Question

A 5.0-m-diameter merry-go-round is initially turning with a 4.0 s period. It slows down and stops in 20 s. Before slowing, what is the speed of a child on the rim (in m/s)? How many revolutions does the merry go round make as it stops (in rev)?

Answer 1

Concepts and reason

The initial angular speed can be calculated by using the relation of angular speed with the time. Then the speed on the rim can be calculated by the relation of speed with the radius and the angular speed.

The angular acceleration can be calculated by using expression for angular acceleration, and then find the number of revolutions (angular displacement) made by merry-go-round before it stops by using kinematic equation.

Fundamentals

The relation between angular speed ${\omega _0}$ and time period T is given as follows:

${\omega _0} = \frac{{2\pi }}{T}$

The relation between the speed and the angular speed is,

$v = r{\omega _0}$

Here, v is the speed, r is the radius, and ${\omega _0}$is the initial angular speed.

The angular acceleration is defined as the rate of change in angular speed, and mathematically it can be expressed as follows:

$\alpha = \frac{{\omega - {\omega _0}}}{t}$

Here, $\alpha$ is the angular acceleration, $\omega$ is the final angular speed, ${\omega _0}$ is the initial angular speed, and t is the time.

According to kinematic equations, the relation between angular displacement, time, initial angular speed and angular acceleration is,

$\theta = {\omega _0}t + \frac{1}{2}\alpha {t^2}$

Here, $\theta$ is the displacement, $\alpha$ is the angular acceleration, ${\omega _0}$ is the initial angular speed, and t is the time.

(a)

The initial angular speed is,

${\omega _0} = \frac{{2\pi }}{T}$

Substitute 4.0 s for T.

$\begin{array}{c}\\{\omega _0} = \frac{{2\pi }}{{4.0{\rm{ s}}}}\\\\ = 1.57{\rm{ rad/s}}\\\end{array}$
‎

The speed of child on the rim is,

$v = r{\omega _0}$

Substitute $\frac{5}{2}\,{\rm{m}}$ for r and $1.57{\rm{ rad/s}}$ for ${\omega _0}$.

$\begin{array}{c}\\v = \left( {\frac{5}{2}{\rm{ m}}} \right)\left( {1.57{\rm{ rad/s}}} \right)\\\\ = 3.93{\rm{ m/s}}\\\end{array}$

The speed of the child on the rim is 3.93 m/s.

(b)

The angular acceleration is,

$\alpha = \frac{{\omega - {\omega _0}}}{t}$

Substitute 20 s for t, 0 rad/s for $\omega$, and 1.57 rad/s for ${\omega _0}$.

$\begin{array}{c}\\\alpha = \frac{{\left( {0{\rm{ rad/s}}} \right) - \left( {1.57{\rm{ rad/s}}} \right)}}{{20{\rm{ s}}}}\\\\\alpha = - 0.0785{\rm{ rad/}}{{\rm{s}}^2}\\\end{array}$

The angular acceleration is $- 0.0785{\rm{ rad/}}{{\rm{s}}^2}$.

The distance travelled by the merry-go-round is,

$\theta = {\omega _0}t + \frac{1}{2}\alpha {t^2}$

Substitute 20 s for t, $- 0.0785{\rm{ rad/}}{{\rm{s}}^2}$ for $\alpha$, and 1.57 rad/s for ${\omega _0}$.

$\begin{array}{c}\\\theta = \left( {1.57{\rm{ rad/s}}} \right)\left( {20{\rm{ s}}} \right) + \frac{1}{2}\left( { - 0.0785{\rm{ rad/}}{{\rm{s}}^2}} \right){\left( {20{\rm{ s}}} \right)^2}\\\\ = 15.7{\rm{ rad}}\\\end{array}$

Convert radian to revolutions.

$\begin{array}{c}\\\theta = 15.7{\rm{ rad}}\left( {\frac{{{\rm{rev}}}}{{{\rm{2}}\pi {\rm{ rad}}}}} \right)\\\\ = 2.5{\rm{ rev}}\\\end{array}$

The number of the revolution is 2.5 rev.

Ans: Part a

The speed of the child on the rim is 3.93 m/s.

Part b

The number of the revolution made by the merry-go-round before it stops is 2.5 rev.