A block of mass m is placed at the lowest point of a smooth vertical track of radius R . In this position, the block is given a horizontal velocity u such that the block is just able to perform a comp

Question

A block of mass m is placed at the lowest point of a smooth vertical track of radius R . In this position, the block is given a horizontal velocity u such that the block is just able to perform a complete vertical circular motion. The acceleration of block, when its velocity is vertical is

TestHub · Accepted Answer

The initial velocity of the block to undergo a complete vertical circular motion is $u = \sqrt{5 g R}$

Let us assume that the speed of the block when it is moving in the vertical direction is $v$ , then using conservation of mechanical energy we get

$\frac{1}{2} m {(\sqrt{5 g R})}^{2} = \frac{1}{2} m v^{2} + m g R$

$\Rightarrow v^{2} = \sqrt{3 g R}$

The tangential and centripetal accelerations of the block are

$a_{t} = g (↓)$ and $a_{c} = 3 g (\leftarrow)$

$a_{net} = \sqrt{a_{c}^{2} + a_{t}^{2}} = g \sqrt{10}$

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