# Ball rolling down an inclined plane going in to a loop

I apologize if this question is not up to par. When I was doing exercises in basic mechanics I checked the answers and I can't seem to find what I'm doing wrong. Suppose we have a ball with mass $m$ and radius $r$ on an inclined plane with height $h$. At the end of the inclined plane is a loop with a radius of $R$ and we can assume that $r<<R$. We are asked what the minimum height is the inclined plane should have so as to let the ball complete the loop. Here is my reasoning:

We have $U_1\geq K_{rot,2}+K_{trans,2}+U_2$.

For the ball rotating we have $I=\frac{1}{2}mr^2$ and $\omega=\frac{v}{r}$.

So $mgh\geq \frac{1}{2}I\omega^2+\frac{1}{2}mv^2+mg\cdot2R=\frac{3}{4} mv^2+mg\cdot 2R$.

The minimal speed to complete the loop implies $F_{centripetal}=\frac{mv^2}{R}= mg$

So $v^2= Rg$ and we have $gh\geq \frac{3}{4}Rg+g\cdot 2R$ which means $h\geq 2\frac{3}{4}R$, while the book says that the answer should be $h\geq 2.70R$. Can you explain what I am doing wrong?

Thank you

EDIT: Moment of inertia corrected.

For the ball rotating we have $I=\frac{2}{5}mr^2$ and $\omega=\frac{v}{r}$.

So $mgh\geq \frac{1}{2}I\omega^2+\frac{1}{2}mv^2+mg\cdot2R=\frac{7}{10} mv^2+mg\cdot 2R$.

After the edit with correction of the moment of inertia I am getting the right answer. We get $gh\geq\frac{7}{10}Rg+g\cdot 2R$ so $h\geq2\frac{7}{10}R$ in accordance with the book.

First thing, for a rotating ball, $I=\frac{2}{5}mR^2$. You also need to be clear on what $\omega$ you are talking about.
The kinetic energy of a rotating ball is $\frac12 I_{cm}\omega_{cm}^2 + \frac12 mv_{cm}^2$. Here, $v_{cm}=v$. But, $\omega_{cm}=v_{cm}\times \frac{r}{R}$. Since $r<<R$, we can take the net kinetic energy to be just $\frac12 mv^2$; the $\frac12 I_{cm}\omega_{cm}^2$ term becomes too small to matter.
Basically, for a ball of center of mass moment of inertia $I$, mass $m$, radius $r$, rotating about itself with $\omega_cm$, revolving in a circle of radius $R$ with $\omega'$ , the energy is NOT $\frac12 I\omega^2+ \frac12(I+mR^2)\omega'^2+\frac12 mv^2$, it is $\frac12 I\omega^2+ \frac12 mv^2=\frac12 I\omega^2+ \frac m (\omega'R)^2$.
• Thank you for correcting the moment of inertia. If you are still of the opinion that I have done something incorrectly, could you please elaborate as I don't really understand the rest of your answer and how it applies to my approach to this question. I think you might have gotten a bit mixed up with $R$ and $r$. I will accept your answer anyway as it is the only one and it helped me find out what was wrong. – user16228 Nov 23 '12 at 21:03
• @mr.FS: Nope. not mixed up with $r$ and $R$. Though the rest of your answer is completely correct--for some reason I thought you'd made another mistake. Ignore the rest (though it's good to know--it can lead to some confusion) – Manishearth Nov 23 '12 at 21:11