Question

A rollercoaster cart starts from rest at the top of a 100m

high hill. It moves around a frictionless track before entering a vertical, circular loop at ground level. What is the maximum radius that the loop can have if the cart is to clear the top of the loop without slipping?

Answer 1

The cart starts from rest at the top of a 100 m high hill.

The track is frictionless, so all initial gravitational potential energy will convert to kinetic energy.

The gravitational potential energy at the top is: PE = mgh

PE = (mass of cart) x (gravitational field strength) x (height)

= mc x g x 100

= mc x 10 x 100 = mc x 1000 Joules (assuming g = 10 m/s^2)

All this potential energy converts to kinetic energy at the bottom:

KE = 1/2mv^2

KE = PE

1/2mv^2 = mc x 1000

v = sqrt(2 x 1000/m) = sqrt(2000/m)

For the cart to clear the top of the circular loop of radius r, its velocity must be:

v >= 2πr/T

where T is the time period of one revolution.

Equating the velocity equations from step 4 and 5:

sqrt(2000/m) >= 2πr/T

Solving for the maximum radius r:

r <= (m/2000) x (T x sqrt(2000/m))^2

= (m/2000) x (T x sqrt(2000))

= mT^2/2000

So the maximum radius of the loop that the cart can clear without slipping is directly proportional to the mass of the cart and the square of the time period of one revolution.