E
(e) 4E

Answer: D

Relevant Equations: KE= 1/2 I x w^2, I=MR^2

Attempt at Solving: Io=MR^2, In=4MR^2, thus KE=4 (1/2 I x w^2)= 4E

Can someone please explain why the answer is 3E? I really appreciate your assistance! Thanks!

 

[rl.bhat]

Hi Rittted, welcome to PF.
Moment of inertia of a uniform disc about an axis through cm perpendicular to the plane is 1/2*M*R^2.
Using theorem of parallel axis, you can find the moment of inertia about an axis on the edge of the disk.

 

[CompuChip]

Note that for a uniform disk, I = (1/2) m r^2, not m r^2.

Furthermore, the parallel axis theorem may come in handy.

Note that the kinetic energy is linear in I ($E = \tfrac{1}{2} I \omega^2$) so you only need to show that $I = 3 I_\mathrm{cm}$, where I is the moment of inertia about the axis on the edge, and $I_\mathrm{cm} = \tfrac{1}{2} m r^2$

 

[Ritted]

Thanks for all of the help guys! I really appreciate it!

So I= 1/2mr^2 + mr= 3/2mr^2=3I

and

KE= 3E

Great!