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Rotational Dynamics and Rolling Motion - Class XI
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A solid sphere of mass 0.5 kg and diameter 1 m rolls without sliding with a constant velocity of 5 m/s, the ratio of the rotational K.E. to the total kinetic energy of the sphere is :
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A
$\cfrac{2}{7}$
💡 Explanation:
Total kinetic energy for a rolling body is the sum of translational and rotational kinetic energy, given by K.E.(total) = (1/2)mv^2 + (1/2)Iomega^2. For a solid sphere, the moment of inertia I = (2/5)mr^2 and v = r omega. Substituting this in gives K.E.(rotational) = (1/5)mv^2 and K.E.(total) = (7/10)mv^2, leading to a ratio of 2/7.