JEE Main201910 Apr 2019Morning ShiftPhysicsMotion in Two DimensionsActual
A particle of mass m is moving along a trajectory given by x = x 0 + a c o s ω 1 t y = y 0 + b s i n ω 2 t The torque, acting on the particle about the origin, at t = 0 is:
Options
- A+ m y 0 a ω 1 2 k ^
- B- m x 0 b ω 2 2 - y 0 a ω 1 2 k ^
- CZero
- Dm - x 0 b + y 0 a ω 1 2 k ^
Correct answer
A. + m y 0 a ω 1 2 k ^
Step-by-step solution
x = x 0 + a c o s ω 1 t y = y 0 + b s i n ω 2 t a x = - a ω 1 2 c o s ω 1 t a y = - b ω 2 2 s i n ω 2 t At t = 0 , F → = m a → = - a ω 1 2 i ^ And r → = x 0 + a i ^ + y 0 j ^ Torque, τ = r → × F → τ = x 0 + a i ^ + y 0 j ^ × - a ω 1 2 i ^ = m y 0 a ω 1 2 k ^