Tackling a Classical Mechanics Problem at Pisa University

[Pigkappa]

Homework Statement

See figure below. A homogeneous semicylinder rests on a flat plane. A particle fallse from height h striking point A, then it remains attached. Assume the semicylinders will rotate without slipping. Call $\mu$ the ratio of cylinder mass to particle mass, R the cylinder radius.
(i) What is the minimum h for point A to touch the ground?
(ii) If h has exactly that value, what is the minimum friction coefficient k for the non-slipping condition to be true?

Relevant Equations

Equation of motion; geometrical constraints.

This problem is hard. It found it listed among problems discussed in a classical mechanics course for physicists at the university of Pisa and don't have a full solution. It's not 100% guaranteed that there's a nice close-form solution, but probably yes; and if not, there should be some trick to make a numerical solution somewhat tractable.

I solved point (i) but not (ii).

Apologies for the typos ("fallse") in the question, I can't find out how to edit there.

To solve (i), first find the distance $d$ between the center C of the semicylinder and the centre of mass G. A reasonably short calculation shows $\overline{CG} = \frac{4}{3\pi} R$.

Then find the moment of inertia of the semi-cylinder around O. The trick is that the moment of inertia around the center C is $\frac{1}{2} M R^2$; this must be $I_C = I_G + M * \overline{CG}^2$. Then also use $I_O = I_G + M \times \overline{OG}^2$. After some algebra you'll find that $I_O = M R^2 \times \frac{9\pi - 16}{6\pi}$.

Now find the angular velocity of cylinder + particle around O after the impact by imposing conservation of angular momentum around O in the impact. Call $\alpha = \frac{9\pi - 16}{6\pi}$ to save some typing, and $v_0$ the particle speed before impact. Result is:
$w_0 = \frac{v_0}{R} \frac{1}{\alpha \mu + 2}$

Now use conservation of energy to see if the body arrives with A on the ground.
The available kinetic energy is:
$K = \frac{1}{2} m (\sqrt 2 R w_0)^2 + \frac{1}{2} I_O w_0^2 = ... = m v_0^2 \frac{1}{2(2+\alpha \mu)}$
The difference in potential energy is:
$\Delta U = M g \frac{4R}{3\pi} - mgR$
The result is that if $\Delta U < 0$, A will always touch the ground; this occurs if $\mu \le \mu_{0} \equiv \frac{3\pi}{4}$. Otherwise, the minimum required height is:
$h_0 = R (\frac{4\mu}{3\pi} - 1) (2 + \frac{9\pi - 16}{6 \pi} \mu)$

To solve (ii), I am stuck. We need to calculate the components of the force between table and cylinder, and require the sideways force to be at most given by what friction can provide. We can parametrise the system by the angle $\theta$ between a vertical line and the line between C and G.

The angular momentum of the system is: $L = (I_O + m \overline{AO}^2) \dot{\theta}$. Here $I_O, \overline{AO}$ both depend on $\theta$. I find $\overline{AO} = 2 R^2 (1 - \sin \theta)$. We can get $I_O$ with the same trick as in part (i) though the result will be a bit more complicated. The derivative of $\vec L$ will not be pretty and the torque will also depend on $\theta$ so we are looking at a differential equation that's unlikely to be tractable. At this point I am a bit exhausted.

I wonder if I am going at this in an inefficient way; or if this is going to simplify miraculously down the line and I just don't see it.

 

[Pigkappa]

(This was an answer to a post that has been deleted and suggested checking for non-slipping just during the collision)
I interpreted it as requiring the non slipping condition to be true during the entire motion, not just right after the collision

 

[haruspex]

I don't think you need to solve a differential equation.
You will have four unknowns, sin(theta) and theta's first three derivatives.
You can write down the KE conservation in terms of these, and differentiate twice to get two more equations. You can write the horizontal acceleration of the mass centre in terms of these, differentiate that and set the result to zero for max acceleration.
You now have four equations in those four unknowns. But there are cubic terms, and because of the trig functions you could end up with a polynomial degree six.

Edit: two mistakes in the above. I forgot about the continued involvement of mass m, which just makes it a bit more complicated, and about the changing normal force. It is not sufficient to look for max horizontal acceleration; need to look for max ratio of that to normal force.

Also, this does not address the question of slip during the impulse. That is much easier since we can ignore g and just look at the ratio of normal impulse to horizontal impulse. I get that the min coefficient is $\frac{M(R-d)}{mR+M(\frac 32R-2d)}$ (regardless of the height from which the mass is dropped). Note this tends to a maximum as m tends to zero.

 

[Pigkappa]

Note this tends to a maximum as m tends to zero.

Is this expected? Shouldn't it be the opposite - if the extra mass is negligible, non-slipping should be easier I'd say?

I get a slightly different result. The calculation is quite messy, not sure if you found a quick way to do it. I'll try to keep it short. Angular velocity after impact was derived in my solution above: $w_0 = \frac{v_0}{R} \frac{1}{\alpha \mu + 2}$, where $\alpha = (9\pi - 16) / (6\pi)$. Momentum before impact: $\vec P_0 = -m v_0 \hat y$. Momentum after impact: $\vec P_1 = -w_0(mR + (R-d) M) \hat x - m w_0 R \hat y$. Difference is $\Delta P = \vec P_1 - \vec P_0$. The friction coefficient must be $k = |\Delta P_x| / |\Delta P_y|$. The result I find is eerily close to yours:
$k = \frac{mR + M(R-d)}{mR + M(\frac{3}{2}R - 2d)}$

 

[haruspex]

Shouldn't it be the opposite - if the extra mass is negligible, non-slipping should be easier I'd say?

Bear in mind that the greater the mass m the greater the vertical impulse reaction from the ground, allowing a greater horizontal frictional impulse.
But I just found a mistake in my working. I omitted the horizontal impulse m needs.

If m hits the half cylinder at speed u, and the resulting rotation is $\omega$ then taking moments about the point of contact with the ground
$muR=m(\sqrt 2R)^2\omega+M\omega(\frac 12R^2-d^2+(R-d)^2)=2mR^2\omega+M\omega R(\frac 32R-2d)$.
$mu=2mR\omega+M\omega (\frac 32R-2d)$.
If no slip, immediately after impact, the vertical and horizontal components of m's velocity are each $R\omega$. If J is the impulsive reaction from the ground, $J=m(u-R\omega)=mu(1-\frac{mR}{2mR+M(\frac 32R-2d)})$.
The horizontal impulse needed is $K=mR\omega+M(R-d)\omega$. So the min coefficient is $\frac KJ=\frac{(mR+M(R-d))\frac{mu}{2mR+M(\frac 32R-2d)}}{mu(1-\frac{mR}{2mR+M(\frac 32R-2d)})}$
$=\frac{2mR+M(2R-2d)}{2mR+M(3R-4d)}$, as you found.
Since R>2d, this is less than .5, and yes, as m tends to zero it tends to its minimum.