NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice
A small ring of mass m is constrained to slide along a horizontal wire fixed between two rigid supports. The ring is connected to a particle of same mass by an ideal string & the whole system is released from rest as shown in the figure. If the coefficient of friction between ring A and wire is 3 5 , the ring will start sliding when the connecting string will make an angle θ with the vertical, then θ will be (particl
Options
- A30 o
- B45 o
- C60 o
- DNone of these
Correct answer
B. 45 o
Step-by-step solution
Using energy conservation, we get mgl cos θ = mv 2 2 ⇒ mv 2 l = 2 mgcosθ As the mass is moving in a circular motion, applying circular motion equation, T - mg cos θ = 2 mg cos θ T = 3 mg cos θ Tsinθ = μ mg + Tcosθ 3 mg cos θ sin θ = μ mg + 3 mg cos 2 θ θ = 4 5 ∘