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← JEE Advanced 2014 · Paper 1
PhysicsInteger+3Question 19 of 60

JEE Advanced 2014 · Paper 1, Physics question 19

A horizontal circular platform of radius 0.5 m and mass 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 centre 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 m s−19\ \mathrm{m\,s^{-1}} with respect to the ground. The rotational speed of the platform in rad s−1\mathrm{rad\,s^{-1}} after the balls leave the platform is

A horizontal disc on a vertical axle; two toy guns on a diameter, equidistant from the centre, fire balls in opposite horizontal directions perpendicular to that diameter.
Answer and worked solution

Official answer: 4

Angular momentum about the axis is conserved and starts at zero. Each ball leaves with 9 m s⁻¹ (relative to the ground) at 0.25 m from the axis, both in the same rotational sense:
Lballs=2×0.05×9×0.25=0.225 kg m2 s−1.L_\text{balls}=2\times0.05\times9\times0.25=0.225\ \mathrm{kg\,m^2\,s^{-1}}.

Platform. I=12×0.45×0.52=0.05625I=\tfrac12\times0.45\times0.5^2=0.05625 kg m², turning the opposite way:
ω=0.2250.05625=4 rad s−1.\omega=\frac{0.225}{0.05625}=4\ \mathrm{rad\,s^{-1}}.

Answer: 4

Topics:Angular momentum and its conservation
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