Question

A spherical marble that has a mass of 45.0 g and a radius of 0.500 cm rolls without slipping down a loop-the-loop track that has a radius of 30.0 cm. The marble starts from rest and just barely clears the loop to emerge on the other side of the track.

1)

What is the minimum height that the marble must start from to make it around the loop? (Express your answer to three significant figures and in cm.)

Answer 1

During the motion, the marble gains translational and rotational kinetic energy:

Etrans = 1/2 m v^2
Erot = 1/2 I w^2
= 1/2 (2/5 m r^2) (v/r)^2
= 1/5 m v^2

So the energy balance, when starting from rest at height H and going to height 2R (at the top of the loop) is:

m g H = m g (2 R) + 1/2 m v^2 + 1/5 m v^2

m g H = 2 m g R + 7/10 m v^2
which equates the initial potential energy on the left (kinetic energy zero initially) to the total mechanical energy on the right (potential energy + translational kinetic energy + rotational kinetic energy).

To stay on the loop at the highest point, the velocity must be such that the gravity alone is not enough to provide the centripetal force (there has to be a nonzero normal force helping too):

m v^2 / R > m g

m v^2 >m g R

Using this on the right hand side of the energy balance we have

m g H > 2 m g R + 7/10 ( m g R)

m g H > 27/10 m g R

H > 27/10 R = 2.7 * 0.3 = 0.81 m