Physics - Rotation Question with Solution | TestHub

PhysicsRotationTranslational + RotationalMedium2 minQB

PhysicsMediumnumerical range

Passage / Comprehension

A small sphere of mass $1 k g$ is rolling without slipping on a rough stationary base with linear speed $v = \frac{200}{7} m / s$ . It leaves the inclined plane at point C.

Find ratio of rotational and translational kinetic energy of the sphere when it strikes the ground after leaving from point C :

Answer:

0.16

Solution:

only v change & $ω$ remains same $K_{Rot} = \frac{1}{2} \frac{2}{5} mR^{2} ω^{2} = \frac{1}{5} mv_{c}^{2} = \frac{1}{5} m \frac{100}{7}$ $mgl = \frac{1}{2} mv_{D}^{2} + \frac{1}{2} \frac{2}{5} mR^{2} ω^{2} - \frac{1}{2} mv_{c}^{2} - \frac{1}{2} \frac{2}{5} mR^{2} ω^{2}$

∴ 20 = v_{D}^{2} - \frac{100}{7}

$∴ K_{T} = \frac{1}{2} m \frac{240}{7} = \frac{120}{7} m$

∴ ratio $= \frac{1}{6}$

Stream:JEESubject:PhysicsTopic:RotationSubtopic:Translational + Rotational

⏱ 2mℹ️ Source: QB

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