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A cylinder of mass m and radius r rests on two supports of the same height. One support is stationary, while the other slides from under the cylinder at a velocity $v$ . Determine the force of normal pressure $N$ exerted by the cylinder on the stationary support at the moment when the distance between the points $A$ and $B$ of the supports is $A B = r \sqrt{2}$ , assuming that the supports were very close to each other at the initial instant. The friction between the cylinder and the supports should be neglected.

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$\frac{m g}{2} - \frac{m v^{2}}{2 r}$

$\frac{m g}{\sqrt{2}} - \frac{m v^{2}}{2 r}$

$\frac{m g}{\sqrt{2}} - \frac{m v^{2}}{2}$

$\frac{m g}{\sqrt{2}} - \frac{m v}{r}$

answer is A.

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Detailed Solution

The velocity of the point B in horizontal direction is $v$ .

The centre of the cylinder is half way between $A$ and $B$ , so its velocity will be $u = \frac{v}{2}$

The velocity perpendicular to line $A O$ ,

$u' \cos θ = u = \frac{v}{2}$

$∴$ $u' = \frac{v}{2 \cos θ},$

$\cos θ = \frac{\sqrt{2 r / 2}}{r} = \frac{1}{\sqrt{2}}$

Now $m g \cos θ - N = \frac{m u'^{2}}{r}$

$N = m g \cos θ - \frac{m u'^{2}}{r}$

$= m g \cos θ - \frac{m {(\frac{v}{2} \cos θ)}^{2}}{r}$

$= \frac{m g}{\sqrt{2}} - \frac{m v^{2}}{2 r}$

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