Question

A merry-go-round is a common piece of playground equipment. A 2.8-m-diameter merry-go-round with a mass of 260 kg is spinning at 22 rpm. John runs tangent to the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John's mass is 39 kg. What is the merry-go-round's angular velocity, in rpm, after John jumps on?

Answer 1

R = 1.4 m, M = 260 kg, v = 5 m/s

Using conservation of angular momentum.

L_i = I_i * w_i + m V R   

Final momentum L_f = I_f * w_f

Moment of inertia for a disk: I_i = 1/2 * M * R²

=1/2 * 260kg * (1.4 m)² = 254.8 kg.m²

Final moment of inertia can be determined using parallel axis theorem:

I_f = I_i + mR²

= 254.8 kg.m² + 39 kg * (1.4 m)²

= 331.24 kg.m²

Converting revolutions per minute to radians per second:

(22 rpm) * ( 2? rad / rev) * (1 min / 60s) = 2.30 rad/s

L_i = L_f

I_i * w_i + mVR=  I_f * w_f   

(254.8 kg*m²) * 2.30 rad/s + 39kg * 1.4 m * 5m/s = 331.24 kg.m² * ?_f

?_f = 2.59 rad/s

?_f = 24.73 rpm

Answer 2

from law of conservation of energy

(1/2)*I1*w1^2 + (1/2)*mj*v^2 = (1/2)(I1+I2)*W^2

here I1 = (1/2)*m*R^2 = 0.5*260*1.4^2 =254.8 kgm^2...

I2 = M*r^2 = 39*1.4^2 = 76.44 kg m^2...
w1 = 22rpm
mj   = 39 kg...
v = 5 m/sec..

w^2 = [(1/2)*I1*w1^2 + (1/2)*mj*v^2] / [(1/2)(I1+I2)]

w^2 = [(254.8*22^2 )+ (39*25)]/(254.8+76.44) = 375.25

w = 19.4 rpm