Rollercoaster physics problem?

Question

A car on a rollercoaster starts at the top of a big hill at rest and rolls without friction. It enters the bottom of the 10 m radius loop with a speed of 20 m/s.

A. What is the force the seat bottom exerts on a 50 kg passenger at the bottom of the loop?


B. What is the force the seat bottom exerts on the passenger at the top of the loop? (Hint: consider conservation of energy)


C. What is the force the seat bottom exerts on the passenger at a point halfway between the top and bottom of the loop?


D. After exiting the loop, the car continues on and passes over the hill of radius R as shown. As the designer, you must calculate the minimum radius of the hill so that the car just maintains contact with the track as it passes over the top.


Thanks in advance everyone!! :D

Answer 1

The centrifugal force at the bottom of the loop is m*v^2/r, and to this must be added the gravity force, m*g, so the force is m*(g+v^2/r).

The velocity of the car at the top of the loop will be determined by the loss of kinetic energy from rising from bottom to top. The kinetic energy at the bottom is .5*m*v^2; the energy at the top is .5*m*vt^2, the difference is the potential energy loss = m*2*r. Therefore find vt from .5*m*(v^2-vt^2) = m*2*r. When you get vt, the centrifugal force at the top is m*vt^2/r. At the top, this is subtracting from gravity so the net force is m*g - m*vt^2/r

Halfway up, the potential energy loss is m*g*r, and from this find the velocity (as previously) and from the velocity find the cetrifugal force. Since the force from the seat is horizontal, gravity does not come into play.

You must find the kinetic energy and potential energy at the point where the car leaves the loop. In order for the car to stay on track at the top of the hill, the centrifugal force must not be more than the gravity force. So we have the relation m*vh^2/R = m*g; Now you need to find vh. The difference in height between the hill and the point where the car left the loop times m*g is the potential energy change, and that must equal the kinetic energy change (KE on leaving the loop minus KE at the top of the hill, which is .5*m*vh^2)

Answer 2

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