Question

A child stands on the edge of a merry-go-round as shown below. The rotational inertia of the merry-go-round about its rotation axis is $150{\text{kg-m}}^2$. The child catches a ball of mass $1.0\text{kg}$ thrown by a friend. What is the angular speed of the merry-go-round just after the ball is caught?

• A. $0.05\text{rad/s}$
• B. $0.07\text{rad/s}$
• C. $0.03\text{rad/s}$
• D. $0.09\text{rad/s}$

Answer 1

The correct option is B $0.07\text{rad/s}$


In this case, there is no external torque is applied.

So, angular momentum $(L=mvr=I\omega )$ of the system will be conserved.

Now applying the law of conservation of angular momentum about the rotation axis before and after the ball is caught.
$$L_i=L_f$$
Substituting the given data we get,

$1\times 12\cos {37}^{\circ }\times 2=(150+30\times 2^2+1\times 2^2)\omega$

$\Rightarrow 19.2=274\omega$

$\Rightarrow \omega =0.07\text{rad/s}$

Hence, option (b) is the correct answer.

Why this question? Concept involved -Law of conservation of angular momentum. Tip- First we should break the component of velocity along and perpendicular to the radius of the circular motion. Only momentum due to tangential velocity will be considering as for radial velocity $\theta =0$ and $m\overrightarrow{r}\times \overrightarrow{v}=0$.