The roller coaster in Figure P6.36 starts with a velocity of 16.5 m/s. One of th

Question

The roller coaster in Figure P6.36 starts with a velocity of 16.5 m/s. One of the riders is a small girl of mass 27 kg. At points B and C, the track is circular with the radii of curvature given in the figure. The heights at points A, B, and C are hA= 24.75 m, hB = 34.9 m, and hC=0 m. You may assume the track is frictionless.

Find the velocity of the roller coaster at point B.

Find the velocity of the roller coaster at point C.

Find her apparent weight at B

Find her apparent weight at C

Explanation / Answer

Hi,

In this case we use the conservation of energy to find the velocity at points B and C, since the friction force is not taken into consideration.

EA = EB ::::::::: mghA + (1/2) mvA2  = mghB  + (1/2) mvB2

vB2 = vA2 + 2g(hA - hB) = (16.5 m/s)2 + 2(9.8 m/s2) (24.75 - 34.9) m = 73.31 m2/s2 :::::: vB = 8.56 m/s

EA = EC :::::::: mghA + (1/2) mvA2  = (1/2) mvC2

vC2 = vA2 + 2ghA =  (16.5 m/s)2 + 2(9.8 m/s2) (24.75 m) = 757.35 m2/s2  ::::::::: vC = 27.52 m/s

The 'difference' in weight comes from the effect of the normal force in the different points (B and C) of the roller coster. When we apply Newton's Laws to the person in both points and we remembered that we a an uniform circular movement, we get the following:

mg - N = mv2/r (for the point B)   and N - mg = mv2/r (for the point C)

For the point B: N = (27 kg)(9.8 m/s2) - (27 kg)(8.56 m/s)2 / (10 m) = 66.76 N

This means that the apparent weight of the person would be:

W = mg - N = 197.84 N = 0.75 mg

For the point B:    N = (27 kg)(27.52 m/s)2 / (20 m) - (27 kg)(9.8 m/s2) = 757.82 N

This means that the apparent weight of the person would be:

W = N - mg = 493.22 = 1.86 mg

I hope it helps.

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