Question

Consider a circular vertical loop-the-loop on a roller coaster. A car coasts without power around the loop. Determine the difference between the normal force exerted by the car on a passenger with a mass of m at the top of the loop and the normal force exerted by the car on her at the bottom of the loop.

Express your answer in terms of m and the acceleration due to gravity g.

Answer 1

Part A.

When passenger is at the highest point

Using Force balance on passenger:

Fnet = -N1 - W = Fc

N1 = Normal force at the top direct downwards

W = Weight of person directed downwards

Fc = Centripetal force downwards = m*Vt^2/R

Vt = speed of car at the top

R = radius of hill

W = Weight of driver = m*g

So,

N1 = Fc - W

N1 = m*Vt^2/R - m*g

At the lowest point of the circle:

Using Force balance on passenger:

Fnet = N2 - W = Fc

N2 = Normal force direct upwards at the bottom

W = Weight of person directed downwards

Fc = Centripetal force upwards

N2 = Fc + W

N2 = m*Vb^2/R + mg

Vb = speed of car at the bottom

Now Using energy conservation at the top and bottom

KEt + PEt = KEb + PEb

PEt = Potential energy at the top = m*g*h

h = height difference between top and bottom = 2*R

PEb = Potential energy at the bottom = 0, since h = 0

(1/2)*m*Vt^2 + m*g*2R = (1/2)*m*Vb^2 + 0

divide by m

(1/2)*Vt^2 - (1/2)*Vb^2 = -g*2R

Vt^2 - Vb^2 = -4*g*R

Now Difference between N1 and N2 will be

N2 - N1 = m*Vt^2/R + m*g - (m*Vb^2/R - m*g)

N2 - N1 = (m/R)*(Vb^2 - Vt^2) + 2*m*g

N2 - N1 = (m/R)*(4*g*R) + 2*m*g

N2 - N1 = 4*m*g + 2*m*g

N2 - N1 = 6*m*g

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