A merry-go-round with a a radius of R = 1.62 m and moment of inertia I = 215 kg-

Question

A merry-go-round with a a radius of R = 1.62 m and moment of inertia I = 215 kg-m2 is spinning with an initial angular speed of ? = 1.52 rad/s in the counter clockwise direection when viewed from above. A person with mass m = 57 kg and velocity v = 4.6 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.

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

Explanation / Answer

1) L1=I*w=215*1.52= 326.8 kg-m2/s

2) L2=r x p=r*m*v=1.62*57*4.6= 424.76 kg-m2/s

3) Exactly the same as question 2=  424.76 kg-m2/s

4) w=(L1+L2)/(I+m*r^2)= 2.06 rad/s

5) a=(w^2)*r

a=6.79,

F=m*a= 57* 6.79= 387.03 N

6) v=w*r= 3.33 m/s

7) (L1+L2-L3)/I,

L3=m*v*r=57*3.33*1.62=307.492

(326.8 + 424.76 - 307.492)/215= 2.061 rad/s

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