Question

A mass m = 84 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 15.8 m and finally a flat straight section at the same height as the center of the loop (15.8 m off the ground). Since the mass would not make it around the loop if released from the height of the top of the loop (do you know why?) it must be released above the top of the loop-the-loop height. (Assume the mass never leaves the smooth track at any point on its path.)2)What height above the ground must the mass begin to make it around the loop-the-loop?I already found the speed at the top

Answer 1

The velocity at the top of the hoop must be sufficient to maintain circular motion. That means the slowest it can go, is when the weight of the mass is the only centripetal force.
mg=mv/r^2. Solve for v and it will tell you the minimum speed needed to get over the hoop. To find how high you need to drop the mass from, you need to use the conservation of energy. mgh=1/2mv^2. solve for h with the v from the first part. You must drop the mass from that much higher than the top of the hoop.

Answer 2

Right as you finished I solved it. Thanks for your help! I really appreciate it

Answer 3

good on you.