The linear speed of a uniform spherical shell | vedclass

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Question

(A) For an object rolling down an inclined plane without slipping,the conservation of mechanical energy states that the potential energy at the top eq

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$\sqrt{\frac{6gh}{5}}$

Vedclass · Answer

(A) For an object rolling down an inclined plane without slipping,the conservation of mechanical energy states that the potential energy at the top equals the sum of translational and rotational kinetic energy at the bottom.$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a uniform spherical shell,the moment of inertia about its center is $I = \frac{2}{3}mr^2$.Since the object is rolling without slipping,$\omega = \frac{v}{r}$.Substituting these into the energy equation:$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mr^2)(\frac{v}{r})^2$$mgh = \frac{1}{2}mv^2 + \frac{1}{3}mv^2$$mgh = (\frac{1}{2} + \frac{1}{3})mv^2 = \frac{5}{6}mv^2$$gh = \frac{5}{6}v^2$$v^2 = \frac{6gh}{5}$$v = \sqrt{\frac{6gh}{5}}$

The linear speed of a uniform spherical shell after rolling down an inclined plane of vertical height $h$ from rest is:

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