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Physics - Rotation Question with Solution | TestHub

PhysicsRotationRotational KinematicsMedium2 minPYQ_2007

PhysicsMediumsingle choice

If a solid sphere of mass $1 kg$ and radius $0.1 m$ rolls without slipping at a uniform velocity of $1 m / s$ along a straight line on a horizontal floor, the kinetic energy is

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Solution:

$K . E of rolling = (\frac{1}{2} m v^{2} + \frac{1}{2} I \cdot \frac{v ^{2}}{R ^{2}}) \Rightarrow K \cdot E = \frac{1}{2} m v^{2} + \frac{1}{2} \cdot \frac{2}{5} \cdot \frac{m R ^{2} \cdot v ^{2}}{R ^{2}} K . E = \frac{1}{2} m v^{2} \times \frac{7}{5} m = 1 kg, v = 1 m / s K \cdot E = \frac{1}{2} \times 1 \times 1 \times \frac{7}{5} = \frac{7}{10} J$

Stream:AIIMSSubject:PhysicsTopic:RotationSubtopic:Rotational Kinematics

⏱ 2mℹ️ Source: PYQ_2007

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