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A solid marble of mass m = 40 g and radius r = 2 cm will roll without slipping along the loop-the-loop track if it is released from rest somewhere on the straight section of track. From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop? The radius of the loop-the-loop is R = 0.95 m.

So this problem has to do with Rotation and Inertia and Conservation of energy I think.

I found the Inertia i believe with I=(2/5)mr^2

So the top of the loop-the-loop is 2 times the radius which is .95, so it'd be 1.90 m.

Conservation of mechanical energy says

(1/2)mv_f^2 + mgy_f = (1/2)mv_i^2 + mgy_i ?

How do I arrive to that formula?

So this problem has to do with Rotation and Inertia and Conservation of energy I think.

I found the Inertia i believe with I=(2/5)mr^2

So the top of the loop-the-loop is 2 times the radius which is .95, so it'd be 1.90 m.

Conservation of mechanical energy says

(1/2)mv_f^2 + mgy_f = (1/2)mv_i^2 + mgy_i ?

How do I arrive to that formula?

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