A solid sphere is rolling on a frictionless s | vedclass

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Question

(B) Since the surface is frictionless,there is no torque to change the angular velocity of the sphere. The rotational kinetic energy remains constant

Vedclass · Accepted Answer

$ \ge \sqrt {2gh} $

Vedclass · Answer

(B) Since the surface is frictionless,there is no torque to change the angular velocity of the sphere. The rotational kinetic energy remains constant throughout the motion.Applying the law of conservation of mechanical energy:Total initial energy = Total final energy at height $h$$\frac{1}{2}I{\omega ^2} + \frac{1}{2}m{v^2} = \frac{1}{2}I{\omega ^2} + mgh$Since the sphere must reach at least height $h$,the initial translational kinetic energy must be sufficient to overcome the potential energy gain:$\frac{1}{2}m{v^2} \ge mgh$$v^2 \ge 2gh$$v \ge \sqrt {2gh}$

$A$ solid sphere is rolling on a frictionless surface with a linear velocity $v \; m/s$,as shown in the figure. If the sphere climbs up to a height $h$,then the value of $v$ will be

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