Two boys at the merry go round!

Classical Mechanics Level 5

In Figure above, a $30 kg$ child stands on the edge of a stationary merry-go-round of radius $2.0 m$ . The rotational inertia of the merry-go-round about its rotation axis is $150 kg m^{2}$ • The child catches a ball of mass $1.0 kg$ thrown by a friend. Just before the ball is caught, it has a horizontal velocity $\vec{v}$ of magnitude $12 m/s$ , at angle $\phi= 37°$ with. a line tangent to the outer edge of the merry-go-round, as shown. What is the angular speed (in rad/sec) of the merry-go-round just after the ball is caught?

Liked it try some more

The answer is 0.070.

2 solutions

Gautam Sharma
Nov 19, 2014

AS NET TORQUE ABOUT THE CENTER OF MERRY GO ROUND IS ZERO WE CAN CONSERVE ANGULAR MOMENTUM ABOUT IT. m=mass of ball.

M=mass of child

r=radius of merry go round

${ L }_{ i }={ L }_{ f }$

$mvrsin(90-\phi )=I\omega$

$I=150+m{ r }^{ 2 }+M{ r }^{ 2 }=274\quad kg-{ m }^{ 2 }$

$mvr\sin { (90-\phi ) } =1\times 12\times 2sin53=19.2$

19.2=274 $\omega$

$\omega$ =0.070

In case of collisions it is better to talk in terms of angular impulse rather than torque.

Abhishek Sharma - 6 years, 2 months ago

How come it's a level 5 problem!!!!

saptarshi dasgupta - 2 years, 4 months ago

Ayon Ghosh
Jul 19, 2017

Yes $L = Iω$ $=$ $constant$ we have $L_{initial}$ $=$ $(m_{ball})$ $(v cos theta)$ $(R)$ note v sin theta is irrelevant as it is along radius.And expression for $I$ is $I_{total}$ $=$ $(m_{ball} + m_{child})R^2$ $+$ $I_{merry go round}$ .There now we have all the information.just substitute to get $ω$ .

Two boys at the merry go round!

The answer is 0.070.

2 solutions

0 pending reports