Question

A heavy particle is suspended by a 1⋅5 m long string. It is given a horizontal velocity of $\sqrt{57}\mathrm{m}/\mathrm{s}$. (a) Find the angle made by the string with the upward vertical when it becomes slack. (b) Find the speed of the particle at this instant. (c) Find the maximum height reached by the particle over the point of suspension. Take g = 10 m/s².

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{string},L=1.5\mathrm{m} \\ \mathrm{Initial}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{particle},u=\sqrt{57}\mathrm{m}/s \\ \left(\mathrm{a}\right)mg\cos \mathrm{\theta }=\frac{m{\nu }^{\mathit{2}}}{L} \\ {\mathrm{\nu }}^2=Lg\cos \mathrm{\theta }...\left(\mathrm{i}\right) \end{gathered}$$

Change in K.E. = Work done

$$\begin{gathered} \frac{1}{2}m{\nu }^2-\frac{1}{2}m{u}^2=-mgh \\ \Rightarrow {\nu }^2-57=-2\times 1.5g\left(1+\cos \mathrm{\theta }\right) \\ \Rightarrow {\nu }^2=57-3g\left(1+\cos \mathrm{\theta }\right)...\left(ii\right) \end{gathered}$$



Putting the value of $\nu$ from equation (i),

$$\begin{gathered} 15\cos \mathrm{\theta }=57-3g\left(1+\cos \mathrm{\theta }\right) \\ \Rightarrow 15\cos \mathrm{\theta }=57-30-30\cos \mathrm{\theta } \\ \Rightarrow 45\mathrm{\theta }=27 \\ \Rightarrow \cos \mathrm{\theta }=\frac{3}{5} \\ \Rightarrow \mathrm{\theta }={\cos }^{-1}\frac{3}{5}=53° \end{gathered}$$
(b) From equation (ii),

$$\begin{gathered} \nu =\sqrt{57-3g\left(1+\cos \mathrm{\theta }\right)} \\ =\sqrt{9}=3\mathrm{m}/\mathrm{s} \end{gathered}$$

(c) As the string becomes slack at point P, the particle will start executing a projectile motion.

$$\begin{gathered} h=\mathrm{OF}+\mathrm{F}C \\ =1.5\cos \mathrm{\theta }+\frac{u^2{\sin }^2\mathrm{\theta }}{2g} \\ =\left(1.5\right)\times \frac{3}{5}+\frac{9\times {\left(0.8\right)}^2}{2\times 10} \\ =1.2\mathrm{m} \end{gathered}$$