Physics - Circular Motion Question with Solution | TestHub

PhysicsCircular MotionDynamicsMedium2 minQB

PhysicsMediumnumerical

A ring of mass $2 π kg$ and of radius 0.25 m is making 300 rpm about an axis through its centre perpendicular to its plane. The tension (in newtons) developed in the ring is : (take $π^{2} = 10$ )

Answer:

250.00

Solution:

As $2 T sin \frac{θ}{2} = dm ω^{2} r$ (for small angle $sin \frac{θ}{2} \to \frac{θ}{2}$ )

but $dm = \frac{m}{ℓ} θ r$

As $ℓ = 2 π r ∴ T = m ω^{2} r /2 π$

Put $m = 2 π kg ω = 10 π radian / s$

and $r = 0.25 m$

Stream:JEESubject:PhysicsTopic:Circular MotionSubtopic:Dynamics

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...

Physics - Circular Motion Question with Solution | TestHub

Doubts & Discussion

JEE Main 2028 (Class 11)

JEE Advanced 2028 (Class 11)