Question

A metal ring of mass m and radius R is placed on a smooth horizontal table and is set rotating about its own axis in such a way that each part of the ring moves with a speed v. Find the tension in the ring.

• A. N
• B. N
• C. N
• D. N

Answer 1

The correct option is D

N

Consider a small part ACB of the ring that subtends an angle
$\Delta$ $\theta$ at the centre as shown in figure. Let the tension in the ring be T.

The forces on this small part ACB are

(a) tension T by the part of the ring left to A,

(b) tension T by the part of the ring right to B,

(c) weight $\left(\Delta m\right)g$ and

(d) normal force N by the table.

The tension at A acts along the tangent at A and the tension at B acts along the tangent at B. As the

small part ACB moves in a circle of radius Rat a constant speed v, its acceleration is towards the centre

(along CO) and has a magnitude $\frac{\left(\Delta m\right)v^2}{R}$

Resolving the forces along the radius CO,
$Tcos(90-\frac{\Delta \theta }{2})+Tcos({90}^{\circ }-\frac{\Delta \theta }{2})=\left(\Delta m\right)\left(\frac{v^2}{R}\right)$
or, $2Tsin\frac{\Delta \theta }{2}=\left(\Delta m\right)\left(\frac{v^2}{R}\right)......\left(i\right)$

The length of the part ACB is R$\Delta \theta$. As the total mass of the ring is m, the mass of the part ACB will be
$\Delta m=\frac{m}{2\pi R}\left(R\Delta \theta \right)=\frac{m}{2\pi }\Delta \theta \therefore 2T\frac{\Delta \theta }{2}=\frac{m}{2\pi }\left(\Delta \theta \right)\frac{v^2}{R}\Rightarrow T=\frac{m{v}^2}{2\pi R}$