A roller-coaster car starts at rest on a frictionless track. It coasts down a lo

Question

A roller-coaster car starts at rest on a frictionless track. It coasts down a long hill (gaining speed as it descends), and then completes a loop-the-loop of radius R (losing speed as it rises, regaining it again as it falls.) If the car starts at rest at an elevation 3R above the lowest part of the loop, what will be the magnitude of the normal force exerted by the track on the car, when it is at point T, at the top of the loop? Express your answer as a pure number times the car's weight, mg What is the car's acceleration (vector), when it is at Point B, at the bottom of the loop? Express your answer as a multiple of the free-fall acceleration g.

Explanation / Answer

Here,

a) at the point T

let the velocity of the car is v

Using conservation of energy

0.5 * m * v^2 = m * g * (3R - R)

v^2 = 4 * g * R

Now , for the normal force at T

normal force at T = m * v^2/R - m *g

normal force at T = 4 * m * g - m * g

normal force at T = 3 * m * g

b)

at the point B ,

let the velocity of the car is v

Using conservation of energy

0.5 * m * v^2 = m * g * (3R)

v^2 = 6 * g * R

acceleration of the car = m * v^2/R

acceleration of the car = 6*g

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