Question

A small ball attached with one end of an inextensible thread is moving in a vertical circle. Ratio of its maximum to minimum velocity is

2 : 1 .

Calculate tension in thread and acceleration of the ball at the moment when velocity of the ball is directed vertically downward

(g = 10 {m s}^{- 2}) .

Accepted Answer

Let the mass of the particle be $m$ and radius of the circular path be $r$ .

We know that, in vertical circular motion, the velocity is minimum when ball is at highest point $A$ and maximum when the ball is at the lowest point $C .$

Let velocity of ball at $A$ be $v_{0}$ . As the ratio given is $2 : 1$ , so the velocity at $C$ must be $2 v_{0} .$

From the conservation of mechanical energy from $C$ to $A$ , we get,

loss in KE $=$ gain in gravitational potential energy

$\Rightarrow \frac{1}{2} m {(2 v_{0})}^{2} - \frac{1}{2} m v_{0}^{2} = m g (2 r)$

$\Rightarrow v_{0}^{2} = \frac{4}{3} g r \dots (i)$

It is clear from the diagram that velocity of the ball is directed vertically downward when the ball is at $B .$ Let velocity at this point be $v .$

Applying conservation of mechanical energy from $A$ to $B$ , we get,

loss in KE $=$ gain in gravitational potential energy

$\Rightarrow \frac{1}{2} m v^{2} - \frac{1}{2} m v_{0}^{2} = m g r$

$\Rightarrow v^{2} = v_{0}^{2} + 2 g r = \frac{4}{3} g r + 2 g r$ [using $(i)$ ]

$\Rightarrow v^{2} = \frac{10}{3} g r \dots (ii)$

Now let us draw $FBD$ at point $B$ .

Along radius:

$T = \frac{m v^{2}}{r} = \frac{10}{3} m g$

Along tangent:

$m g = m a_{t}$

$\Rightarrow a_{t} = g$

Radial acceleration is given by,

$a_{r} = \frac{v^{2}}{r} = \frac{10}{3} g$

Net acceleration of ball at point $B$ is given by,

$a = \sqrt{a_{t}^{2} + a_{r}^{2}} = \frac{g \sqrt{109}}{3}$

Hence, tension in thread and acceleration of the ball at the moment when velocity of the ball is directed vertically downward are $\frac{10}{3} m g$ and $\frac{g \sqrt{109}}{3}$ , respectively.

B M Sharma Solutions for Chapter: Circular Motion, Exercise 49: ILLUSTRATION 9.46

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Circular Motion, Exercise 49: ILLUSTRATION 9.46

Questions from B M Sharma Solutions for Chapter: Circular Motion, Exercise 49: ILLUSTRATION 9.46 with Hints & Solutions

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