Question

A 34kg child runs with a speed of 2.4 m/s tangential to the rim of a stationary merry go round. The merry go round has a moment of inertia of 520 kg*m^2 and a radius of 2.00m. After the child hops on the merry go round it rotates with an angular speed of 0.3 rad/s. Find the initial and final kinetic energy of the system. (merry go round and the child). If a frictional torque of 340 is present to slow down the merry go round, find the angular deceleration and calculate the angle in radians in which the merry go round rotates until it stops

Answer 1

given
m = 34 kg
v = 2.4 m/s

I = 520 kg
R = 2.00 m
w = 0.3 rad/s

Initial kinetic energy = (1/2)*m*v^2

= (1/2)*34*2.4^2

= 97.92 J <<<<<<<------------Answer

final kinetic energy, KEf = (1/2)*(I + I_child)*w^2

= (1/2)*(520 + m*R^2)*w^2

= (1/2)*(520 + 34*2^2)*0.3^2

= 29.52 J <<<<<<<<--------------Answer

angular acceleration of the merry-go-round, alfa = -Torque/(I + I_child)

= -340/(520 + 34*2^2)

= -0.5183 rad/s^2

angular displacement before coming to stop,

theta = (wf^2 - wi^2)/(2*alfa)

= (0^2 - 0.3^2)/(2*(-0.5183))

= 0.0868 radians <<<<<<<<--------------Answer