JEE Challenger
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PhyMedAdv 2014 • Paper 1

Rotational Speed of Circular Platform After Firing Steel Balls

Consider a horizontal circular platform of radius 0.5 m0.5\text{ m} and mass 0.45 kg0.45\text{ kg} that is free to rotate about its central axis. Two massless spring toy-guns, each holding a steel ball of mass 0.05 kg0.05\text{ kg}, are fixed to the platform at a distance of 0.25 m0.25\text{ m} from the centre on opposite sides along a diameter (see figure). Both guns simultaneously launch the balls horizontally and perpendicular to the diameter in opposite directions. After detachment from the platform, each ball has a horizontal speed of 9 ms−19\text{ ms}^{-1} relative to the ground. Find the rotational speed of the platform in rad s−1\text{rad}\text{ s}^{-1} after the balls leave the platform.

Question Diagram 1
Official Numerical Answer4

Topics & Concepts

System of Particles and Rotational MotionAngular MomentumMoment of Inertia

Step-by-Step Solution

To find the rotational speed of the circular platform after the steel balls leave, we use the principle of Conservation of Angular Momentum about the central vertical axis of rotation.

Since no external torque acts on the system (platform + guns + balls) along the axis of rotation, the total initial angular momentum of the system must equal the total final angular momentum.

1. Initial State

Initially, the entire system is at rest. Thus, the total initial angular momentum about the central axis is: Li=0L_i = 0

2. Final State

After the launch, both balls detach simultaneously.

  • Angular Momentum of the Balls: Each ball has a mass m=0.05 kgm = 0.05\text{ kg} and is launched at a distance r=0.25 mr = 0.25\text{ m} from the axis with a horizontal speed v=9 ms−1v = 9\text{ ms}^{-1} relative to the ground. Since the balls are launched in opposite directions at equal distances on opposite sides of the diameter, their individual angular momenta add up constructively: Lballs=2×(mvr)L_{\text{balls}} = 2 \times (m v r)

  • Moment of Inertia of the Platform: The platform is a uniform circular disk of mass M=0.45 kgM = 0.45\text{ kg} and radius R=0.5 mR = 0.5\text{ m}. Since the toy-guns are massless and the balls have detached, the moment of inertia of the remaining platform is: I=12MR2I = \frac{1}{2} M R^2

  • Angular Momentum of the Platform: If the platform rotates with an angular speed ω\omega, its angular momentum is given by: Lplatform=Iω=(12MR2)ωL_{\text{platform}} = I \omega = \left(\frac{1}{2} M R^2\right) \omega

3. Applying Conservation of Angular Momentum

By conservation of angular momentum about the central axis: Li=LfL_i = L_f 0=Lballs−Lplatform0 = L_{\text{balls}} - L_{\text{platform}} Iω=2mvrI \omega = 2 m v r

Substituting I=12MR2I = \frac{1}{2} M R^2: (12MR2)ω=2mvr\left(\frac{1}{2} M R^2\right) \omega = 2 m v r ω=4mvrMR2\omega = \frac{4 m v r}{M R^2}

4. Calculation

Substitute the given values into the formula:

  • m=0.05 kgm = 0.05\text{ kg}
  • v=9 ms−1v = 9\text{ ms}^{-1}
  • r=0.25 mr = 0.25\text{ m}
  • M=0.45 kgM = 0.45\text{ kg}
  • R=0.5 mR = 0.5\text{ m}

ω=4×0.05×9×0.250.45×(0.5)2\omega = \frac{4 \times 0.05 \times 9 \times 0.25}{0.45 \times (0.5)^2} ω=0.450.45×0.25=10.25=4 rad s−1\omega = \frac{0.45}{0.45 \times 0.25} = \frac{1}{0.25} = 4\text{ rad s}^{-1}

Thus, the rotational speed of the platform after the balls leave is 4 rad s−14\text{ rad s}^{-1}.

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