Question

A particle tied to the end of a string oscillates along a circular arc in a vertical plane. The other end of the string is fixed at the centre of the circle. If the string has a breaking strength of twice the weight of the particle; (x) Find the maximum distance that the particle can cover in one cycle of oscillation. The length of the string is 50 cm. (y) Find the tension in the extreme position (z) Find the net acceleration of the particle at extreme

(i) $\frac{2\pi }{3}m$ (ii)$\frac{mg}{4}$ (iii) $g\sqrt{\frac{2}{3}}$ (iv)$\frac{mg}{2}$ (v)$\frac{\pi }{3}m$

(vi) $g\sqrt{\frac{3}{2}}$

• A. x - i, y - ii, z - iii
• B. x - v, y - ii, z - iii
• C. x - v, y - iv, z - iii
• D. x - i, y - iv, z - vi

Answer 1

The correct option is D

x - i, y - iv, z - vi

As the maximum tension occurs at the lowest position, tension at the bottom can be at most 2mg (where m is the mass of particle).

Considering Forces on the particle at the bottom:

Where u : velocity at bottom

2mg - mg = $\frac{m{u}^2}{l}\Rightarrow u=\sqrt{lg}$

(x) Let $\theta$ = angular amplitude from bottom to the extreme, loss in KE = Gain in GPE

$\frac{1}{2}m{u}^2=mg(l-lcos\theta )\Rightarrow \frac{1}{2}mlg=mg(l-lcos\theta )$

$(1-cos\theta )=\frac{1}{2}\Rightarrow \theta ={60}^{\circ }$

Length of are covered in one cycle = $4\left(l\theta \right)=4\left(\frac{1}{2}\frac{\pi }{3}\right)=\frac{2\pi }{3}m$

(y) At extremes, the speed is zero and hence radial acceleration is zero. Balancing radial forces we get:$T_E=mgcos\theta =\frac{mg}{2}$

(z)

At extremes:

$a_r=0m/s^2$ and

$a_t=\frac{F_i}{m}=\frac{mgsin{60}^{\circ }}{m}=g\sqrt{\frac{3}{2}}$ $\Rightarrow$ Net acceleration = $g\sqrt{\frac{3}{2}}$