Question

A uniform metallic chain in a form of circular loop of mass m = 3 kg with a length $\ell$ = 1 m rotates at the rate of n = 5 revolutions per second. Find the tension T (in Newton) in the chain.

Answer 1

75

Answer 2

The tension in a rotating circular chain is found by considering a small segment of the chain. The centripetal force required for this segment is provided by the net inward component of the tension forces acting on its ends.

1. Calculate the radius of the loop ($R = \ell/(2\pi)$) and the angular velocity ($\omega = 2\pi n$).
2. Consider a small mass element $dm = (m/(2\pi))d\theta$.
3. The net inward force due to tension on this element is $T d\theta$.
4. Equate this force to the centripetal force required for the element: $T d\theta = dm \omega^2 R$.
5. Substitute $dm$ and solve for $T$: $T = \frac{m \omega^2 R}{2\pi}$.
6. Plug in the numerical values to get $T = 75$ N.