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JEE Advanced 2014 Paper 1

Section: Physics

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Question : 19 of 60

Marks: +1, -0

A horizontal circular platform of radius

0.5

0.45

kg is free to rotate about its axis. Two massless spring toy-guns, each carrying a steel ball of mass

0.05

kg are attached to the platform at a distance

0.25

m from the center on its either sides along its diameter (see figure). Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the balls have horizontal speed of

9\,\text{m/s}

with respect to the ground. The rotational speed of the platform in rad

\text{s}^{-1}

after the balls leave the platform is

[JEE Adv 2014 P1]

Your Answer:

Solution:

Since net torque about center of rotation is zero, so we can apply conservation of angular momentum of the system about center of disc

L_i=L_f

0=I\omega+2mv\left(\frac{r}{2}\right)

,comparing magnitude

\therefore\left(\frac{0.45\times 0.5\times 0.5}{2}\right)\omega=0.05\times 9\times \frac{0.5}{2}\times 2

\therefore\omega=4

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