Question

A simple pendulum of length L with a bob of mass m is deflected from its rest position by an angle θ and released (following figure). The string hits a peg which is fixed at a distance x below the point of suspension and the bob starts going in a circle centred at the peg. (a) Assuming that initially the bob has a height less than the peg, show that the maximum height reached by the bob equals its  initial height. (b) If the pendulum is released with \[\theta = 90^\circ \text{ and x = L}/2\] , find the maximum height reached by the bob above its lowest position before the string becomes slack. (c) Find the minimum value of x/L for which the bob goes in a complete circle about the peg when the pendulum is released from \[\theta = 90^\circ \]

Answer 1

(a) When the bob has an initial height less than the distance of the peg from the suspension point and the bob is released from rest (Fig.(i)), 
let body travels from A to B then by the principle of conservation of energy (total energy should always be conserved)
Total energy at A = Total energy at B

\[\text{ i . e }. \left( \text{ K . E .} \right)_A + \left( P . E . \right)_A = \left( K . E . \right)_B + \left( P . E . \right)_B \]
\[ \Rightarrow \left( P . E . \right)_A = \left( P . E . \right)_B \]
\[\text{ because }\left( K . E . \right)_A = \left( K . E . \right)_B = 0\]

So, the maximum height reached by the bob is equal to the initial height of the bob.

(b) When the pendulum is released with θ \[= 90^\circ \text{ and  x }= \frac{L}{2},\]

Let the string become slack at point C, so the particle will start making a projectile motion.

Applying the law of conservation of emergy

\[\left( \frac{1}{2} \right) \text{ m}\nu_c^2 - 0 = \text{ mg }\left( \frac{L}{2} \right) \left( 1 - \cos \alpha \right)\]

[because, distance between A and C in the vertical direction is \[\frac{L}{2} \text{ and }\] \[\frac{L}{2} = \left( 1 - \cos \alpha \right)\]
\[ \Rightarrow V_c^2 = gL \left( 1 - \cos \alpha \right) . . . (i)\]

 

Again, from the free-body diagram (fig. (ii)),

\[\frac{m \nu_c^2}{L/2} = \text{ mg } \cos \alpha . . . (\text{ii})\]

[because, T_c = 0]
From equations (i) and (ii),

\[gL \left( 1 - \cos \alpha \right) = \frac{gL}{2} \cos \alpha\]
\[ \Rightarrow 1 - \cos \alpha = \frac{1}{2} \cos \alpha\]
\[ \Rightarrow \frac{3}{2} \cos \alpha = 1\]
\[ \Rightarrow \cos \alpha = \left( \frac{2}{3} \right) . . . (\text{iii})\]

To find highest position C₁ upto which the bob can go before the string becomes slack.(as we have found out the value of α   so now we want to find the distance of the highest point upto which the bob goes before the string becomes slack,using this value of α  .

\[BF = \frac{L}{2} + \frac{L}{2} \cos \theta\]
\[ = \frac{L}{2} + \frac{L}{2} \times \frac{2}{3}\]
\[ = L \left( \frac{1}{2} + \frac{1}{3} \right)\]
\[\text{ So, BF }= \left( \frac{5L}{6} \right)\]

(c) If the particle has to complete a vertical circle at the point C,

\[\frac{\text{m}\nu_c^2}{\text{L - x}} = \text{mg} . . . (\text{i})\]

Again, applying energy conservation principle between A and C

\[\left( \frac{1}{2} \right) \text{m}\nu_c^2 - 0 = \text{mg}\left( OC \right)\]

\[ \Rightarrow \left( \frac{1}{2} \right) \text{m}\nu_c^2 = \text{mg}\left\{ L - 2 \left( \text{L - x} \right) \right\}\]

\[ = mg \left( 2\text{x - L} \right)\]

\[ \Rightarrow \nu_c^2 = 2g \left( 2\text{x - L} \right) . . . (\text{ii})\]

From equations (i) and (ii),

\[g \left( L - \text{ x }\right) = 2g \left( 2\text{x - L} \right)\]
\[ \Rightarrow\text{ L - x = 4x - 2L}\]
\[ \Rightarrow 5x = 3L\]
\[ \therefore \frac{x}{L} = \frac{3}{5} = 0 . 6\]
\[\text{ So, the minimum value of } \left( \frac{x}{L} \right) \text{ shoule be }0 . 6 .\]