Question

A cart moves with negligible friction or air resistance along a roller coaster track. The cart starts from rest at the top of a hill of unknown height. It then goes down that first hill and up another circularly shaped hill of height 10 meters and radius 15 meters. The ride is designed so that the cart just barely loses contact with the track at the top of the second hill (10 meters above ground level). What is the height of the starting hill relative to ground

Answer 1

Answer:

hinit = 17.5 m

Explanation:

• Assuming no friction present, the mechanical energy must be conserved, which means that at any point of the trajectory, the sum of the gravitational potential energy and the kinetic energy must keep the same.
• At the top of the hill, since it starts from rest, all the energy must be potential, and we can express it as follows:

       $E_{o} = U_{o} = m*g*h_{init} (1)$

• When the car arrives to the top of the second hill, as we know that it is lower than the first one, the energy of the car, must be part gravitational potential energy, and part kinetic energy.
• We can express this final energy as follows:

       $E_{f} = U_{f} + K_{f} = m*g* h_{2} + \frac{1}{2} *m*v_{f} ^{2} (2)$

• In order to find hinit, we need to make (1) equal to (2), and solve for it.
• In (2) we have the value of h₂ (10 m), but we still need the value of the speed at the top of the second hill, vf.
• Now, when the car is at the top of the hill, there are two forces acting on it, in opposite directions: the normal force (upward) and the weight (downward).
• We know also that there is a force that keeps the car along the circular track, which is the centripetal force.
• This force is just the net downward force acting on the car (it's vertical at the top), and is just the difference between the weight and the normal force.
• If the cart just barely loses contact with the track at the top of the second hill, this means that at that point the normal force becomes zero.
• So, the centripetal force must be equal to the weight.
• The centripetal force can be expressed as follows:

       $F_{c} = m*\frac{v_{f} ^{2}}{R} (3)$

• We have just said that (3) must be equal to the weight:

       $F_{c} = m*\frac{v_{f} ^{2}}{R} = m*g (4)$

• Simplifying, and rearranging, we can solve for vf², as follows:

       $v_{f}^{2} = R*g (5)$  

• Replacing (5) in (2), simplifying and rearranging in (1) and (2) we finally have:

      $h_{init} = h_{2} + \frac{1}{2} R = 10m + 7.5 m = 17.5 m (6)$