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The coordinates of a particle of mass 'm...

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 :

A

zero

B

`m(-x_(0)b + y_(0)a)omega_(1)^(2) hatk`

C

`+my_(0)aomega_(1)^(2)hatk`

D

`-m(x_(0)b omega_(2)^(2) - y_(0)a omega_(1)^(2))hatk`

Text Solution

Verified by Experts

The correct Answer is:
C
`barr =xhati + y hatj, bartau = barr xx barF` (At t=0)
So, `a_(s) =(d^(2)x)/(dt^(2)) =-omega_(1)^(2)ahati`, `barr_("at t=0") =(x_(0) +a)hati + y_(0)hatj`
So, `bartau = barr xx barF = m omega_(1)^(2)ay_(0)hatk`
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