Question

A merry-go-round is a common piece of playground equipment. A 3.0-m-diameter merry-go-round with a mass of 250 kg is spinning at 20 rpm. John runs tangent to the merry-go-round
at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. John’s mass is 30 kg. What is the merry-goround’s angular velocity, in rpm, after John jumps on?

Short answer

After John jumps on the angular velocity is 25.37 rpm

Step-by-step solution

Given information

Mass of merry go round (M) =250 kg

Speed of merry go round =20 rpm

Diameter of merry go round = 3m

So radius =1.5 m
Mass of John =30 kg

Speed of John = 5m/s

Explanation

First find the Moment of inertia of merry go round

$I=\frac{1}{2}M{R}^2..........................................\left(1\right)$

So the momentum of the merry go round is

$\Rightarrow L_m=I\omega =\frac{1}{2}M{R}^2\left(\omega \right)...............................\left(2\right)$

Substitute the given value to get the momentum

$L_m=I\omega =\frac{1}{2}(250\mathrm{kg})(1.5\mathrm{m}{)}^2(2.093\mathrm{rad}/\sec )=588.656{\mathrm{kgm}}^2/\sec ...........(3)$

Angular momentum of John

$L_j=mV{R}^2........................\left(4\right)$

Substitute the values we get

$L_j=mV{R}^2=(30\mathrm{kg})(5\mathrm{m}/\mathrm{s})(1.5\mathrm{m}{)}^2=337.5{\mathrm{kgm}}^2/\sec ..........(5)$

Now find total angular momentum
Total angular momentum = Angular momentum of merry go round + angular momentum of John

$$\begin{gathered} L_{\text{total}}=L_m+L_j \\ \Rightarrow L_{\text{total}}=\frac{588.656{\mathrm{kgm}}^2}{\sec }+\frac{337.5{\mathrm{kgm}}^2}{\sec } \\ \Rightarrow L_{\text{total}}=926.55{\mathrm{kgm}}^2/\sec \end{gathered}$$

We know L = Iω

Lets find inertia

$I_{\text{total}}=\frac{1}{2}M{R}^2+m{R}^2...........................\left(6\right)$

We can find the angular velocity by substituting values in equation (6)

$$\begin{gathered} {\omega }_{\text{total}}=\frac{L_{\text{total}}}{\frac{1}{2}M{R}^2+m{R}^2} \\ {\omega }_{\text{total}}=\frac{\frac{926.55{\mathrm{kgm}}^2}{\sec }}{\frac{1}{2}(250\mathrm{kg})(1.5\mathrm{m}{)}^2+(30\mathrm{kg}\left)\right(1.5\mathrm{m}{)}^2} \\ {\omega }_{\text{total}}=2.656\mathrm{rad}/\sec \\ {\omega }_{\text{total}}=25.37\mathrm{rpm} \end{gathered}$$