A particle of mass m is moving along a trajectory given by x = x₀ + a cos ω ₁t y = y₀ + b sin ω ₂t The torque, acting on the particle about the origin, at t = 0 is:

(C) - m(x₀bω ₂² - y₀aω ₁²) k

(D) m (–x₀b + y₀a) ω ₁² k

Rotational Motion: JEE Main 2019 Physics question

JEE Main 2019 · 10 Apr, Morning shift · Q8108

JEE Main 201910 Apr, Morning shiftPhysicsEasyRotational Motion140× since 2002

A particle of mass m is moving along a trajectory given by x = x₀ + a cos

\omega

₁t y = y₀ + b sin

\omega

₂t The torque, acting on the particle about the origin, at t = 0 is :

JEE Main 2019 · 10 Apr, Morning shift · Q8108

JEE Main 201910 Apr, Morning shiftPhysicsEasyRotational Motion140× since 2002

A particle of mass m is moving along a trajectory given by x = x₀ + a cos

\omega

₁t y = y₀ + b sin

\omega

₂t The torque, acting on the particle about the origin, at t = 0 is :