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NEET PHYSICSRotational motionMedium

Question

A disc of radius 2 m2 \text{ m} and mass 100 kg100 \text{ kg} rolls on a horizontal floor. Its centre of mass has a speed of 20 cm/s20 \text{ cm/s}. How much work is needed to stop it?

A

1 J1 \text{ J}

B

3 J3 \text{ J}

C

30 J30 \text{ J}

D

2 J2 \text{ J}

Step-by-Step Solution

According to the work-energy theorem, the work required to stop the rolling disc is equal to its total initial kinetic energy. For a disc rolling without slipping, its total kinetic energy is the sum of its translational and rotational kinetic energies: Ktotal=Ktrans+Krot=12mv2+12Iω2K_{total} = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 The moment of inertia of a disc about its central axis is I=12mr2I = \frac{1}{2}mr^2. In pure rolling, the relation between linear speed and angular speed is v=rωv = r\omega, so ω=vr\omega = \frac{v}{r}. Substituting these values: Ktotal=12mv2+12(12mr2)(vr)2=12mv2+14mv2=34mv2K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2 Given: Mass, m=100 kgm = 100 \text{ kg} Velocity, v=20 cm/s=0.2 m/sv = 20 \text{ cm/s} = 0.2 \text{ m/s} Ktotal=34×100×(0.2)2=75×0.04=3 JK_{total} = \frac{3}{4} \times 100 \times (0.2)^2 = 75 \times 0.04 = 3 \text{ J} Therefore, the work needed to stop the disc is 3 J3 \text{ J}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Rotational motion. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRotational motionradiushorizontalcentreneeded

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