Animated Solution for Physics - Kinematics: At a particular instant of time, position vector $\vec{r}$, velocity vector $\vec{v}$ and angular position $\theta$ of a particle traversing a path AB are shown in the figure. Here $\varphi$ is the angle made by the velocity vector with the positive x-axis. Which of the following statements is/are correct?

When a particle traverses a general curvilinear path, tracking its motion requires a deep understanding of vectors and their rates of change. In this problem, we are given a snapshot of a particle's journey. We have its position vector $\vec{r}$, which tells us where it is relative to the origin, and its velocity vector $\vec{v}$, which tells us where it is going and how fast.

Crucially, these vectors are defined by their angles with the positive x-axis: $\theta$ for the position vector and $\varphi$ for the velocity vector. The interplay between these two angles unlocks the entire kinematic profile of the particle.

Radial and Transverse

To understand angular velocity, we must first dissect the velocity vector $\vec{v}$. Angular velocity, denoted by $\omega$ or $\frac{d\theta }{dt}$, is a measure of how fast the position vector is sweeping through an angle.

Imagine holding a string with a stone attached to it. If you pull the string straight towards you, the stone moves, but the angle of the string doesn't change. The angle only changes when the stone moves perpendicular to the string. Similarly, the angular velocity $\frac{d\theta }{dt}$ depends exclusively on the component of velocity that is perpendicular to the position vector $\vec{r}$.

From the geometry of the figure, the position vector is at an angle $\theta$ and the velocity vector is at an angle $\varphi$. The angle between them is simply $\varphi -\theta$. Therefore, we can resolve the velocity $v$ into a perpendicular component:

Now, applying the fundamental relationship between linear and angular variables ($v=r\omega$), we can isolate the angular velocity. Since it is driven entirely by the perpendicular velocity component, we have:

This elegant expression perfectly matches option (a), confirming it as a correct statement.

Moving on to acceleration, we must abandon the simplistic formulas of pure circular motion. In a general curve, acceleration $\vec{a}$ is the total rate of change of the velocity vector. Because velocity is a vector, it can change in two distinct ways: its magnitude (speed) can change, and its direction can change. This naturally splits acceleration into two orthogonal components: tangential (${\vec{a}}_t$) and normal (${\vec{a}}_n$).

The Speed Changer

The tangential component of acceleration is strictly responsible for changing the speed of the particle. It is defined simply as the time derivative of the speed:

Option (b) suggests that $a_t=r\frac{d^2\theta }{d{t}^2}$. This is a classic trap! This formula is only valid for rigid circular motion where the radius $r$ is a constant. For a general path, $r$ is changing, making this expression incorrect.

The Direction Changer

The normal (or centripetal) component of acceleration is responsible for changing the direction of the velocity vector. It always points perpendicular to the velocity, towards the center of curvature.

To find its magnitude, think about what defines the direction of the velocity vector. It is the angle $\varphi$. Just as the rotation of the position vector ($\frac{d\theta }{dt}$) creates a perpendicular velocity $v_{\perp }=r\frac{d\theta }{dt}$, the rotation of the velocity vector itself creates the normal acceleration. The rate at which the velocity vector is turning is $\frac{d\varphi }{dt}$. Therefore, the normal acceleration is the speed $v$ multiplied by this rate of rotation:

Looking at the remaining options, option (c) incorrectly proposes $v\frac{d\theta }{dt}$ for normal acceleration. This mixes up the rotation of the position vector with the rotation of the velocity vector. Option (d), however, correctly identifies $v\frac{d\varphi }{dt}$ as the modulus of the normal component of acceleration.

Thus, through careful vector resolution and an understanding of intrinsic coordinates, we conclude that the correct statements are indeed (a) and (d).