Question

Twins Bob and Sam are on a large merry-go round which is coated with ice making it frictionless. Bob and Sam each have a mass of 47.6 kg respectively. The merry-go round spins at a constant rate of revolution. Bob sits 2.00 m from the center and has to hold on to a metal post with a horizontal force of 80.0 N to stay on the merry-go round. If Sam is sitting 4.00 m from the center, with what force must he hold on to keep from falling off?

Answer 1

See that both twins are sitting on same merry-go-round so both will take same time to complete one revolution, but their linear speed and distance from center will be different.

So We know that centripetal force is given by:

F = m*V^2/R

for Bob

Fb = m*Vb^2/Rb

Vb = sqrt (Fb*Rb/m) = sqrt (80.0*2.00/47.6) = 1.833 m/sec

Now time taken by Bob to complete one revolution will be:

T1 = 2*pi*R1/V1 = 2*pi*2.00/1.833 = 6.855 sec

Now in this same time Sam Also completes one revolution, So

T2 = 2*pi*R2/V2

V2 = 2*pi*4.00/6.855 = 3.666 m/sec

Now Centripetal force on Sam will be:

F2 = m*V2^2/R2 = 47.6*3.666^2/4.00

F2 = 159.93 N = 160.0 N

So sam should hold on with 160 N force to keep himself from falling off

Method 2:

Since Fc = m*V^2/R

V = w*R & w = 2*pi/T, So

Fc = m*w^2*R = m*(2*pi/T)^2*R

Since m and T are constant for both, So

F2/F1 = R2/R1

F2 = F1*(R2/R1) = 80.0*(4.00/2.00) = 160.0 N

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