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Stuck here, help me understand: - Motion of System Of Particles and Rigid Body - NEET

If \vec{F} is the force acting on a particle having position vector \vec{r} and \vec{t} be the torque of this force about the origin, then:

  • Option 1)

    \vec{r}.\vec{t}>0\: and \: \vec{F}.\: \vec{t}<0

  • Option 2)

    \vec{r}.\vec{t}=0\: and\: \vec{F}.\vec{t}=0

  • Option 3)

    \vec{r}.\vec{t}=0\: and\: \vec{F}.\vec{t}\neq 0

  • Option 4)

    \vec{r}.\vec{t}\neq 0 and \vec{F}.\vec{t}=0

 
Answers (1)
113 Views

As discussed in

Torque -

underset{	au }{
ightarrow}= underset{r}{
ightarrow}	imes underset{F}{
ightarrow}   

 

- wherein

This can be calculated by using either  	au=r_{1}F; or; 	au=rcdot F_{1}

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

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

 

 

 

 \bar \tau= \bar r \times \bar f\:\:\:\:\Rightarrow \bar r.\bar \tau=0\:\:\:\:\bar F. \bar \tau=0

Since, \tau is perpendicular to the plane of \bar r and \bar F

Hence the dot product of \bar \tau and \bar r and \bar F is zero.


Option 1)

\vec{r}.\vec{t}>0\: and \: \vec{F}.\: \vec{t}<0

This option is incorrect.

Option 2)

\vec{r}.\vec{t}=0\: and\: \vec{F}.\vec{t}=0

This option is correct.

Option 3)

\vec{r}.\vec{t}=0\: and\: \vec{F}.\vec{t}\neq 0

This option is incorrect.

Option 4)

\vec{r}.\vec{t}\neq 0 and \vec{F}.\vec{t}=0

This option is incorrect.

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