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)

 

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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