A girl of mass 50.6 kg stands on the edge of a frictionless merry-go-round of mass 827 kg and radius 3.72 m that is not moving. She throws a 1.13-kg rock in a horizontal direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground, is 7.82 m/s. Calculate (a) the angular speed of the merry-go-round and (b) the linear speed of the girl after the rock is thrown. Assume that the merry-go-round is a uniform disk.

(a). The value of angular speed of the merry-go-round = - 5.82 ×

(b). The linear speed of the girl after the rock is thrown V = -1.89 ×

Explanation:

Given data

Mass of the girl = 50.6 kg

Mass of merry-go-round = 827 kg

Radius r = 3.72 m

The speed of the rock relative to the ground = 7.82

(a). The angular speed of the merry-go-round is given by

Put all the values in above formula

= - 5.82 ×

This is the value of angular speed of the merry-go-round.

(b). The liner speed of the girl is given by

⇒ V = r ×

⇒ V = - 3.72 × 5.82 ×

⇒ V = -1.89 ×

This is the linear speed of the girl after the rock is thrown.

the answer would be a

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explanation: