First slide

Rotational motion

Question

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 $g = 9.8 m s^{- 2}$ ]

Moderate

Solution

The cylinder will topple when the torque mgr equals the torque $ma \frac{h}{2}$ (see figure)
or $a = \frac{2 gr}{h} = \frac{g}{2} (∵ h = 4 r)$ …….(i)
Now $v = 2 .45 t^{2}$
$∴ a = \frac{dv}{dt} = \frac{d}{dt} (2 .45 t^{2}) = 4 .9 t$ ……..(ii)
Equating (i) and (ii), we get $t = \frac{g}{2 \times 4 .9} = \frac{9 .8}{9 .8} = 1 s$

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