# Conservation of energy and vertical circular motion

Thanx! Appreciate that!

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Doc Al
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Realize that the block must have a certain amount of kinetic energy (at point B) to make it around the loop if there was no friction. (Hint: How fast must it be going at the very top to stay in contact with the track?) Then realize that the block uses up additional energy doing work against friction. (How much? Consider the definition of work.)

I agree that it's to have sufficient enrgy to overcome resistance to continue the loop at its highest pt, but what has energy got to do with force? I only remember that force * displacement of force = energy. But in this case. the frictional force doesnt travel on a st line, making it hard to calculate the ans. I cant possibly take the diameter as the displacement right? Neither can I take the perimeter of half a circle as the displacement? I am seriously lost here. The second equation seems really hard to form, especially involving the circular motion.

Doc Al
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tubworld said:
I agree that it's to have sufficient enrgy to overcome resistance to continue the loop at its highest pt, but what has energy got to do with force? I only remember that force * displacement of force = energy.
You just answered your own question. Work = Force x Displacement (parallel to the force).

But in this case. the frictional force doesnt travel on a st line, making it hard to calculate the ans. I cant possibly take the diameter as the displacement right? Neither can I take the perimeter of half a circle as the displacement?
The problem tells you the average frictional force is F. So all you need is the distance the block travels in getting up to the top. (Yes... it's half the circumference. It's that simple.)

Ohh... ... I see... but now that i have this value, where does the centripetal acceleration come to place? i dont seem to have any use for it in this question...

Doc Al
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You'll need to use centripetal acceleration to figure out the minimum speed the block must have to maintain contact with the track as it reaches the very top. (That minimum speed is not zero!) Apply Newton's 2nd law.

Just one question that is bugging me though, is there a need for calculus in this question?

Doc Al
Mentor
No calculus is needed to solve this problem.

thanx!
i got it solved! appreciate that!