Question

I will give a medal to the best response!

What is the distance from axis about which a uniform, balsa-wood sphere will have the same moment of inertia as does a thin-walled, hollow, lead sphere of the same mass and radius R, with the axis along a diameter, to the center of the balsa-wood sphere?

The program I'm using wants the answer as a ratio of d/R

Answer 1

I know that moment of inertia for a solid sphere = \[\frac{ 2 }{ 5 }MR^2\]
and for a hollow sphere = \[\frac{ 2 }{ 3 }MR^2\]

Answer 2

I know I'm probably going to have to set both moments of inertia equal to each other. But where do I get d from?

Answer 3

Also, I don't think I'm understanding the question correctly.

Answer 4

@AllTehMaffs?

Answer 5

This problem sounds easy. I just don't understand it (maybe I'm too overworked).

Answer 6

Parallel axis thm says that about a point of rotation not the center of mass a distance d away from the center of mass
\[I=I_{cm} + Md^2\]
You know that the hollow lead sphere has
\[I_H=\frac{2}{3}MR^2\]
and the moment of inertia about the center of mass for the balsa sphere is
\[I_{cm}=\frac{2}{5}MR^2\]
So to set them equal to each other
\[I_H = I_{cm}+Md^2\]
\[\frac{2}{3}MR^2=\frac{2}{5}MR^2+Md^2\]
\[M\left(\frac{2}{3}R^2\right)=M\left(\frac{2}{5}R^2+d^2\right)\]

Answer 7

Hmmm...

well, so far so good I'm understanding. So do I just solve for the ratio d/R?

Answer 8

Yeah, you solve for d in terms of R, and then d/R will just be the coefficient in front of d

Answer 9

\[d^2=(\frac{2}{3}-\frac{2}{5})R^2\]
\[d=\frac{2\sqrt{15}}{15}R\]

Answer 10

Is it \[\sqrt{\frac{ 4 }{ 15 }}\]

Answer 11

THANK YOU SOOOO MUCH!!!

Answer 12

I hope someone else gives you a medal too! (you really earned it!)

Answer 13

^_^ welcome. Glad I could help!