1)What is the magnitude of the initial angular momentum of the merry-go-round? 3

Question

1)What is the magnitude of the initial angular momentum of the merry-go-round? 315.18 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? 523.38  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? 523.38 kg-m2/s

4) What is the angular speed of the merry-go-round after the person jumps on? 1.97 rad/s

5) Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on? 461.6 N

I WANT TO SOLVE JUST THESE

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

A merry-go-round with a a radius of R = 1.83 m and moment of inertia I-206 kg-m2 is spinning with an initial angular speed of -1.53 rad/s in the counter clockwise direection when viewed from above. A person with mass m 65 kg and velocity v-4.4 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.

Explanation / Answer

6)The linear velocity is given by angualr velocity*radius = 1.97*1.83=3.61m/s

7)The angular velocity of the merry-go-round remain same because since no force acted on the person so no force acted on the merry-go-round.

Angualr speed = 1.97rad/s

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