- 35

- 0

- Summary
- Can this problem be solved? If so, is this the right way to go about it?

Hello,

So I've been toying with the following problem and I'd really appreciate some feedback and advice.

I am contemplating an approach based on the method of successive approximations and Approach 2 above: 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. However, there is also the issue of how the lower piece gets arrested. Should I assume its angular momentum also vanishes at this instant? 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.

I'd appreciate any advice on how to hone the method I'm currently considering, as well as answering questions I posed above. It's also entirely possible I'm overlooking something obvious (maybe I've been blinkered by my latest approach?). If you think I have, I welcome your opinion. Please advise.

Thanks,

QM

So I've been toying with the following problem and I'd really appreciate some feedback and advice.

**The problem statement (as unambiguous as I can make it):**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?**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 [itex]\frac{3\pi}{64}R\omega_0[/itex] where [itex]\omega_0=\sqrt{\frac{15g}{8R}}[/itex] 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 [itex]\omega_1=\frac{3\pi}{64}\omega_0[/itex]. 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 [itex]\sin\theta[/itex] 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.**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 [itex]v_1=\frac{3}{8}R\omega_0[/itex] and [itex]\omega_1=\frac{45}{166}\omega_0[/itex]. Here you can already see the problem--the linear velocity of the lowest point of the hemisphere has negative horizontal component!**

Approach 2:Approach 2:

I am contemplating an approach based on the method of successive approximations and Approach 2 above: 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. However, there is also the issue of how the lower piece gets arrested. Should I assume its angular momentum also vanishes at this instant? 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.

I'd appreciate any advice on how to hone the method I'm currently considering, as well as answering questions I posed above. It's also entirely possible I'm overlooking something obvious (maybe I've been blinkered by my latest approach?). If you think I have, I welcome your opinion. Please advise.

Thanks,

QM