You are looking down at a horizontal merry-go-round with a radius of 10.0 meters
ID: 2303030 • Letter: Y
Question
You are looking down at a horizontal merry-go-round with a radius of 10.0 meters, rotating about point P as shown below. Two people apply forces on opposite sides of the merry-go-round, one of 20.0 Newtons and the other of 10.0 Newtons. The moment of inertia of the merry-go-round is 100 kg-m2. Ignore friction. N.B. The numbers here are chosen so that the answers can be obtained without a calculator! What is the net torque acting on the merry-go-round (include the direction)? What is the angular acceleration of the merry-go-round (include the direction)? If it starts at rest, what is the angular velocity after 2.00 seconds (include the direction)? What is the angular displacement after 2.00 seconds (include the direction)? What is its rotational kinetic energy in joules at this time (2.00 s)? What is the work done by the net torque? If the answers to parts e and f are the same, explain. If they are different, explain the difference.Explanation / Answer
A) n et torque is T = T1+T2...
T1 = 20*10 = 200 N.m
T2 = 10*10 = -100 N.m..
T = 200-100=100 N.m...along positive Z axis...perpendicular to the paper i.e out side the paper...
B) angular accelaration alpha = T/I = 100/100 = 1 rad/s^2...in the ditection of torque i.e perpendicular to the paper i.e out side the paper...
C) wf = wi+(alpha*t) = 1*2 = 2 rad/sec...
D) theta = wi*t +(1/2)*alpha*t^2 = (1/2)*1*2^2 = 2 rad from the starting point
E) Rotational K.E = (1/2)*I*wf^2 = (1/2)*100*2^2 = 200 J..
F) W = T*theta = 100*2 = 200 J...
g) answers are same work done by torque is stored in the meery go round as the Kinetic energy
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