Question

A mass m = 87 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius R = 19.9 m and finally a flat straight section at the same height as the center of the loop (19.9 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.) 1)What is the minimum speed the block must have at the top of the loop to make it around the loop-the-loop without leaving the track?m/s

Answer 1

v=13.97m/s

Step-by-step explanation:

The mass would not make it around the loop if released from the height of the top of the loop because it would reach the top of the loop with speed 0m/s (spending all its gravitational potential energy in achieving the same height, leaving nothing for the kinetic energy).

For the minimal speed the mass would just not have a contact with the track, which means that the only force acting on it would be its weight, and this weight must be the centripetal force, at that point, for that circular motion. So we write
`W=F_(cp)`, which means:

`mg=mv^2/r`

Solving for v we get:

`v=√(gr)`

So we just substitute the values we have:

`v=√((9.8m/s^2)(19.9m))=13.97m/s`