Question

The direction of angular velocity is along:
A. the tangent to the circular path
B. the inward radius
C. the outward radius
D. the axis of rotation

Answer 1

Hint: The angular velocity of a body only present during the circular motion of a body. For a body with tangential velocity $\left( {{v}_{\tan }} \right)$ and radius of the circular motion (r), the magnitude of the angular velocity $\left( \omega \right)$ is given by: $\omega =\dfrac{v}{r}$, also given by the rate of change of angular displacement $\left( \theta \right)$. That is: $d\omega =\dfrac{d\theta }{dt}$. The direction of the angular velocity is given by the right hand thumb rule.

Complete step by step answer:
Let’s make a diagram of a body executing a circular motion.

Since, the motion is circular in nature, the circular path followed by the body will have a centre or axis of the path followed. The distance from the axis of the path till the body will be the radius of the circular path, given by (r). The vector form will be: $\bar{r}$.
Now a tangent is drawn to the circular path. The velocity of the body along the direction of the tangent is known as the tangential velocity of the body, given by: \[{{\bar{v}}_{\tan }}\]. This quantity is also a vector quantity.

Initially, the body was at position (1) and then, later on it reached a position (2). Since the particle is executing circular motion, the distance from the axis will still remain (r) and the magnitude of the tangential velocity will also remain constant. However, the direction of the tangential velocity will change. The angular displacement caused is given by $\left( \theta \right)$.

As per circular motion, the angular velocity term is given by the rate of change of the angular displacement. That is: $\bar{\omega }=\dfrac{d\bar{\theta }}{dt}$.

The particle is travelling in anticlockwise direction, when travelling from position (1) to position (2). Hence, using the right hand thumb rule, if we consider curling our fingers along the anticlockwise direction then, the direction of thumb will give us the direction of angular velocity.
Hence, in this case, the direction of angular velocity is outside the plane of the direction of motion given by the axis of rotation of the body.
Hence, the correct answer is option D.

Note:
The relation between the angular velocity\[\left( {\bar{\omega }} \right)\] of the body and the linear tangential velocity$\left( {{{\bar{v}}}_{\tan }} \right)$ of the body will be: ${{\bar{v}}_{\tan }}=\bar{\omega }\times \bar{r}$. This can be found from, the basic formula of circular motion, given by: $\bar{\omega }=\dfrac{d\bar{\theta }}{dt}$, which is; the angular velocity of a body is the rate of change of angular displacement. The angular displacement of the body is given by: $\bar{\theta }\approx \dfrac{{\bar{s}}}{{\bar{r}}}$, where (s) is the linear displacement of the body. Therefore, $\bar{\omega }=\dfrac{d\bar{\theta }}{dt}\Rightarrow \bar{\omega }=\dfrac{1}{{\bar{r}}}\dfrac{d\bar{s}}{dt}\Rightarrow \bar{\omega }=\dfrac{{\bar{v}}}{{\bar{r}}}$. Here, linear velocity: $\bar{v}={{\bar{v}}_{\tan }}$.