A particle is whirled in a vertical circle of radius R . If its velocity at the lowest point is just enough to complete the circle ( u = 5 g R ), then the magnitude of its total acceleration when th

Question

A particle is whirled in a vertical circle of radius R . If its velocity at the lowest point is just enough to complete the circle ( u = 5 g R ​ ), then the magnitude of its total acceleration when the string becomes horizontal is ( g is acceleration due to gravity).

TestHub · Accepted Answer

Let the velocity at the lowest point be $u = 5 g R$ .

When the string becomes horizontal, the particle has ascended a height $h = R$ .

Using conservation of mechanical energy between the lowest point and the horizontal position:

$\frac{1}{2} m u^{2} = \frac{1}{2} m v^{2} + m g R$

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

$v^{2} = 3 g R$

At the horizontal position, the particle has two components of acceleration:

Centripetal acceleration ( $a_{c}$ ) directed towards the center:

$a_{c} = \frac{v ^{2}}{R} = \frac{3 g R}{R} = 3 g$

Tangential acceleration ( $a_{t}$ ) due to gravity acting downwards:

$a_{t} = g$

Since $a_{c}$ and $a_{t}$ are perpendicular to each other, the magnitude of the total acceleration is:

$a = a_{c}^{2} + a_{t}^{2}$

$a = (3 g)^{2} + g^{2}$

$a = 9 g^{2} + g^{2} = g 10$

Answer: $g 10$

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