Question

One end of an inextensible string of length 3m is fastened to a fixed point O, 2m above horizontal ground. A small particle is attached to the other end of the string. The particle describes a horizontal circle 1m below O. Find in terms of g, the tension in the string and the angular velocity of the particle.

Answer 1

Explanations & Calculations

• The described situation can be illustrated by the following figure.

• In such a situation what happens is the thread provides the needed forces to describe the motion in a circle.
• Because for an object to move in a circle, a force called 'centripetal force' is needed & if an object describes a circular motion then there should exist some means of providing the needed centripetal force.
• Consider the thread makes an angle $\small \theta$ with the vertical.

• Applying F =ma both vertically & horizontally, those needed parameters can be found.

• Vertically,

$\qquad\qquad \begin{aligned} \small \uparrow F &= \small ma\\ \small T\cos\theta-mg&= \small 0\\ \small T\cos\theta&= \small mg\cdots(1) \end{aligned}$

• Horizontally,

$\qquad\qquad \begin{aligned} \small \leftarrow F&= \small ma \\ \small T\sin\theta&=\small mr\omega^2\cdots(2) \end{aligned}$

a)

• By (1),

$\qquad\qquad \begin{aligned} \small T&= \small \frac{mg}{\cos\theta}\\ &= \small \frac{mg}{\frac{1}{3}}\\ &= \small \bold{3mg} \end{aligned}$

• Value of m should be known.

b)

• By (2),

$\qquad\qquad \begin{aligned} \small (3mg)\sin\theta&= \small m(3\sin\theta)\omega^2\\ \small \omega^2&= \small g\\ \small \omega&= \small \bold{\sqrt{g}} \end{aligned}$

• Note that $\small r$ is not the length of the thread but the projection.