A cylinder, released from the top of an inclined plane, rolls without sliding and reaches the bottom with speed vr. Another identical cylinder, released from the top of the same inclined plane, slides without rolling and reaches the bottom with speed vs. Then

Question

Infinity Learn · Accepted Answer

When the cylinder rolls without sliding, the acceleration down the plane is $$ar=23gsin$$⁡θ ∵$$a=gsin$$θ$$1+$$β and β$$=12When$$ the cylinder slides without rolling, the acceleration $$isas=gsin$$⁡θwhere θ is the inclination of the plane.If h is the height of the inclined plane, their speeds on reaching the bottom are given by $$vr=2arh$$ and $$vs=2ashSince$$ as>ar, it follows that vs>vr, which is choice (b).

A cylinder, released from the top of an inclined plane, rolls without sliding and reaches the bottom with speed v_r. Another identical cylinder, released from the top of the same inclined plane, slides without rolling and reaches the bottom with speed v_s. Then

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