Question

The bob of a pendulum at rest is given a sharp hit to impart a horizontal velocity

\sqrt{10 g l},

where

l

is the length of the pendulum. Find the tension in the string when
(a) the string is horizontal,
(b) the bob is at its highest point,
(c) the string makes an angle of

60 °

with the upward vertical.

Accepted Answer

(a) Given speed at the bottom point $A$ is $v_{1} = \sqrt{10 g l}$ at $A$ . Let the speed at point $B$ (when the string is horizontal) be $v_{2}$ and tension in the string be $T_{B}$ .
Using conservation of mechanical energy between $A$ and $B$ , we get,

Loss in kinetic energy $=$ gain in gravitational potential energy

$\Rightarrow \frac{1}{2} m v_{1}^{2} - \frac{1}{2} m v_{2}^{2} = m g l$
$\Rightarrow \frac{1}{2} m (10 g l) - \frac{1}{2} m v_{2}^{2} = m g l \Rightarrow v_{2}^{2} = 8 g l$

Now, from Newton's second law in radial direction, we get,

$T_{B} = \frac{m v_{2}^{2}}{R} = \frac{m (8 g l)}{l} = 8 m g$

(b) Let the velocity at the highest point $C$ be $v_{3}$ and tension in the string be $T_{C}$ .

Using conservation of mechanical energy between $A$ and $C$ , we get

Loss in kinetic energy $=$ gain in gravitational potential energy

$\frac{1}{2} m v_{1}^{2} - \frac{1}{2} m v_{3}^{2} = m g (2 l)$

$\Rightarrow \frac{1}{2} m (10 g l) = \frac{1}{2} m v_{3}^{2} + 2 m g l$
$\Rightarrow v_{3}^{2} = 6 g l$

Now, from Newton's second law in radial direction, we get,

$T_{C} + m g = \frac{m v_{3}^{2}}{l}$

$\Rightarrow T_{C} = \frac{m v_{3}^{2}}{l} - m g = 5 m g$

(c) Let the velocity at point $D$ where the string makes an angle of $60 °$ with the upward vertical be $v_{4} .$

Now using conservation of mechanical energy between $A$ and $D,$ we get,

Loss in kinetic energy $=$ gain in gravitational potential energy

$\Rightarrow (\frac{1}{2} m v_{1}^{2} - \frac{1}{2} m v_{4}^{2}) = [m g l (1 + \cos 60 °)]$

$\Rightarrow v_{4}^{2} = 7 g l$

Now, from Newton's second law in radial direction, we get,

$T_{D} + m g \cos 60 ° = \frac{m v^{2}}{l}$

$\Rightarrow T_{D} = \frac{m v^{2}}{l} - m g \cos 60 °$

$\Rightarrow T_{D} = m \frac{(7 g l)}{l} - 0.5 m g$
$\Rightarrow T_{D} = 7 m g - 0.5 m g = 6.5 m g$

B M Sharma Solutions for Chapter: Circular Motion, Exercise 47: ILLUSTRATION 9.44

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Circular Motion, Exercise 47: ILLUSTRATION 9.44

Questions from B M Sharma Solutions for Chapter: Circular Motion, Exercise 47: ILLUSTRATION 9.44 with Hints & Solutions

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