First slide

Vertical circular motion

Question

A small glass marble of mass ‘m’ oscillates between the two edges, inside a hemispherical glass bowl of radius ‘r’.If ‘V’ is the speed of marble at the lowest position, the normal reaction at that position is

Moderate

A

$\large \frac {mv^2}{r}$
B

$\large \frac {2mv^2}{r}$
C

$\large \frac {3mv^2}{r}$
D

$\large \frac {3mv^2}{2r}$

Solution

Conserving energy between A and B, $\large \frac 12mv^2=mgr\Rightarrow \frac {mv^2}{r}=2mg$
At B, N - mg = centripetal force = mv²/r
$\large \therefore$ N = mg + 2mg = 3mg = 3mv²/2r

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