Question

A simple pendulum of length $L$ having a bob of mass $m$ is deflected from its rest position by an angle $θ$ and released. 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 centered 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 $θ = 90 °$

Accepted Answer

Given-
Length of the string = $m$ ,
Mass of the bob = $m$ ,
Initial angle made by the string with downward vertical = $θ$
(a)
After hitting the peg, the bob will continue to move upward until its velocity become zero. As mechanical energy is conserved, and initial and final kinetic energies are zero the initial and final potential energies must also be the same. Therefore, we can see that the maximum height reached by the Bob after hitting the peg will be the same as the initial height from which the Bob was released.
(b)
Given-
$θ = 90^{0}, x = L / 2$
Let the angle made by the sting with horizontal be $α$ and the velocity be $v$ when the string slacks, as shown in the figure.

At the instant, the string slacks, the tension in the string will be zero.
Applying equation for centripetal force-
$m g \cos (90 - α) + T = \frac{m v^{2}}{L / 2} m v^{2} = \frac{m g L \sin α}{2} . . . (1)$

Applying conservation of mechanical energy between the lowermost position and the position where the strings slacks-
$ΔU + ΔK \cdot E = 0 - m g (\frac{L}{2} - \frac{L}{2} \sin α) + \frac{1}{2} m v^{2} - 0 = 0 . . . (2)$
From equation $(1)$ and equation $(2)$ -
$- \frac{m g L}{2} + \frac{m g L}{2} \sin α + \frac{m g L \sin α}{4} = 0 \Rightarrow \sin α = \frac{2}{3}$
Height of the Bob above the lowest position when the string becomes slack-
$H = \frac{L}{2} + \frac{L}{2} \sin α = \frac{L}{2} + \frac{L}{2} \cdot (\frac{2}{3}) = \frac{5 L}{6}$

(c)
After hitting the peg, The Bob moves in a circular part of radius $L - x$ . A particle connected to a string of length $R$ needs minimum velocity of $\sqrt{5 g R}$ At the lowermost position to complete its vertical motion.in this case as the value of $x$ decreases length of the string $L - x$ increases. So the requirement of minimum velocity also increases.
For the minimum value of $x$ The velocity acquired by The bob (let us assume $v$ ) must be equal to the minimum velocity required.

Applying conservation of mechanical energy between position $A$ and lowermost position-
$- m g L + \frac{1}{2} m v^{2} - 0 = 0 v = \sqrt{2 g L} . . . (3)$
Minimum velocity required-
$v_{m i n} = \sqrt{5 g (L - x)} . . . (4)$
Equating equation $(3)$ and $(4)$ -
$x = 0.6 L$

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