Question

A merry-go-round with a a radius of R = 1.62 m and moment of
inertia I = 215 kg-m^{2} is spinning with an initial
angular speed of ω = 1.52 rad/s in the counter clockwise direection
when viewed from above. A person with mass m = 56 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.

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

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 magnitude of 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

Solution :-

Given that :

Radius (R) = 1.62m

Moment of inertia (I) = 215Kg/m2

Initial angular speed = 1.53 rad/s

Mass of the person = 56Kg

Velocity of the person = 4.6m/s

Formula to calculate the initial angular momentum of A is,

Formula to calculate the angular momentum of person before jumps (B) is,

= 417.3Kg.m2/s

Part 1 :

Formula to calculate the angular speed of the mercy-g0round is,

After jump condition :

Substitute all the values in equation (1)

= 2.055rad/s

Therefore, the angular speed of the merry-go-round is = 2.055 rad/s

Part 2 :

Formula to calculate the force is,

Calculate the acceleration :

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