# [Rotational dynamics] cube sliding on a dish

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1. Apr 24, 2016

### FranzDiCoccio

1. The problem statement, all variables and given/known data

• A small cube is sliding on a round dish (see attached figure) .
• The cube is always in contact with the (vertical) edge of the dish (which prevents the cube from falling outside the dish itself).
• There is friction between the cube and the dish.
• The dish can rotate around a vertical axis going through its center, but it is initally "blocked".
• At some point the block is released (when the cube crosses some angular position in the frame of reference of the dish). At this moment the cube velocity is v.
• The dish starts rotating dragged by the cube. The cube starts slowing down w.r.t. the dish, and it stops after one round (that is, it stops at the same angular position where it was when the brake was released). After this moment the cube and the dish rotate together at the same angular velocity.

The data are

• the velocity of the cube when the brake is released: v = 3,10 m/s.
• the moment of inertia of the dish I=0,75 kg⋅m2.
• the radius of the dish: r=0,65 m.
• the mass of the cube: 0,082 kg.

The question is: what is the friction force? (no further detail)
Solution: 0.055 N.

2. Relevant equations

1) $\vec{L}_f = \vec{L}_i$
2) $K_f = K_i$ (not sure about this)
3) $\Delta K = W$

3. The attempt at a solution

I am assuming that the question is about the friction between the floor of the dish and the cube. Not sure about this though (see below).
I'd solve it using eq. 1) to find the final velocity of the dish (in the rhs of the equation I'd use the total angular momentum). This would allow me to find the decrease in kinetic energy. This decrease should come from the work done by friction force. Since I know the displacement (one circumference), I can calculate the friction force.
So, since
$$m v r = (m r^2+I) \omega$$
I get
$$\omega =\frac{m v r}{m r^2+I}\approx 0.21 rad/s$$
Next
$$\Delta K = \frac{1}{2}(I+mr^2) \omega^2 - \frac{1}{2} m v^2\approx - 0.38 J$$
Taking into account that the work done by the friction force is negative it shoul then be
$$F= -\frac{\Delta K}{2 \pi r} \approx 9.2\cdot 10^{-2} N$$

This differs from the result given by the book.

I am not entirely sure about my solution. Specifically, I am not sure about whether conservation of kinetic energy applies. On the one hand I'd say so, because whatever force the cube exerts on the dish, there is an opposite force exerted by the dish on the cube.
However equation 1 and 2 does not seem to be compatible. The dish and the cube rotating at the same final velocity makes me think of a sort of "inelastic collision", and I won't expect kinetic energy to be the same as it was at the beginning.

I had a look at the "official" solution to this problem (in the "teacher's website" of the book) and it does not make a lot of sense in my opinion, for many reasons. According to the book, the friction force is ultimately equal to the centripetal force acting on the cube: $F_a = m \omega^2 r$.
There are other bits that do not convince me at all in the book's solution, but at this point my questions are:

1) does my solution make any sense, assuming that the friction is between the floor of the dish and the cube?
2) Is there a scenario where the book's solution makes sense?

2. Apr 24, 2016

### Nathanael

I think your book really messed up... The point of the vertical wall around the dish is to provide the centripetal force. It's still possible for the frictional force to have some centripetal component, and in this sense the problem is a bit ambiguous, but, I think the problem creator intended the entire centripetal force to be provided by the wall, so that friction is entirely tangential. (Notice you implicitly assumed this, because only the tangential component of friction will do work.)

I solved the problem by different means (I considered the kinematics instead of work) and I arrived at the same answer as you, 0.092 Newtons.

As for your second equation, it does not make sense... Equal and opposite forces do not imply conservation of kinetic energy. (You never used this equation in your solution though, so I still agree with your method.)

3. Apr 24, 2016

### FranzDiCoccio

Hi Nathanael,
thanks a lot for your help. I included equation 2) just because the "official" solution used that (although in a very unconvincing way). I did not use it in my solution because, as I say, it seems to me that it is incompatible with eq. 1), which should be more fundamental.

As I say, in my opinion the "official" solution does not make sense on many levels. Even if in principle one could think of a friction with the edge and not with the floor, I would not call the resulting friction force "the friction force". Since it would be proportional to the velocity of the cube w.r.t. the dish, it would vary with time. So there is not one single value of it.
I mean, it would still be possible to solve the problem, but it would be much more complex. Definitely outside the scope of the book, unless I'm missing some clever shortcut.

Also, it is strange that the book's solution would not depend on the friction coefficient (the centripetal force is just the normal force for the edge... One needs a friction coefficient, right?) or on the path that the cube travelled on the dish.

Last edited: Apr 24, 2016
4. Apr 24, 2016

### Nathanael

It's also incompatible with eq. 3), because Ki=Kf ⇒ ΔK=0 ⇒ Worknet=0 which of course is not the case.
That's a good point; I just assumed friction was with the ground of the dish, but if it was instead with the wall of the dish then the frictional magnitude will vary in time. That would make the problem a bit more interesting... In that case, we should be able to solve for the frictional coefficient. (But of course we would have to know that friction is purely with the wall. If it is some combination of with the wall and with the ground then we can't solve for anything.)
Did the book's solution treat the friction as being with the wall? Anyway I'm not sure what you mean by "depend on the path of the cube."

5. Apr 24, 2016

### FranzDiCoccio

Sure, of course

Yes, that's what I thought, and why I assumed the simplest situation, i.e. friction with the ground. This should be at least constant, since it depends on the weight of the cube.

It gives no information about the details of the friction. I thought it might be with the wall just because the detailed solution concludes "at equilibrium only the centripetal force and the friction force act on the cube, so that $F_a=m w_f r^2$".

The book's solution only depends on the final angular velocity, which is determined by the conservation of the angular momentum. I mean, that velocity is the same irrespective of how long the cube runs before stopping. Suppose the friction coefficient is doubled. I expect that the cube runs half of the circumference before stopping.
But my soluton depends on the length of the path of the cube. So either it is correct, or I'm missing something, and that length is useless.

6. Apr 24, 2016

### Nathanael

This is wrong on multiple levels... First off, the centripetal force is not an acting force. Secondly there are other forces involved. Thirdly equilibrium never occurs...

It sounds to me that whoever wrote up the solution is deeply confused! You, on the other hand, seem to have a firm grasp of the problem, and I agree with your ideas over the solution-writer's ideas.

(And yes the answer indeed depends on the length of the path of the cube relative to the dish.)