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The height of a solid cylinder is four times its radius, lt is kept vertically at time t = 0 on a belt which is moving in the horizontal direction with a velocity v = 2.45 t2 where v is in ms-1 and t is in second. If the cylinder does not slip, it will topple over at time t (in seconds) equal to…….[ Take g=9.8 ms-2 ]

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Detailed Solution

The cylinder will topple when the torque mgr equals the torque mah2 (see figure)or a=2grh=g2 (∵h=4r) …….(i)Now v=2.45t2∴ a=dvdt=ddt2.45t2=4.9t ……..(ii)Equating (i) and (ii), we get t=g2×4.9=9.89.8=1s

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The height of a solid cylinder is four times its radius, lt is kept vertically at time t = 0 on a belt which is moving in the horizontal direction with a velocity v = 2.45 t2 where v is in ms-1 and t is in second. If the cylinder does not slip, it will topple over at time t (in seconds) equal to…….[ Take g=9.8 ms-2 ]