KINEMATICS O17

1. Analyzing Angular Velocity

The angular velocity $\omega = \frac{d\theta}{dt}$ represents the rate at which the position vector $\vec{r}$ rotates.

Consider the particle at a specific point on its path (Point P). The velocity vector $\vec{v}$ makes an angle $\phi$ with the positive x-axis, and the position vector $\vec{r}$ makes an angle $\theta$ with the positive x-axis. The angle between the directions of $\vec{v}$ and $\vec{r}$ is $(\phi – \theta)$.

The component of the velocity $\vec{v}$ perpendicular to the position vector $\vec{r}$ (transverse velocity, $v_{\perp}$) is what causes the change in angle $\theta$. From the diagram, this component is $v \sin(\phi – \theta)$.

Using the relation $v_{\perp} = r\omega$, we get:

$$ \omega = \frac{d\theta}{dt} = \frac{v_{\perp}}{r} = \frac{v \sin(\phi – \theta)}{r} $$

Statement (a) is Correct.

2. Analyzing Acceleration Components

Tangential Acceleration ($a_t$):

The tangential acceleration represents the rate of change of the speed (magnitude of velocity), so $a_t = \frac{dv}{dt}$. Statement (b) gives the formula $r \frac{d^2\theta}{dt^2}$, which corresponds to the tangential acceleration ($r\alpha$) only in circular motion where the radius $r$ is constant. In general curvilinear motion, this expression is incorrect.

Statement (b) is Incorrect.

Normal Acceleration ($a_n$):

The normal acceleration is directed towards the center of curvature and is responsible for changing the direction of the velocity vector. Its magnitude is given by $a_n = \frac{v^2}{R_c}$, where $R_c$ is the radius of curvature.

We can relate this to the rate of change of the angle $\phi$ (the angle of the velocity vector). Using the chain rule:

$$ a_n = \frac{v^2}{R_c} = v^2 \frac{d\phi}{ds} = v^2 \left( \frac{d\phi}{dt} \cdot \frac{dt}{ds} \right) $$

Since speed $v = \frac{ds}{dt}$, we have $\frac{dt}{ds} = \frac{1}{v}$. Substituting this back:

$$ a_n = v^2 \left( \frac{d\phi}{dt} \cdot \frac{1}{v} \right) = v \frac{d\phi}{dt} $$

This confirms that the modulus of the normal component of acceleration is $v \frac{d\phi}{dt}$. Statement (c) incorrectly uses $\frac{d\theta}{dt}$.

Statement (d) is Correct.

Correct Options: (a) and (d)