# Toppling Hemisphere

#### {???}

Homework Statement
A solid hemisphere is placed on a horizontal ledge so that its center is directly above the edge, fixed there by a delicate thread. It is given a push of negligible velocity and begins to topple over the edge. When it collides perfectly inelastically with the vertical face of the ledge, the thread breaks. What is the kinetic energy at this moment?
Homework Equations
CM of a hemisphere: $\frac{3}{8}R$
Moment of inertia about geometric center: $\frac{2}{5}MR^2$
Moment of inertia about CM: $I_0=\frac{83}{320}MR^2$ (can be derived via parallel-axis theorem)
Angular momentum: $J=I_0\omega+Mv_\mathrm{CM}r_\mathrm{CM}$ (where $r_\mathrm{CM}$ is relative to the point about which angular momentum is measured)
Kinetic energy: $K=\frac{1}{2}Mv_\mathrm{CM}^2+\frac{1}{2}I_0\omega^2$
I emphasise again that this is a problem I created myself to better my understanding of moments of inertia and angular momentum.
Approach 1: I recognise immediately that energy, momentum, and angular momentum are all nonconserved during the collision. My first strategy was to use brute-force integration to determine "how much negative momentum" was in the quarter-sphere that had negative horizontal component of momentum. I then demanded that all of this negative horizontal momentum be erased in the collision, leaving an equal (and opposite, of course) momentum in the forward direction. This gave me a velocity of $\frac{3\pi}{64}R\omega_0$ where $\omega_0=\sqrt{\frac{15g}{8R}}$ is the angular velocity immediately before the collision. Next, I specified that immediately after the collision, the lowest point of the hemisphere should have no horizontal velocity due to the inelasticity of the collision. Being careful to realise that there are both horizontal and vertical components to the linear velocity produced by rotation about the CM, I conclude that the final angular velocity is $\omega_1=\frac{3\pi}{64}\omega_0$. The problem is that I require the velocities of the upper quarter-sphere to redistribute throughout the hemisphere and I am concerned this affects the result; moreover, I should be able to apply similar "arresting" reasoning to the angular momentum of the hemisphere about its geometric center and cannot figure out my limits of integration. If I demand that only a quarter-sphere of angular momentum is arrested by the ledge, then I need a good reason why the domain of applicability does not extend further; there's no $\sin\theta$ factor here and thus no reason that is apparent to me why I shouldn't keep going. However, integrating over the whole hemisphere means it completely stops rotating, and that just doesn't seem instinctively physically accurate. (My instinct here is that neglecting the angular momentum effects here means my linear momentum is lower than it should be.)

Approach 2: I dissect the hemisphere into two identical quarter-spheres and demand that, immediately after the collision, neither quarter-sphere is rotating, the upper quarter-sphere maintains its original momentum vector, and the horizontal velocity of the lower quarter-sphere is arrested. I then imagine that the quarter-spheres momentarily "reconnect," in a way which conserves momentum and angular momentum (but not kinetic energy; c.f. two blocks colliding perfectly inelastically). Doing this gives me $v_1=\frac{3}{8}R\omega_0$ and $\omega_1=\frac{45}{166}\omega_0$. Here you can already see the problem--the linear velocity of the lowest point of the hemisphere has negative horizontal component!
My latest approach: The method of successive approximations - Separate the hemisphere into a part with positive horizontal velocity and a part with negative horizontal velocity. Arrest the negative horizontal velocity of the lower piece. Then, "reconnect" the pieces in a way that conserves linear and angular momentum, and then repeat until there is no negative horizontal velocity portion. Now, I am expecting that this process should take infinitely many iterations to reach the answer, since the reasoning of Approach 2 seems like it will always produce a portion of the sphere with negative horizontal velocity. That I'm prepared to deal with. I'm expecting the piece with negative horizontal velocity gets smaller each time due to the increase in horizontal momentum of the hemisphere at each step; however, the angular momentum would then also decrease, though by a smaller amount each time. In any event, my instinct suggests it should not go to zero. This leads me to conclude the angular velocity of the top piece should remain the same before "reconnecting." (If it were set to zero, at a much later step, the hemisphere would be nearly stationary after collision, which seems nonphysical.)

Does my latest approach seem reasonable? If not, can Approach 1 be salvaged?

Thanks,
QM

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#### haruspex

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In the absence of a diagram, a more thorough description of the set-up is in order.
What is the initial orientation of the hemisphere, flat up, flat down or something else?
How exactly is the thread connected? I don't understand how it can be fixed in place by the thread if, after a nudge, it doesn't break till some time later.

#### kuruman

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When it collides perfectly inelastically with the vertical face of the ledge, the thread breaks. What is the kinetic energy at this moment?
Which "moment" is this, an infinitesimal time $dt$ just before or an infinitesimal time $dt$ immediately after the inelastic collision is completed?

#### {???}

In the absence of a diagram, a more thorough description of the set-up is in order.
What is the initial orientation of the hemisphere, flat up, flat down or something else?
How exactly is the thread connected? I don't understand how it can be fixed in place by the thread if, after a nudge, it doesn't break till some time later.
The initial orientation of the hemisphere is flat down. The thread is of infinitesimal length and connects the geometric center to the ledge. The purpose of the thread is just to ensure the flat surface of the hemisphere is pressed against the ledge when it collides - otherwise only the bottom edge touches the ledge because the hemisphere will have nonzero horizontal velocity when it loses contact.
Which "moment" is this, an infinitesimal time $dt$ just before or an infinitesimal time $dt$ immediately after the inelastic collision is completed?
Infinitesimal time $dt$ after.

#### kuruman

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I think I get the picture now, but I must still be missing something. I assume that the ledge has infinite mass, i.e. it is firmly attached to the Earth. A perfectly inelastic collision means that the colliding masses stick together and move as one thereafter. In this case the collision stops the rotation and the hemisphere's kinetic energy immediately after the collision is zero to a very good approximation. Why do you think it is not or what am I missing? I think the suggestion by @haruspex to provide a diagram is appropriate at this point.

#### {???}

I do disagree that the hemisphere stops rotating altogether. What I am imagining is that the lowest point of the hemisphere has no horizontal linear velocity immediately after the collision.

Here's why I believe the hemisphere and the ledge do not stick together after the collision. Immediately before the collision, mentally divide the hemisphere into two "quarter-spheres." The upper quarter-sphere has positive horizontal momentum while the lower quarter-sphere has negative horizontal momentum. The nature of the collision should be such that the horizontal momentum of the lower quarter-sphere is "arrested," i.e., set to zero. That means the total momentum should be positive, since nothing happened to the momentum of the upper quarter-sphere.

If you treat the quarter-spheres as separate dynamical objects up to and including the moment of collision, then the upper quarter-sphere is instantaneously moving forward and down while the lower quarter-sphere is just moving down. If you then imagine that they "reconnect," the resulting hemisphere (which, we're assuming, is the original hemisphere) will have both some linear velocity and some angular velocity. The horizontal linear velocity will be greater than it was immediately before the collision (namely, zero), and the angular velocity will be less than it was immediately before the collision (namely, $\sqrt{\frac{15g}{8R}}$). So I do not believe the hemisphere comes completely to rest apart from vertical CM velocity. The question really is, what are those linear and angular velocities? (Truthfully, one is enough, since $v_x=R\omega$ here.)

#### kuruman

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Here's why I believe the hemisphere and the ledge do not stick together after the collision. Immediately before the collision, mentally divide the hemisphere into two "quarter-spheres." The upper quarter-sphere has positive horizontal momentum while the lower quarter-sphere has negative horizontal momentum. The nature of the collision should be such that the horizontal momentum of the lower quarter-sphere is "arrested," i.e., set to zero. That means the total momentum should be positive, since nothing happened to the momentum of the upper quarter-sphere.
Let's backtrack a bit to the point where the hemisphere is overhanging the ledge and is about to start toppling over. I mentally divide the hemisphere into two "quarter-spheres" as you suggest. The overhanging quarter-sphere has no normal force acting on it to cancel the force of gravity therefore it acquires momentum down. Does the other quarter-sphere stay where it is on the ledge while the other quarter rotates away? It does not. Why not? Answer: Because you have a rigid body and that means that the angular velocity $\omega$ of any point about the axis of rotation is the same as that of any other point. Therefore if the angular velocity about a given axis of one quarter-sphere goes to zero for one point on the hemisphere, it must go to zero for all points on the entire hemisphere.

#### haruspex

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It can happen that such a problem is not well-posed, i.e. it depends on details of the surfaces and materials. If so, the way to solve it is to take it as having some elasticity, model it accordingly as a springy surface, then let the elasticity tend to zero.

But first I would try a slightly simpler version. Instead of a hemisphere consider a uniform rod length 2r which is set rotating at rate ω0 about its mid point, where it is attached to the corner of the ledge. On impact it experiences an impulse J at (effectively) distance y below the corner.
You can write equations relating this to the gained horizontal velocity and the new rotation rate. You can also write a constraint representing the fact that the rod does not penetrate the surface, and an expression for the remaining KE.
Perfectly inelastic means the KE is minimised, subject to the constraint and applicable conservation laws. That might give you enough.

Edit: after more thought, I think I can show y=r. For the uniform rod that leads to $\omega=\omega_0/4$.

Edit2: my y=r comes from assuming that it is all over in a single impact. In the real world that might not be true. Imagine a slightly curved surface, represented as an arc of pegs. Each time it hits a peg, that becomes the instantaneous centre of rotation.
I've not done the algebra, but I suspect that as we let the number of pegs tend to infinity we get a rolling motion with no loss of energy. Next, we let the radius of curvature tend to infinity, but the result is constant, so the limit is also no loss of energy.

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#### {???}

Let's backtrack a bit to the point where the hemisphere is overhanging the ledge and is about to start toppling over. I mentally divide the hemisphere into two "quarter-spheres" as you suggest. The overhanging quarter-sphere has no normal force acting on it to cancel the force of gravity therefore it acquires momentum down. Does the other quarter-sphere stay where it is on the ledge while the other quarter rotates away? It does not. Why not? Answer: Because you have a rigid body and that means that the angular velocity $\omega$ of any point about the axis of rotation is the same as that of any other point. Therefore if the angular velocity about a given axis of one quarter-sphere goes to zero for one point on the hemisphere, it must go to zero for all points on the entire hemisphere.
I think that you are misunderstanding some of my "quarter-sphere" argument, performing reductio ad absurdum, or both. The argument I am forming is as follows:
1. Separate the hemisphere into two quarter-spheres immediately before the collision.
2. Have the two quarter-spheres reconnect immediately after the collision, conserving linear and angular momentum while rotating as a rigid body with uniform angular velocity $\omega$.

You have neglected step 2, thus leading to a clear impossibility. There is also the important difference that the normal force exerted by the ledge on the supported quarter-sphere is exerted for a non-infinitesimal amount of time, in contrast to the impulsive force exerted by the vertical face of my ledge in this scenario.
It can happen that such a problem is not well-posed, i.e. it depends on details of the surfaces and materials. If so, the way to solve it is to take it as having some elasticity, model it accordingly as a springy surface, then let the elasticity tend to zero.

But first I would try a slightly simpler version. Instead of a hemisphere consider a uniform rod length 2r which is set rotating at rate ω0 about its mid point, where it is attached to the corner of the ledge. On impact it experiences an impulse J at (effectively) distance y below the corner.
You can write equations relating this to the gained horizontal velocity and the new rotation rate. You can also write a constraint representing the fact that the rod does not penetrate the surface, and an expression for the remaining KE.
Perfectly inelastic means the KE is minimised, subject to the constraint and applicable conservation laws. That might give you enough.

Edit: after more thought, I think I can show y=r. For the uniform rod that leads to $\omega=\omega_0/4$.

Edit2: my y=r comes from assuming that it is all over in a single impact. In the real world that might not be true. Imagine a slightly curved surface, represented as an arc of pegs. Each time it hits a peg, that becomes the instantaneous centre of rotation.
I've not done the algebra, but I suspect that as we let the number of pegs tend to infinity we get a rolling motion with no loss of energy. Next, we let the radius of curvature tend to infinity, but the result is constant, so the limit is also no loss of energy.
When it was suggested (in the other thread) to start with a brick rather than a sphere, my mind also jumped to the uniform rod as a simple test case for the method that might not involve too much rigid-body integration. I'll see where I get (in particular, if I get $\frac{1}{4}\omega_0$), and write up a solution in my next post to this thread. Thanks for your help!

#### {???}

The uniform rod of length $2r$ has moment of inertia $I_0=\frac{1}{3}Mr^2$ and kinetic energy $\frac{1}{6}Mr^2\omega_0^2$ at the moment of collision.

We separate the rod into two rods of length $r$ and moment of inertia $\frac{1}{12}(\frac{1}{2}M)r^2=\frac{1}{24}Mr^2$ about their centers. The horizontal momentum of each just before the collision is $\frac{1}{2}Mr\omega_0$. Immediately after the collision, the bottom half-rod is arrested. Meanwhile, the top rod is rotating with angular velocity $\omega_0$ and its CM is moving with velocity $v=\frac{1}{2}r\omega_0$, so it has angular momentum about the center $J=(\frac{1}{12}\frac{M}{2}r^2)\omega_0+(\frac{1}{2}M)(\frac{1}{2}r\omega_0)(\frac{1}{2}r)=\frac{1}{6}Mr^2\omega_0$, so for the entire system $p=\frac{1}{2}Mr\omega_0$ and $J=(\frac{1}{6}Mr^2)\omega_0$.

We now assume the rod reconnects immediately after the collision, conserving linear and angular momentum. Then $Mv_\mathrm{CM}=\frac{1}{2}Mr\omega_0$ and $\frac{1}{3}Mr^2\omega=\frac{1}{6}Mr^2\omega_0$, i.e., $v_\mathrm{CM}=\frac{1}{2}r\omega_0$ and $\omega=\frac{1}{2}\omega_0$. In particular, $v_\mathrm{CM}=r\omega$, so the lowest point of the rod is instantaneously at rest, as desired.

If I may attempt to reverse-engineer your solution now: You are assuming that the impulse $\Delta p$ (I'm calling it that since $I$ and $J$ are taken) is effectively delivered to a single point. If we designate the distance of this point below the corner by $y$, then you obtain
$$p_0+\Delta p=p,\qquad J_0-y\Delta p=J.$$
Rewriting these in terms of the given and desired variables gives
$$\Delta p=Mv_\mathrm{CM},\qquad\tfrac{1}{3}Mr^2\omega_0-y\Delta p=\tfrac{1}{3}Mr^2\omega.$$
Substituting the first equation into the second eliminates the impulse $\Delta p$ from the problem:
$$\tfrac{1}{3}Mr^2\omega_0-yMv_\mathrm{CM}=\tfrac{1}{3}Mr^2\omega.$$
The condition of perfect inelasticity necessitates $v_\mathrm{CM}=r\omega$:
$$\tfrac{1}{3}Mr^2\omega_0-Mry\omega=\tfrac{1}{3}Mr^2\omega.$$
You assume or can otherwise prove that $y=r$:
$$\tfrac{1}{3}Mr^2\omega_0-Mr^2\omega=\tfrac{1}{3}Mr^2\omega$$
$$\Rightarrow\tfrac{1}{3}Mr^2\omega_0=\tfrac{1}{3}Mr^2\omega+Mr^2\omega=\tfrac{4}{3}Mr^2\omega$$
$$\Rightarrow\omega=\tfrac{1}{4}\omega_0.$$

Here's the rub: I'm not convinced that $y=r$. This would suggest that all of the impulse is delivered perpendicular to the endpoint of the rod, whereas I maintain that (in the idealised world of spherical cows etc.) the impulse is applied continuously along the lower half of the rod. This would dilute the effect of that impulse in reducing angular momentum, leading to a higher $\omega$. Can you provide a similarly reasoned argument for why the impulse should be applied only to the endpoint? Moreover, if you make the same assumption I do, do you obtain $\omega=\frac{1}{2}\omega_0$ instead?

The above problem is illustrative, but in fact so simplified that it lacks several of the characteristic features of my problem. Notice that, the separating-reconnecting hypothesis does not in this case produce a piece which is moving in the wrong direction. Dealing with this piece was essentially my whole problem.

I think that at this point I need to work the problem for a rectangular brick. I'm not expecting integrals to show up, but any that do should be simpler than the corresponding one for a hemisphere.

EDIT: I've spotted a mistake in my work:
We separate the rod into two rods of length $r$ and moment of inertia $\frac{1}{12}(\frac{1}{2}M)r^2=\frac{1}{24}Mr^2$ about their centers. The horizontal momentum of each just before the collision is $\frac{1}{2}Mr\omega_0$.
The $M$ should be $\frac{M}{2}$; there is an additional (rather, multiplicative?) factor of 2 from the motion of the CM of the half-rod, so this should be $\frac{1}{4}Mr\omega_0$. As far as I can tell, the angular velocity analysis I performed is free of such errors, which means I do obtain $v=\frac{1}{4}\omega_0$. I think my method of successive approximations should now be applicable since $v_x<0$ below some portion of the rod.

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#### {???}

Okay, MSA got me nowhere—literally, to the location of nothing. I’m getting $v=\omega=0$, which not only seems unphysical, but also suggests I made an arithmetic error since $v(n+1)>v(n)$. I’d state my recurrence relations for the record, but it’s 2:00 AM and I need some sleep. See you in the morning.

#### haruspex

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not convinced that y=r
Consider, instead of a flat surface, a peg at y below the rotation axis.
Maximum loss of KE would mean that the peg becomes the new axis of rotation. With y<r, the bottom end of the rod would then be moving back below the peg.

#### {???}

Consider, instead of a flat surface, a peg at y below the rotation axis.
Maximum loss of KE would mean that the peg becomes the new axis of rotation. With y<r, the bottom end of the rod would then be moving back below the peg.
So you're saying that to minimize $K$ one needs to have $y=r$. Okay.

Now, using both our methods (assuming I got yours more or less correct) we both find $v=\frac{1}{4}R\omega_0$. In your case, this comports with $\omega=\frac{1}{4}\omega_0$ since $v=R\omega$. In my case, I obtain instead $\omega=\frac{1}{2}\omega_0$ on angular momentum conservation grounds. I've been arguing that failure to conserve angular momentum necessitates an impulse, which would increase linear momentum. Now I'm actually not sure if I've already accounted for this increase. Should I instead assume this linear momentum is not further affected by the change in angular momentum?

(Thanks again for your help, and I'm sorry if I'm giving you a hard time; it's just that this problem seems to me conceptually rich and I want to make sure I don't miss anything.)

#### haruspex

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failure to conserve angular momentum necessitates an impulse
Angular momentum only means anything in the context of a chosen axis.
If the impulse from the wall is at a point y below the top then that impulse has no moment about that point, so angular momentum is conserved about it.
With the rod and y=r, the angular momentum about that axis before impact was $\frac 13mr^2\omega_0$; after impact it is $\frac 43mr^2\omega$.

#### {???}

All right, I think I have almost everything I need to solve my original problem.
My last question is purely a mathematical one: Given the sequence defined recursively by
$$a_{n+1}=\frac{1}{4}\left(1+\frac{a_n}{b_n}\right)^2b_n,$$
$$b_{n+1}=\frac{1}{8}b_n\left(1+\frac{a_n}{b_n}\right)^3+\frac{3}{8}\left(1+\frac{a_n}{b_n}\right)^2\left(1-\frac{a_n}{b_n}\right)b_n,$$
and the initial conditions $a_0=0,b_0=1$, I find for large $N$ the mysterious property that $a_N\approx b_N\approx\frac{1}{\pi}$, with the approximation getting ever better for larger $N$. Indeed, using Mathematica, after a million iterations I find the approximation holds to six decimal places. Can anyone help me prove that $\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n=\frac{1}{\pi}$?

#### haruspex

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All right, I think I have almost everything I need to solve my original problem.
My last question is purely a mathematical one: Given the sequence defined recursively by
$$a_{n+1}=\frac{1}{4}\left(1+\frac{a_n}{b_n}\right)^2b_n,$$
$$b_{n+1}=\frac{1}{8}b_n\left(1+\frac{a_n}{b_n}\right)^3+\frac{3}{8}\left(1+\frac{a_n}{b_n}\right)^2\left(1-\frac{a_n}{b_n}\right)b_n,$$
and the initial conditions $a_0=0,b_0=1$, I find for large $N$ the mysterious property that $a_N\approx b_N\approx\frac{1}{\pi}$, with the approximation getting ever better for larger $N$. Indeed, using Mathematica, after a million iterations I find the approximation holds to six decimal places. Can anyone help me prove that $\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n=\frac{1}{\pi}$?
That said, if I substitute $\alpha_n=a_n/b_n$ and $\beta_n=b_{n+1}/b_n$ I end up with $\alpha_{n+1}(2-\alpha_n)=1$.

#### {???}

That said, if I substitute $\alpha_n=a_n/b_n$ and $\beta_n=b_{n+1}/b_n$ I end up with $\alpha_{n+1}(2-\alpha_n)=1$.
Actually, it is the last part of the original (simplified) problem. You see, if one demands that the collision is not perfectly inelastic, but just inelastic enough that the lowest point is stationary, then this arises as the recurrence relation which converges on the final linear and angular velocity. Indeed, taking $a_n=v_n$ and $b_n=R\omega_n$, and replacing the IC by $a_0=0,b_0=R\omega_0$, you find that $v_1=1/4R\omega_0,\omega_1=1/2\omega_0$.

The problem is that, unlike linear recurrence relations, the value it approaches is IC-dependent. We expect, of course, that $v_n\approx R\omega_n\approx R\omega_{n+1}$, with exact equality in the limit, so $a_{n+1}/a_n,b_{n+1}/b_n,a_n/b_n\rightarrow1$ as $n\rightarrow\infty$. That’s not insane, when you think about it. But what is absolutely mental is that the sequences individually tend to $R\omega_0/\pi$. And I have absolutely no explanation why. The limit clearly must be between $a_n$ and $b_n$ for every $n$, and to six decimal places and a million iterations it is.

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#### haruspex

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Actually, it is the last part of the original (simplified) problem. You see, if one demands that the collision is not perfectly inelastic, but just inelastic enough that the lowest point is stationary, then this arises as the recurrence relation which converges on the final linear and angular velocity. Indeed, taking $a_n=v_n$ and $b_n=R\omega_n$, and replacing the IC by $a_0=0,b_0=R\omega_0$, you find that $v_1=1/4R\omega_0,\omega_1=1/2\omega_0$.

The problem is that, unlike linear recurrence relations, the value it approaches is IC-dependent. We expect, of course, that $v_n\approx R\omega_n\approx R\omega_{n+1}$, with exact equality in the limit, so $a_{n+1}/a_n,b_{n+1}/b_n,a_n/b_n\rightarrow1$ as $n\rightarrow\infty$. That’s not insane, when you think about it. But what is absolutely mental is that the sequences individually tend to $R\omega_0/\pi$. And I have absolutely no explanation why. The limit clearly must be between $a_n$ and $b_n$ for every $n$, and to six decimal places and a million iterations it is.
With your initial values, the recurrence relation I got in post #16 has the solution $\alpha_n=\frac n{n+1}$. From that I deduce $b_n=\displaystyle\prod_{r=0}^{n-1}\frac{(n+\frac 12)^2(n+2)}{(n+1)^3}$
That is starting to look like some of the product expansions for $\pi$.

#### {???}

In the the new thread I started I outline this exact method. Regrouping products and canceling, I got precisely one-half times the inverse of the Wallis product for $\pi/2$. That is, $\omega=\omega_0/\pi$. And that is the answer I was looking for.

Even though I came upon the answer before you posted this, I still want to thank you for your help - it was very...helpful!

Cheers,
QM

#### {???}

Sorry to bump this thread, but I was able to apply the ideas discussed for the toppling rod (in the event of a perfectly inelastic collision) back to the hemisphere and would like to confirm my answer for reasons that will become obvious.

Does this appear correct? If so, it's completely bonkers. I've never seen such big numbers in a problem which did not specify numbers to begin with.

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