I think I missed something in this mechanics problem.

We're given a polished (no friction) and homogeneous hemicircle which has mass $M$ and a particle of mass $m$ laying on the top of it.

enter image description here

There is also no friction between the hemicircle and the ground. Find the equation for the angle θ in which the particle abandons the hemicircle surface.

I can't find my mistake.

As we have no external horizontal forces acting on the system particle + hemicircle we must have conservation of the horizontal position of the center of mass.

I made this horrible drawing to try to understand the movement:

enter image description here

we would have $x=\frac{mRsin(θ)}{M+m}$, making $M=k⋅m⟺k=\frac Mm$, we would have:

$$x=\frac{R\sin(θ)}{1+k}⟹\dot x=\frac{R\dot θ\cos(θ)}{1+k}$$

from this, I found that the velocity of the particle with respect to the ground was:

$$R\dotθ(\cos(θ)(\frac k{k+1})u_x−\sin(θ)u_y)=\vec v_P$$

And conservation of energy

$$mgR=mgR\cos(θ)+\frac{mv^2_P+M\dot x^2}2$$

gave me:

$$2gR(1−\cos(θ))(1+k)=\dot θ^2[\sin^2(θ)+k]$$

Finally, I used the abandonment equation: $\frac gR \cos(θ)=\dot θ^2$

Which led me to:


but I feel something is wrong because $k=0$ should give me $cos(θ)=\frac23$. Any insights on my mistakes?

I think on the line of the energy conservation the term $\sin^2(θ)$ is the one causing trouble. I'm very confident about the expression for the velocity of the particle.


1 Answer 1


I don't think this is wrong. $k=0$ represents the case where the large hemisphere has zero mass, so no inertia. I think you are looking for the case where the large hemisphere is "fixed", so much heavier than the particle. This is $k\to\infty$, which gives you the result you desire.

Note that $k\to0$ gives a single solution of $\cos\theta = 1$, i.e. the particle leaves the sphere immediately. This is what you'd expect as a small motion of the particle causes the hemisphere to shoot off into the distance as it has almost zero mass.

  • $\begingroup$ well, you might be right because gravity can not affect a massless object. But I tend to think that, regardless of how small the little mass is, the limit $k \rightarrow 0$ should lead to $\cos(\theta) = \frac 23$ which is the official answer of the original question: tutorbrasil.com.br/forum/viewtopic.php?f=9&t=99618 $\endgroup$ May 11 at 16:29
  • $\begingroup$ @hellofriends I'm afraid I can't read the page. What is it that makes you think $k\to0$ should lead to $\cos \theta = \frac 2 3$? I am assuming the original question has a fixed hemisphere. $\endgroup$
    – Robbie
    May 11 at 16:41
  • $\begingroup$ ah, must be the server location. The problem is exactly this one that I wrote. I think like that because as the mass $m$ approaches zero, we would have the same effect as letting $M$ going to infinity. Here is the answer in the page: $[3.(\frac{m}{M})^3+3(\frac{m}{M})^2-2(\frac{m}{M})-3]cos^3\theta +2(1-\frac{m}{M})(1+\frac{m}{M})^2cos^2\theta -2(\frac{m}{M})^2(1+\frac{m}{M})cos\theta +(\frac{m}{M})^2(1+\frac{m}{M})=0$ $\endgroup$ May 11 at 16:45
  • $\begingroup$ @hellofriends $M\to\infty$ is equivalent to $k\to\infty$, not $k\to0$, as I stated in my original answer. Or am I not understanding you? $\endgroup$
    – Robbie
    May 11 at 16:47
  • 1
    $\begingroup$ @hellofriends Yes, because $k=10^4$ in both cases. What does this have to do with $k\to 0$? $\endgroup$
    – Robbie
    May 11 at 16:53

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