Question

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.

Answer 1

Answer:

(a). The value of angular speed of the merry-go-round $\omega$ = - 5.82 × $10^{-3}$ $\frac{rad}{s}$

(b). The linear speed of the girl after the rock is thrown V = -1.89 × $10^{-2}$ $\frac{m}{s}$

Explanation:

Given data

Mass of the girl $m_{g}$ = 50.6 kg

Mass of merry-go-round $m_{m}$ = 827 kg

Radius r = 3.72 m

The speed of the rock relative to the ground $V_{r}$ = 7.82 $\frac{m}{s}$

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

$\omega = - [\frac{m_{r}v_{r} }{r} ] \frac{2}{m_{m} + 2m_{g} }$

Put all the values in above formula

$\omega = \frac{(1.13)(7.82)}{3.27} \frac{2}{827 + (2)50.6}$

$\omega$ = - 5.82 × $10^{-3}$ $\frac{rad}{s}$

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

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

⇒ V = r × $\omega$

⇒ V = - 3.72 × 5.82 × $10^{-3}$

⇒ V = -1.89 × $10^{-2}$ $\frac{m}{s}$

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