System Of Particles And Rotational Motion Ques 2
2. If $\overrightarrow{\mathrm{F}}$ is the force acting on a particle having position vector $\overrightarrow{\mathbf{r}}$ and $\vec{\tau}$ be the torque of this force about the origin, then:
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(a) $\overrightarrow{\mathrm{r}} \cdot \vec{\tau}>0$ and $\overrightarrow{\mathrm{F}} \cdot \vec{\tau}<0$
(b) $\overrightarrow{\mathrm{r}} \cdot \vec{\tau}=0$ and $\overrightarrow{\mathrm{F}} \cdot \vec{\tau}=0$
(c) $\overrightarrow{\mathrm{r}} \cdot \vec{\tau}=0$ and $\overrightarrow{\mathrm{F}} \cdot \vec{\tau} \neq 0$
(d) $\overrightarrow{\mathrm{r}} \cdot \vec{\tau} \neq 0$ and $\overrightarrow{\mathrm{F}} \cdot \vec{\tau}=0$.
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Answer:
Correct Answer: 2.(b)
Solution: (b) $\quad \vec{\tau}=\vec{r} \times \vec{F} \Rightarrow \vec{r} \cdot \vec{\tau}=0 \quad $
$\vec{F} \cdot \vec{\tau}=0$
Since, $\vec{\tau}$ is perpendicular to the plane of $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{F}}$, hence the dot product of $\vec{\tau}$ with $\overrightarrow{\mathrm{r}}$ and $\overrightarrow{\mathrm{F}}$ is zero.
Radial component of force makes no contribution to the torque. Only transverse component of force contributes to the torque.
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