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 (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 $\mathrm{\theta }=90°\mathrm{and}x=\mathrm{L}/2$, 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 $\mathrm{\theta }=90°$.

Figure

Answer 1

(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

$$\begin{gathered} i.e.{\left(K.E.\right)}_{\mathrm{A}}+{\left(P.E.\right)}_{\mathrm{A}}={\left(K.E.\right)}_{\mathrm{B}}+{\left(P.E.\right)}_{\mathrm{B}} \\ \Rightarrow {\left(P.E.\right)}_{\mathrm{A}}={\left(P.E.\right)}_{\mathrm{B}} \\ \mathrm{because}{\left(K.E.\right)}_{\mathrm{A}}={\left(K.E.\right)}_{\mathrm{B}}=0 \end{gathered}$$

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 $=90°\mathrm{and}x=\frac{L}{2},$



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



Applying the law of conservation of emergy

$\left(\frac{1}{2}\right)m{\nu }_c^2-0=mg\left(\frac{\mathrm{L}}{2}\right)\left(1-\cos \alpha \right)$

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

$$\begin{gathered} \frac{\mathrm{L}}{2}=\left(1-\cos \alpha \right) \\ \Rightarrow {\mathrm{V}}_c^2=g\mathrm{L}\left(1-\cos \alpha \right)...\left(i\right) \end{gathered}$$

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

$\frac{m{\nu }_c^2}{\mathrm{L}/2}=mg\cos \alpha ...\left(ii\right)$

[because, T_c = 0]
From equations (i) and (ii),
$$\begin{gathered} g\mathrm{L}\left(1-\cos \alpha \right)=\frac{g\mathrm{L}}{2}\cos \alpha \\ \Rightarrow 1-\cos \alpha =\frac{1}{2}\cos \alpha \\ \Rightarrow \frac{3}{2}\cos \alpha =1 \\ \Rightarrow \cos \alpha =\left(\frac{2}{3}\right)...\left(iii\right) \end{gathered}$$
To find highest position C₁ upto which the bob can go before the string becomes slack.(as we have found out the value of $\alpha$ 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 $\alpha$.

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

$\frac{m{\nu }_c^2}{\mathrm{L}-x}=mg...\left(i\right)$

Again, applying energy conservation principle between A and C

$$\begin{gathered} \left(\frac{1}{2}\right)m{\nu }_c^2-0=mg\left(\mathrm{OC}\right) \\ \Rightarrow \left(\frac{1}{2}\right)m{\nu }_c^2=mg\left\{\mathrm{L}-2\left(\mathrm{L}-\mathrm{x}\right)\right\} \\ =mg\left(2x-\mathrm{L}\right) \\ \Rightarrow {\nu }_c^2=2g\left(2x-\mathrm{L}\right)...\left(ii\right) \end{gathered}$$

From equations (i) and (ii),

$$\begin{gathered} g\left(\mathrm{L}-x\right)=2g\left(2x-\mathrm{L}\right) \\ \Rightarrow \mathrm{L}-x=4x-2\mathrm{L} \\ \Rightarrow 5x=3\mathrm{L} \\ \therefore \frac{x}{\mathrm{L}}=\frac{3}{5}=0.6 \\ \mathrm{So},\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{of}\left(\frac{x}{\mathrm{L}}\right)\mathrm{shoule}\mathrm{be}0.6. \end{gathered}$$