# Find the equation for the angle $\theta$ in which the particle leaves the semicircle. No Friction [closed]

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.

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:

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:

$$3\cos(θ)−\frac{\cos^3(θ)}{1+k}=2$$

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.

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.
• 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 May 11 at 16:29
• @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. May 11 at 16:41
• 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: $[3.(\frac{m}{M})^3+3(\frac{m}{M})^2-2(\frac{m}{M})-3]cos^3\theta +2(1-\frac{m}{M})(1+\frac{m}{M})^2cos^2\theta -2(\frac{m}{M})^2(1+\frac{m}{M})cos\theta +(\frac{m}{M})^2(1+\frac{m}{M})=0$ May 11 at 16:45
• @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? May 11 at 16:47
• @hellofriends Yes, because $k=10^4$ in both cases. What does this have to do with $k\to 0$? May 11 at 16:53