- #1

FranzDiCoccio

- 342

- 41

**1. Homework Statement**

- 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⋅m
^{2}. - 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.

## Homework Equations

1) [itex]\vec{L}_f = \vec{L}_i[/itex]

2) [itex]K_f = K_i[/itex] (not sure about this)

3) [itex]\Delta K = W[/itex]

## 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

[tex]m v r = (m r^2+I) \omega[/tex]

I get

[tex]\omega =\frac{m v r}{m r^2+I}\approx 0.21 rad/s[/tex]

Next

[tex]\Delta K = \frac{1}{2}(I+mr^2) \omega^2 - \frac{1}{2} m v^2\approx - 0.38 J[/tex]

Taking into account that the work done by the friction force is negative it shoul then be

[tex]F= -\frac{\Delta K}{2 \pi r} \approx 9.2\cdot 10^{-2} N [/tex]

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: [itex]F_a = m \omega^2 r [/itex].

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?