A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the merry-go-round along a path tangent to the rim of the at a speed of 5.0 m/s. Determine the angular velocity of the system after the boy hops on the merry-go-round.

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

Where;

= The tangential force

I =  The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

= The rotational inertia of the merry-go-round

= The angular acceleration of the merry-go-round

= The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

Where;

= The rotational inertia of the boy

= The angular acceleration of the boy

= The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

Which gives;

From which we have;

The velocity of the merry-go-round, , after the boy hops on the merry-go-round = 1.5 m/s.

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