Question

In figure, a $30kg$ child stands on the edge of a stationary merry-go-round of. radius $2m$ The rotational inertia of the merry-go-round about its rotation axis is $196kg{m}^2$ The child catches a ball of mass $1kg$ thrown by a friend. Just before the ball is caught, it has a horizontal velocity $\overrightarrow{v}$ of magnitude $10m/s,$ at angle $\varphi ={37}^{\circ }$ with a line tangent to the outer edge of the merry-go-round, as shown. What is the angular speed of the merry-go-round just after the ball is caught?

• A. $0.05rad/s$
• B. $0.25rad/s$
• C. $0.75rad/s$
• D. $0.10rad/s$

Answer 1

The correct option is A $0.05rad/s$

Formula used : $L=I\omega and{L}_{ball}=mvr$

Given:
$m_{man}=30kg,R=2m,I_{platform}=196kg{m}^2,m_{ball}=1kg,v_{ball}=10m/s\varphi ={37}^{\circ }$

Let $m_{\text{ball}}$ and $v$ be the mass and initial speed of the ball and $R$ the radius of the merry-go-round. The initial angular momentum is given by

$L_i=m_{ball}\left(v\cos \varphi \right)R=1\times 10\cos {37}^{\circ }\times 2=16units$

$=(196+30\times 2^2+1\times 2^2)\omega$

$L_i=L_f$

$16=320\omega$

$\therefore \omega =0.05rad/s$

FINAL ANSWER: (a)