Question

A particle of mass m is moving along a trajectory given by
x = x₀ + a cos ω₁t
y = y₀ + b sin ω₂t
The torque, acing on the particle about the origin, at t = 0 is:
(A) my₀aω₁² kˆ
(B) m(-x₀b + y₀a)ω₁² kˆ
(C) -m(-x₀bω₂² - y₀aω₁²)kˆ
(D) Zero

Answer 1

Answer:

D ) ZERO

Explanation:

the sin and cos

is = o

x.y

is also 0