The coordinates of a particle of mass `'m'` as function of time are given by `x=x_(0)+a_(1) cos(omegat)` and `y=y_(0)+a_(2)sin(omega_(2)t)`. The torque on particle about origin at time `t=0` is :

Options

A. zero
B. `m(-x_(0)b + y_(0)a)omega_(1)^(2)k`
C. `+my_(0)a omega_(1)^(2)k`
D. `-m(x_(0)bomega_(2)^(2) - y_(0)a omega_(1)^(2))k`

Answer

C

Explanation

`vecr = xhati + yhatj, vectau = vecr xx vecF` (At t=0)
So, `a_(w)=(d^(2)x)/(dt^(2)) =-omega_(1)^(2)a cos omega_(1)^(2)a hati`, `a_(y) = (d^(2)Y)/(dt^(2)) = -omega_(2)^(2)b sin omega_(2)t`
`vecF_(at t=0) = mveca=-momega_(1)^(2) a hati, vecr_(at t=0) = (x_(0) + a)hati + y_(0)hatj` So, `vectau = vecr xx vecF = momega_(1)^(2) a y_(0)hatk`