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.

The main thing is is that you need to remember that the formula "Kinetic energy=rotational energy + translational energy" works only when you consider all rotations about center of mass. You cannot just keep tacking on terms for each motion you see. Even though the ball is revolving around the center of the loop, we still classify this as translational motion. If you don't do this, you can easily get confused while building the expression for KE.

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 $\frac12 I\omega^2+ \frac12 mv^2=\frac12 I\omega^2+ \frac m (\omega'R)^2$.

  • $\begingroup$ 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. $\endgroup$ – user16228 Nov 23 '12 at 21:03
  • $\begingroup$ @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) $\endgroup$ – Manishearth Nov 23 '12 at 21:11

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