Kinematics

JEE Main level, single correct, medium

307 of 308

Question 307 of 308

A particle of mass mm is moving along a trajectory given by
x=x0+acosω1tx={x}_{0}+acos{\omega }_{1}t
y=y0+bsinω2ty={y}_{0}+bsin{\omega }_{2}t
The torque, acting on the particle about the origin, at t=0t=0 is:

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Answer: (A) +my0aω12k^+m{y}_{0}a{\omega }_{1}^{2}\hat{k}

x=x0+acos(ω1t)x={x}_{0}+acos\left({\omega }_{1}t\right)
y=y0+bsin(ω2t)y={y}_{0}+b sin \left({\omega }_{2}t\right)
ax=−aω12cos(ω1t){a}_{x}=-a{\omega }_{1}^{2}cos \left({\omega }_{1}t\right)
ay=−bω22sin(ω2t){a}_{y}=-b{\omega }_{2}^{2}sin \left({\omega }_{2}t\right)
At t=0t=0 ,
F⃗=ma⃗=(−aω12i^)\vec{F}=m\vec{a}=\left(-a {\omega }_{1}^{2} \hat{i}\right)
And
r⃗=(x0+a)i^+y0j^\vec{r}=\left({x}_{0}\text{+} a\right) \hat{i}+{y}_{0} \hat{j}
Torque, τ=r⃗×F⃗\tau =\vec{r}\times \vec{F}
τ=((x0+a)i^+y0j^)×(−aω12i^)\tau =\left(\left({x}_{0}\text{+} a\right) \hat{i}+{y}_{0} \hat{j}\right)\times \left(-a {\omega }_{1}^{2} \hat{i}\right)
=my0aω12k^=m{y}_{0} a{\omega }_{1}^{2} \hat{k}

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