Suppose you have some object (which can roll like a ball,cylinder,wheel,etc) rolling down an incline without slipping (moment of intertia $I=kmr^2$. I want to find the accleration of the ball as it rolls, and I calculate this in two different ways, and I want to know if these ways are equivalent.

First I consider torques about a pivot point at the point of contact between the ball and incline. The only torque is due to gravity so $\tau=I\alpha=Ia/r=mgr\sin(\theta)$ which simplifies to $a=g\sin(\theta)/k$.

Alternatively I consider torques about a pivot point at the center of the object. The only torque is due to friction so $\tau=fr=Ia/r$. I sense that this alternative method is flawed for some reason. If it is not, it can be used to calculate $f$, using the $a$ we derived via the first method, right?


Rather, I think your first method is flawed. Because your $\alpha$ is always about the pivot you select. Since you select the contact point as pivot, then $\alpha$ should be about the contact point. So $\alpha$ is not $a/r$ of course.

Here is my approach for this problem.enter image description here

First, I select the CM (center of mass) as the pivot. Let $f$ be the friction at the contact point. Every moment the total torque about the pivot is $$\tau= fR= I \alpha$$ Here is another matter. How to determine $I$ ? My opinion is that since the shell is completely filled with frictionless fluid, then the fluid will not rotate with the shell at all since there is literally no force to push it to. So we ought to count only the shell's part. (I'm not definitely sure here, though) And thus $I=\frac{2}{3}MR^2$. Hence $$f=\frac{I\alpha}{R}=\frac{2}{3}M R\alpha=\frac{2}{3} Ma$$

Then, consider the famous theorem of kinetic energy (which doesn't require conservation !). Let $l$ be the distance the shell (or the CM, to be exact) has moved from the point where it is released, $v$ be the speed of the CM and $\omega$ be the angular speed. Note that the gravity and the friction are the only net forces acting on the shell together with the fluid contained, we have $$(2Mg\sin\theta-f)l=\frac{1}{2}(2M)v^2+\frac{1}{2} I \omega^2 $$ Note that for a rotation without sliding, $v=R\omega$ always holds, thus the above equation simplifies to $$(2Mg\sin\theta-f)l=\frac{4}{3}Mv^2$$ Differentiate both sides by $dt$ $$(2Mg\sin\theta -f)v=\frac{8}{3}Mv \frac{dv}{dt}=\frac{8}{3} Mva$$ Put $f=\frac{2}{3}Ma$ into it and note that $v$ cancels out, thus $$a=\frac{3}{5}g\sin\theta$$

  • $\begingroup$ You are right. But then how can I solve the problem by taking torqu about the contact point (suppose we did not know the friction force $f$, then this approach would be nessecary). In addition to using the appropriate moment of inertia (as per the parallel axis theorem), we need to relate the angular accleration to the linear accleration. I don't see how to do this. Edit: actually if you look at #12 here: aapt.org/physicsteam/2013/upload/exam1-2013-solutions.pdf. Then the solution to this problem seems to assume that $\alpha$ is equal to $a/r$. I thought this was incorrect?!?!! $\endgroup$ – Joshua Benabou Jan 22 '15 at 17:58
  • $\begingroup$ To be honest I am also confused with the first method, I don't think it is proper to choose the contact point as pivot. In fact, even the second method, I'm afraid, is also problematic, because it simply applies the conservation without even considering the friction force! Although in this case it does not produce heat, it is still not conservative and thus the "conservation" most likely doesn't hold! My answer is $\frac{3}{5}g\sin\theta$, I'll add it to my post and you can have a look at it to check if it is correct. $\endgroup$ – Vim Jan 23 '15 at 3:39
  • $\begingroup$ ignore the frictionless fluid part, as it makes the problem very tricky. Just assume the moment of inertia is the usual $0.5MR^2$. I found the same conservation of energy solution, but I am interested in the torque solution. I know there exists a solution with taking torques about the contact point, and also there is one with taking torques about the center of mass, but both have the problems we stated above. Someone need to clear this up. $\endgroup$ – Joshua Benabou Jan 23 '15 at 4:50
  • $\begingroup$ I have to say if you must choose the contact point as pivot the calculation may be formidable. Because the relationship between $a$ and $\alpha$ (about the pivot) is not very easy to establish. And , why $0.5MR^2$? If we count the shell alone and ignore the fluid, it should be $\frac{2}{3}MR^2$, isn't it? $\endgroup$ – Vim Jan 23 '15 at 5:03
  • $\begingroup$ And the "conservation solution" as I have shown you is, in my opinion, problematic. Because $(U+T)$ doesn't seem to conserve, considering the friction is not conservative here. In fact, it will do some negative work to the ball as it slides down the incline. $\endgroup$ – Vim Jan 23 '15 at 5:08

Crap I am stupid. This problem is not hard at all.

Consider pivot point at the center of mass. By torques, $fR=I\alpha=I(a/R)$. By linear acceleration, $mgsin(\theta)-f=ma$. Now solve for $a$ and $f$.

  • $\begingroup$ Oh yes, this will surely do. But I think there is a mistake. It should be $2Mg\sin\theta-f=2Ma$\. If we suppose $I$ doesn't include the fluid's part. Yeah, the answer should be $\frac{4}{3}g\sin\theta$. I'm wrong. I'll tryna find the mistake. $\endgroup$ – Vim Jan 25 '15 at 14:45

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