Question

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

• A. $\vec{r}\cdot \vec{\tau} > 0 $ and $ \vec{F}\cdot \vec{\tau} < 0$
• B. $\vec{r}\cdot \vec{\tau} = 0$ and $\vec{F}\cdot \vec{\tau} = 0$
• C. $\vec{r}\cdot \vec{\tau} = 0$ and $ \vec{F}\cdot \vec{\tau} \ne 0$
• D. $\vec{r}\cdot \vec{\tau} \ne 0 $ and $ \vec{F}\cdot \vec{\tau} = 0$

Answer 1

The Correct Option is B

 To solve this problem, we must understand the concepts of force, position vector, and torque in the context of physics. The question provides us with a force vector \(\vec{F}\), a position vector \(\vec{r}\), and a torque \(\vec{\tau}\).

**Torque and Dot Product:***

Torque \(\vec{\tau}\) is defined as the cross product of the position vector \(\vec{r}\) and the force vector \(\vec{F}\):

\[\vec{\tau} = \vec{r} \times \vec{F}\]

**Properties of Dot Product and Cross Product:***

1. The dot product of two vectors \(\vec{A}\) and \(\vec{B}\) is given by \(\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta\), where \(\theta\) is the angle between the vectors.
2. The dot product of a vector with itself is always zero when it is perpendicular to the vector it's being crossed with, due to the properties of the cross product resulting in a vector perpendicular to both.

**The Meaning of \(\vec{r} \cdot \vec{\tau} = 0$**\)

The position vector \(\vec{r}\) and the torque \(\vec{\tau}\) (which is perpendicular to \(\vec{r}\) and \(\vec{F}\)) are perpendicular, leading to:

\[\vec{r} \cdot \vec{\tau} = 0\]

**The Meaning of \(\vec{F} \cdot \vec{\tau} = 0$**\)

Similarly, the force vector \(\vec{F}\) is also perpendicular to the torque vector \(\vec{\tau}\), hence:

\[\vec{F} \cdot \vec{\tau} = 0\]

**Conclusion:***

The correct answer is that both \(\vec{r} \cdot \vec{\tau} = 0\) and \(\vec{F} \cdot \vec{\tau} = 0\) are true because the torque is perpendicular to both \(\vec{r}\) and \(\vec{F}\).

Thus, the correct answer is:
\(\vec{r} \cdot \vec{\tau} = 0\) and \(\vec{F} \cdot \vec{\tau} = 0\)