# Centripetal force of rolling disc

1. Jun 9, 2016

### danny20051

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

http://moodle.telt.unsw.edu.au/pluginfile.php/1602233/question/questiontext/1933161/1/1188262/Screen%20Shot%202014-04-03%20at%2011.21.21%20am.png

A disc has radius r<<R and mass m = 7.8 g. When released, it rolls on its edge without skidding on the track in the sketch. The circular section has radius R = 6.5 cm. The initial height of m above the bottom of the track is h = 39.0 cm. When the ball is rolling at the top of the circular region, what is the magnitude of the force the disc exerts on the track? (Hints: how fast is it going? and don't forget to draw a free body diagram)

There is a sketch with this that is a disc positioned on a vertical inclination at the bottom of which is a circular loop of radius R.

2. Relevant equations
U=mgh
K=0.5mv^2
F=(mv^2)/r

3. The attempt at a solution
So my understanding is that the disc will initially have a potential energy of mgh. This will then be converted into kinetic energy once this disc is at the bottom. Once the disc reaches the top of the circular region some of this kinetic energy will be converted into potential energy. This being mgh where in this case h=0.13m(diameter).
The remaining energy must be in the form of kinetic energy. I believe some energy would be in the rotational energy of the disc however i assume the statement r<<R is indicating its negligible.

So (0.0078*9.8*0.39)-(0.0078*9.8*0.13) = 0.5mv^2

So (2*KE)/r is the centripetal force. Where r=0.065m

This results in the answer 0.61152N

I am not too sure what i am missing. Any help is greatly appreciated.

2. Jun 9, 2016

### Delta²

The centripetal force is equal to (the force the track applies to the disc plus the weight of the disc). That is, part of the centripetal force is due to the normal force and part due to the weight. The problem basically asks for the opposite of the normal force.

However this doesn't seem to be the only mistake, seems that totally ignoring the rotational kinetic energy is an oversimplification.

3. Jun 9, 2016

### danny20051

mg = 0.07644N So taking this away from the force the disc exerts on the track due to its velocity still doesn't bring it anywhere near 0.33N

Would this be due to the rotational energy of the disc? If so i have no clue how to work out the rotational energy of the disc with the info they provided.

4. Jun 9, 2016

### Delta²

Well we cant ignore the rotational energy cause it actually is $E_{rot}=\frac{1}{2}\frac{1}{2}mr^2\omega^2$ where $\omega=\frac{v}{r}$ since the disk rolls without sliding. It is actually equal to half of the translational kinetic energy so it cannot be ignored.

Last edited: Jun 9, 2016