A simple pendulum of length L with a bob of mass m is deflected from its rest position by an angle θ and released (figure 8-E16). 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 $θ = 90 ° and x = L / 2$ , 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 $θ = 90 °$ .

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Solution

(i) (ii)

(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

$i . e . {(K . E .)}_{A} + {(P . E .)}_{A} = {(K . E .)}_{B} + {(P . E .)}_{B} \Rightarrow {(P . E .)}_{A} = {(P . E .)}_{B} because {(K . E .)}_{A} = {(K . E .)}_{B} = 0$

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 $= 90 ° and x = \frac{L}{2},$

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

Applying the law of conservation of emergy

$(\frac{1}{2}) m ν_{c}^{2} - 0 = m g (\frac{L}{2}) (1 - \cos α)$

[because, distance between A and C in the vertical direction is $\frac{L}{2} and$

$\frac{L}{2} = (1 - \cos α) \Rightarrow V_{c}^{2} = g L (1 - \cos α) . . . (i)$

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

$\frac{m ν_{c}^{2}}{L / 2} = m g \cos α . . . (i i)$

[because, T_c = 0]
From equations (i) and (ii),
$g L (1 - \cos α) = \frac{g L}{2} \cos α \Rightarrow 1 - \cos α = \frac{1}{2} \cos α \Rightarrow \frac{3}{2} \cos α = 1 \Rightarrow \cos α = (\frac{2}{3}) . . . (i i i)$
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 $α$ .

$BF = \frac{L}{2} + \frac{L}{2} \cos θ = \frac{L}{2} + \frac{L}{2} \times \frac{2}{3} = L (\frac{1}{2} + \frac{1}{3}) So, BF = (\frac{5 L}{6})$
(c) If the particle has to complete a vertical circle at the point C,

$\frac{m ν_{c}^{2}}{L - x} = m g . . . (i)$

Again, applying energy conservation principle between A and C

$(\frac{1}{2}) m ν_{c}^{2} - 0 = m g (OC) \Rightarrow (\frac{1}{2}) m ν_{c}^{2} = m g \{L - 2 (L - x)\} = m g (2 x - L) \Rightarrow ν_{c}^{2} = 2 g (2 x - L) . . . (i i)$

From equations (i) and (ii),

$g (L - x) = 2 g (2 x - L) \Rightarrow L - x = 4 x - 2 L \Rightarrow 5 x = 3 L ∴ \frac{x}{L} = \frac{3}{5} = 0.6 So, the minimum value of (\frac{x}{L}) shoule be 0.6 .$

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Similar questions

Q. A simple pendulum of length

L

having a bob of mass

m

is deflected from its rest position by an angle

θ

and released (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 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

θ = 90^{\circ}

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

θ = 90^{\circ}

A simple pendulum of length L having a bob of mass m is deflected from its rest position by an angle $θ$ and released (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. If the pendulum is released with $θ = 90^{\circ}$ and x = $\frac{L}{2}$ find the maximum height reached by the bob above its lowest position before the string becomes slack.