Question

Merry Go Round

A merry-go-round with a radius of R = 1.80 m and moment of inertia I = 201 kg-m2 spinning with an initial angular speed of ω = 1.5 rad/s in the counter clockwise direction when viewed from above. A person with mass m = 55 kg and velocity v = 4.5 m/s runs on a path tangent to the merry-go-round. Once at the merry-go-round the person jumps on and holds on to the rim of the merry-go-round.

What is the magnitude of the initial angular momentum of the merry-go-round about the center?

What is the angular speed of the merry-go-round after the person jumps on?

What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round?

Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?

Once the person gets half way around, they decide to simply let go of the merry-go-round to exit the ride.

What is the linear velocity of the person right as they leave the merry-go-round?

What is the angular speed of the merry-go-round after the person lets go?

Answer 1

a)

Initial Angular momentum of the merry-go-round about the center

L_i,m =IW =201*1.5 =301.5 Kg-m²/s

b)

By Conservation of angular momentum

L_i = L_f

IW+mVR =(I+mR²)W'

301.5 + 55*4.5*1.8 =(201+55*1.8²)*W'

W' = 747/379.2= 1.97 rad/s

c)

the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round

L_i,person=mVR =55*4.5*1.8 =445.5 kg-m²

^d)

The force needed by person to hold on

F=mW'R² =55*1.97*1.8² = 351.04 N

e)

Linear velocity of the person right as they leave the merry-go-round is

V=RW' =1.8*1.97 = 3.546 m/s

f)

It should be same as c

W' =1.97 rad/s