First slide

Rolling motion

Question

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

Moderate

A

$v_{r} > v_{s}$
B

$v_{r} < v_{s}$
C

$v_{r} = v_{s}$
D

$v_{r} = v_{s} = 0$

Solution

When the cylinder rolls without sliding, the acceleration down the plane is
$a_{r} = \frac{2}{3} gsin θ (∵ a = \frac{gsinθ}{1 + β} and β = \frac{1}{2})$
When the cylinder slides without rolling, the acceleration is
$a_{s} = gsin θ$
where $θ$ is the inclination of the plane.
If h is the height of the inclined plane, their speeds on reaching the bottom are given by
$v_{r} = \sqrt{2 a_{r} h} and v_{s} = \sqrt{2 a_{s} h}$
Since $a_{s} > a_{r}$ , it follows that $v_{s} > v_{r}$ , which is choice (b).

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