# Combined linear and rotational motion question

1. Apr 25, 2014

### Dtbennett

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

A small solid disk (r<<R), mass m = 9.3 g, rolls on its edge without skidding on the track shown, which has a circular section with radius R = 9.7 cm. The initial height of the disk above the bottom of the track is h = 30.8 cm. When the ball reaches the top of the circular region, what is the magnitude of the force it exerts on the track? (Hint: how fast is it going then?)

2. Relevant equations

I = 1/2MR^2

F(centripetal) = (mv^2)/r

3. The attempt at a solution

So I'm pretty sure you have to first calculate the velocity of the disk as it enters the circular part. However, I'm confused as to how as we are not provided with a time. Can you assume it is 1 second?

Then the net force must equal the centripetal force at the top of the loop, which will probably be close to zero.
And the speed of the object must match the centripetal force provided by gravity.

so making the centripital force equal to mg gives you

v= sqrt(rg)

#### Attached Files:

• ###### weird question.JPG
File size:
3.2 KB
Views:
67
2. Apr 25, 2014

### voko

You can, but why do you need to? You need the velocity at the top of the circular track, not at its bottom.

This assumption is not based anything substantial, and so best avoided.

Then you can already answer the question in the problem: zero. Does that look right to you?

3. Apr 25, 2014

### haruspex

Use a conservation law.

4. Apr 25, 2014

### Gianf

this problem is really tricky, I am having many problems trying to solve it

Last edited: Apr 25, 2014
5. Apr 25, 2014

### Rellek

6. Apr 25, 2014

### Gianf

Hi,
what I did was saying that mg(h-2r)=1/2mv^2 at the top of the circle. would this be correct?

7. Apr 25, 2014

### haruspex

Yes.

8. Apr 26, 2014

### voko

Since $r$ is taken into account for potential energy, perhaps the kinetic energy due to rotation should also be taken into account?