Combined linear and rotational motion question

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Dtbennett
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Homework Statement



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?)

Homework Equations



I = 1/2MR^2

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

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 centripetal force equal to mg gives you

v= sqrt(rg)

I've gotten this far, but I have no idea where to go from here. Please help!
 

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Dtbennett said:
So I'm pretty sure you have to first calculate the velocity of the disk as it enters the circular part.

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

Then the net force must equal the centripetal force at the top of the loop, which will probably be close to zero.

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

And the speed of the object must match the centripetal force provided by gravity.

Then you can already answer the question in the problem: zero. Does that look right to you?
 
this problem is really tricky, I am having many problems trying to solve it
 
Last edited:
Hi,
what I did was saying that mg(h-2r)=1/2mv^2 at the top of the circle. would this be correct?
 
Since ##r## is taken into account for potential energy, perhaps the kinetic energy due to rotation should also be taken into account?