Question

A metal ring of mass 5 g and radius 1 cm can rotate in horizontal plane about tangent

normal to plane of ring .The ring is initially at rest. Due to constant torque, its angular

velocity becomes 10 rad/sec in 10 seconds. The value of torque is​

Answer 1

torque of the ring is 10^-5 Nm.

It has given that a metal ring of mass , m = 5g and radius , r = 1cm can rotate in horizontal plane about tangent normal to plane of ring. the ring is initially at rest. due to constant torque, its angular velocity, ω = 10 rad/s in t = 10sec.

we have to find the value of torque.

moment of inertia of the metal ring about an axis passing through tangent normal to its plane, I = 2mr²

= 2 × 5 × 10¯³ kg × (1 × 10¯² m)²

= 10¯⁶ kgm²

using formula, ω = ω0 + αt

⇒10 = 0 + 10 × α

⇒α = 1 rad/s²

torque = Iα

= 10^-6 × 10 = 10^-5 Nm.

therefore torque of the ring is 10^-5 Nm.