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A particle of mass m is moving along a trajectory given by

$x=x_{o}+a\: cos\omega _{1}t$

$y=y_{o}+a\: sin\omega _{2}t$

The torque , acting on the particle about the origin, at t = 0 is :

Option 1)
$m(-x_{o}b+y_{o}a) \omega_{1}^{2}\hat{k}$

Option 2)
$+my_{o}a \omega_{1}^{2}\hat{k}$

Option 3)
Zero

Option 4)
$-m(x_{o}b\omega_{2}^{2}-y_{o}a \omega_{1}^{2})\hat{k}^{}$

Answers (1)

S solutionqc

Torque -

$\underset{\tau }{\rightarrow}= \underset{r}{\rightarrow}\times \underset{F}{\rightarrow}$

- wherein

This can be calculated by using either $\tau=r_{1}F\; or\; \tau=r\cdot F_{1}$

$r_{1}$ = perpendicular distance from origin to the line of force.

$F_{1}$ = component of force perpendicular to line joining force.

$x=x_{o}+a\: cos\omega _{1}t$

$\frac{dx}{dt}=-(asin\omega_1t)\omega_1$

$\frac{d^{2}x}{dt^{2}}=-a\omega_1\times \omega_1 cos\omega_1t$

$y=y_{o}+a\: sin\omega _{2}t$

$\frac{dy}{dt}=b\omega_2cos\omega_2t$

$\frac{d^{2}y}{dt^{2}}=-b\omega_2^{2} sin\omega_2t$

So,

$\vec{r}=(x_{o}+a\: cos\omega _{1}t) \hat {i}+$ $(y_{o}+a\: sin\omega _{2}t)\hat{j}$

$\vec{v}=(-a\:sin\omega _{1}t) \hat {i}+$ $(b\omega_{2}\: cos\omega _{2}t)\hat{j}$

$\vec{a}=(-a\omega_{1}^{2}\:cos\omega _{1}t) \hat {i}-(b\omega_{2}^{2}sin\omega_{2}t)\hat{j}$

$F=m\vec{a}$

$\tau =\vec{r}\times \vec{F}$
at t=0,

$\vec{r}=(x_{o}+a)\hat{i}+y_{o}\hat{j}$

$\vec{a}=-a\omega_{1}^{2}\hat{i}$

$\tau =(may_{o}w_{1}^{2})\hat{k}$

Option 1)

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

Option 2)

$+my_{o}a \omega_{1}^{2}\hat{k}$

Option 3)

Zero

Option 4)

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

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