A certain playground merry-go-round has a mass of 200 kg and a radius of 2 m. A

Question

A certain playground merry-go-round has a mass of 200 kg and a radius of 2 m. A child of mass 40 kg stands on the edge of the merry-go-round to ride, while another child pushes the merry-go-round in the most advantageous way, starting from rest, making one complete revolution in 5 s, and then lets go. Assume that the angular acceleration is constant. You may also assume that the merry-go-round is a uniform solid disk. (a)Find the force applied by the child pushing the merry-go-round. (b) What is the linear speed of the person standing on the edge of the merry-go-round at the final angular velocity? (c) What is the linear (tangential) acceleration of the person on the edge? (d) What is the centripetal acceleration of the person at the final angular velocity? (e) What is the angular momentum of the merry-go-round plus the child at the final velocity?

Explanation / Answer

A)

angular accelaration is alpha

then using

theta = 0.5*alpha*t^2

2*3.142 = 0.5*alpha*5^2

alpha = 0.50272 rad/s^2

but torque T = r*F = I*alpha

I =0.5*m1*r^2 + m2*r^2 = (0.5*200*2^2)+(40*2^2) = 560 kg-m^2

force F = (I*alpha)/r = (560*0.50272)/2 = 140.76 N

b) linear speed is v = 2*pi*r/T = (2*3.142*2)/5 = 2.5136 m/s

c) a_tan = r*alpha = 2*0.50272 = 1.00544 m/s^2

d) ac = v^2/r = (2.5136^2)/2 = 3.16 m/s^2

e) L = I*w= 560*(v/r) = 560*(2.5136/2) = 703.808 kg*m^2/s

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