Question

A marble of mass m initially at rest at height h moves along a frictionless rail that forms a hoop of radius R. At the end of the rail the marble hits a spring of elastic constant k. Calculate:

a. The minimum value of h for which the marble hits the spring.
b. The maximum compression of the spring for that value of h.

Answer 1

Answer:

a. The minimum value for h is 2R.

b. The maximum compression of the spring is $\sqrt{\frac{4mgR}{k}}$.

Explanation:

a. If the marble is initially at rest, then its total mechanical energy is equal to its potential energy at that point, mgh. The rail forms a hoop of radius R. In order for the marble to hit the spring, it must reach the top of the hoop, otherwise it falls down.

The height of top of the hoop is 2R. So, in order for the marble to reach the top of the hoop, the height h must be greater than or equal to height of the hoop, 2R.

$h \geq 2R$

In this case, the minimum value for h is 2R.

b. Since the system is frictionless, the compression of the spring depends on the initial potential energy.

$mgh = \frac{1}{2}kx^2\\mg2R = \frac{1}{2}kx^2\\x = \sqrt{\frac{4mgR}{k}}$