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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?

Qmechanic
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math_lover
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2 Answers2

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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 $$a=\frac{3}{5}g\sin\theta$$

Vim
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  • 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: http://www.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?!?!! – math_lover Jan 22 '15 at 17:58
  • 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. – Vim Jan 23 '15 at 03:39
  • 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. – math_lover Jan 23 '15 at 04:50
  • 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? – Vim Jan 23 '15 at 05:03
  • 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. – Vim Jan 23 '15 at 05:08
  • No I mean ignore the fluid and treat the thing as a solid disk/wheel/cylinder. The fluid makes the problem too tricky. Just call the moment of intertia $I$ if you want. And the conservation of energy solution actually is correct. The ball rolls, without slipping, which means friction does no work. Now, as for the torque solution: I know it exists (I've seen it somewhere before, but I can't remember where), but we just have to make it work. – math_lover Jan 23 '15 at 05:38
  • I'm confused why $f$ doesn't work. Clearly when the ball rolls down the incline $f$ doesn't change at all in both magnitude and direction, pointing backwards and thus does negative work to the ball system. I think it is obvious from the formula $$\vec{W}=\vec{F}\vec{s}$$
    And as for the torque solution, I'm working on it later. I sense that it would be a bit tricky if we have to select the contact point as pivot, in that the fluid , now unable to ignore, will take a part in $I$.     – Vim Jan 23 '15 at 05:46
  • @JoshuaBenabou Hi! I've met some new difficulty when I choose the contact point as pivot. It's not about how to derive $a$ but about the $I$. Now the only feasible way I can think of to calculate $I$ is by applying the theorem of parallel axis, which would be quite easy , without the tricky fluid!! Because the fluid presumably doesn't revolve with the shell around the CM, but it does revolve around the pivot! So I think this theorem may not apply here for the fluid, at least not in the usual way. – Vim Jan 23 '15 at 11:05
  • Yes you need the parallel axis to get I when using the pivot at the contact point. Now, looking at the work formula - the work is zero because the distance over which the friction force is applied is 0. This is true because the contact point is instantaneously at rest. This is the distinction between rolling and slipping. In any case, the torque solution is not so hard, this problem is not so hard. We just don't know how to resolve the difficulties we've encountered. I wish we could get more people to look at this question and fix this, because I really need to know the answer. – math_lover Jan 23 '15 at 18:02
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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 $f$.

math_lover
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  • 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. – Vim Jan 25 '15 at 14:45