A particle of mass m is moving along a trajectory given by x = x0 + a cos ω1 t y = y0 + b sin ω2 t The torque, acting on the particle about the origin, at t = 0 is :

Question

A particle of mass m is moving along a trajectory given by x = x0 + a cos ω1 t y = y0 + b sin ω2 t The torque, acting on the particle about the origin, at t = 0 is : (A) m (- x0 b + y0 a) ω12 hatk (B)  + my0 a ω21 hatk (C)  - m (x0 b ω22 - y0 a ω12 ) hatk (D) Zero

Tardigrade Admin · Accepted Answer

F =-m (aω21 cosω1t hati + bω22 sin ω2 t hatj) vecr = (x0 + a cos ω1 t) hati + (y0 +b sinω2 t) hatj vecT = vecr × vecF = - m(x0 +a cosω1 t) b ω22 sinω2 t hatk + m (y0 + b sinω2 t) aω21 cosω1 t hatk = ma ω12 y0 hatk

Q. A particle of mass $m$ is moving along a trajectory given by $x = x_{0} + a cos ω_{1} t$
$y = y_{0} + b sin ω_{2} t$
The torque, acting on the particle about the origin, at $t = 0$ is :

$m (- x_{0} b + y_{0} a) ω_{1}^{2} \hat{k}$

$+ m y_{0} a ω_{1}^{2} \hat{k}$

$- m (x_{0} b ω_{2}^{2} - y_{0} a ω_{1}^{2}) \hat{k}$

Zero

Q. A particle of mass m is moving along a trajectory given by x=x0​+acosω1​t y=y0​+bsinω2​t The torque, acting on the particle about the origin, at t=0 is :

m(−x0​b+y0​a)ω12​k^

+my0​aω12​k^

−m(x0​bω22​−y0​aω12​)k^

Zero

F=−m(aω12​cosω1​ti^+bω22​sinω2​tj^​) r=(x0​+acosω1​t)i^+(y0​+bsinω2​t)j^​ T=r×F=−m(x0​+acosω1​t)bω22​sinω2​tk^+m(y0​+bsinω2​t)aω12​cosω1​tk^ =maω12​y0​k^

Q. A particle of mass $m$ is moving along a trajectory given by $x = x_{0} + a cos ω_{1} t$
$y = y_{0} + b sin ω_{2} t$
The torque, acting on the particle about the origin, at $t = 0$ is :

$m (- x_{0} b + y_{0} a) ω_{1}^{2} \hat{k}$

$+ m y_{0} a ω_{1}^{2} \hat{k}$

$- m (x_{0} b ω_{2}^{2} - y_{0} a ω_{1}^{2}) \hat{k}$