Question

A 40.0 kg child runs with a speed of 2.90 m/s tangential to the rim of a stationary merry-go-round(Figure 1). The merry-go-round has a moment of inertia of 580 kg?m2and a radius of 2.51 m . When the child jumps onto the merry-go-round, the entire system begins to rotate.

A) Calculate the initial kinetic energy of the system. (J)

B)Calculate the final kinetic energy of the system. (J)

Answer 1

Here,

m = 40 Kg

u = 2.9 m/s

I = 580 Kg.m^2

r = 2.51 m

a) for the initial kinetic energy of the system

Initial kinetic energy = 0.50 * 40 * 2.9^2

Initial kinetic energy = 168.2 J

b)

let the final angular speed is wf

Using conservation of angular momentum

40 * 2.9 * 2.51 = (40 * 2.51^2 + 580) * wf

wf = 0.35 rad/s

final kinetic energy of the system = 0.50 * I * wf^2

final kinetic energy of the system = 0.50 * (40 * 2.51^2 + 580) * 0.35^2

final kinetic energy of the system = 51 J

the final kinetic energy of the system is 51 J