Question

If $\overrightarrow{\omega}$ be angular velocity and $\overrightarrow{r}$ be the position vector of a point on circumference of a rotating turntable, where center of table is origin, then

$m\overrightarrow{\omega} \times (\overrightarrow{\omega} \times \overrightarrow{r})$ points towards?

• A. Parallel to axis
• B. Pointing away from axis
• C. Pointing toward axis
• D. Insufficient data

Answer 1

Answer: (C)

Pointing toward axis

Answer 2

To determine the direction of the vector $m\overrightarrow{\omega} \times (\overrightarrow{\omega} \times \overrightarrow{r})$, we can use the vector triple product identity:

$\overrightarrow{A} \times (\overrightarrow{B} \times \overrightarrow{C}) = \overrightarrow{B}(\overrightarrow{A} \cdot \overrightarrow{C}) - \overrightarrow{C}(\overrightarrow{A} \cdot \overrightarrow{B})$

In our case, let $\overrightarrow{A} = \overrightarrow{\omega}$, $\overrightarrow{B} = \overrightarrow{\omega}$, and $\overrightarrow{C} = \overrightarrow{r}$. So, the expression inside the bracket becomes:

$\overrightarrow{\omega} \times (\overrightarrow{\omega} \times \overrightarrow{r}) = \overrightarrow{\omega}(\overrightarrow{\omega} \cdot \overrightarrow{r}) - \overrightarrow{r}(\overrightarrow{\omega} \cdot \overrightarrow{\omega})$

Now, let's analyze the terms:

1. $\overrightarrow{\omega} \cdot \overrightarrow{\omega}$: This is the dot product of the angular velocity vector with itself, which equals the square of its magnitude, $\omega^2$.

   $\overrightarrow{\omega} \cdot \overrightarrow{\omega} = |\overrightarrow{\omega}|^2 = \omega^2$.

2. $\overrightarrow{\omega} \cdot \overrightarrow{r}$: This is the dot product of the angular velocity vector and the position vector. The problem states that $\overrightarrow{\omega}$ is the angular velocity of a rotating turntable, and $\overrightarrow{r}$ is the position vector of a point on its circumference, with the center of the table as the origin. For a turntable rotating about an axis passing through its center, the angular velocity vector $\overrightarrow{\omega}$ is directed along the axis of rotation. The position vector $\overrightarrow{r}$ of a point on the circumference, measured from the center (origin), lies in the plane of the turntable and is therefore perpendicular to the axis of rotation. Since $\overrightarrow{\omega}$ is parallel to the axis and $\overrightarrow{r}$ is perpendicular to the axis, $\overrightarrow{\omega}$ and $\overrightarrow{r}$ are perpendicular to each other. Therefore, their dot product is zero: $\overrightarrow{\omega} \cdot \overrightarrow{r} = 0$.

Substitute these results back into the vector triple product expression:

$\overrightarrow{\omega} \times (\overrightarrow{\omega} \times \overrightarrow{r}) = \overrightarrow{\omega}(0) - \overrightarrow{r}(\omega^2)$ $\overrightarrow{\omega} \times (\overrightarrow{\omega} \times \overrightarrow{r}) = -\omega^2 \overrightarrow{r}$

Finally, multiply by $m$:

$m\overrightarrow{\omega} \times (\overrightarrow{\omega} \times \overrightarrow{r}) = m(-\omega^2 \overrightarrow{r})$ $m\overrightarrow{\omega} \times (\overrightarrow{\omega} \times \overrightarrow{r}) = -m\omega^2 \overrightarrow{r}$

Interpretation of the result:

The vector $\overrightarrow{r}$ points from the origin (center of the turntable) to the point on the circumference. The vector $-\overrightarrow{r}$ therefore points from the point on the circumference directly towards the origin. Since the origin is the center of the turntable and lies on the axis of rotation, the vector $-m\omega^2 \overrightarrow{r}$ points towards the axis of rotation.

This vector represents the centripetal force acting on the particle of mass $m$. Centripetal force is always directed towards the center of the circular path, which in this case is the center of the turntable (on the axis of rotation).