A merry-go-round with a a radius of R = 1.92 m and moment of inertia I = 190 kg-

Question

A merry-go-round with a a radius of R = 1.92 m and moment of inertia I = 190 kg-m2 is spinning with an initial angular speed of = 1.66 rad/s in the counter clockwise direection when viewed from above. A person with mass m = 60 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.

1)What is the magnitude of the initial angular momentum of the merry-go-round? kg-m2/s

2)What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round? kg-m2/s

3) What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round? kg-m2/s

4) What is the angular speed of the merry-go-round after the person jumps on? rad/s

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

6.

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

6. What is the magnitude of the linear velocity of the person right as they leave the merry-go-round? m/s

7) What is the angular speed of the merry-go-round after the person lets go? rad/s

PLEASE, CLEARLY WORK IT OUT CORRECTLY..THNKS

Explanation / Answer

1)

initial angular momentum = moment of inertia * angular speed

L1 = I*w = 190*1.66 = 315.4 kg.m^2/s

2) the angular momentum

L2 = r x p = r*m*v = 1.92*60*4.5 = 518.4 kg.m^2/s

3) the angular momentum

L3 = r x p = r*m*v = 1.92*60*4.5 = 518.4 kg.m^2/s

4) w = (L1+L2)/(I+m*r^2) = (315.4 + 518.4)/(190 + 60*(1.92)^2) = 2.03 rad/s

5) a = (w^2)*r = (2.03)^2*1.92 = 7.91 rad/s^2

and force F = m*a = 60*7.91 = 474.6 N

6) v = w*r = 2.03*1.92 = 3.9 m/s

7) w = (L1+L2-L3)/I

L3 = m*v*r = 60*3.9*1.92 = 449.28 kg.m^2/s

w = (315.4 + 518.4 - 449.28)/190 = 2.024 rad/s

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