Question

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

• A. Modulus of angular velocity is $\frac{d\theta}{dt}=\frac{v\sin(\phi-\theta)}{r}$.
• B. Modulus of tangential component of acceleration is $r\frac{d^2\theta}{dt^2}$.
• C. Modulus of normal component of acceleration is $v\frac{d\theta}{dt}$.
• D. Modulus of normal component of acceleration is $v\frac{d\phi}{dt}$.

Answer 1

Answer: (A), (D)

(a), (d)

Answer 2

Let's analyze each statement based on the principles of kinematics in curvilinear motion.

Given:

• Position vector $\vec{r}$ with magnitude $r = |\vec{r}|$ and angle $\theta$ with the positive x-axis.
• Velocity vector $\vec{v}$ with magnitude $v = |\vec{v}|$ and angle $\phi$ with the positive x-axis.

Analysis of Statement (a): Modulus of angular velocity is $\frac{d\theta}{dt}=\frac{v\sin(\phi-\theta)}{r}$.

The angular velocity of the position vector $\vec{r}$ is $\omega = \frac{d\theta}{dt}$. The velocity vector $\vec{v}$ can be resolved into two components relative to the position vector $\vec{r}$:

1. Radial component ($v_r$): Along the direction of $\vec{r}$.
2. Transverse component ($v_\theta$): Perpendicular to $\vec{r}$.

From the figure, the angle between the position vector $\vec{r}$ (at angle $\theta$ with x-axis) and the velocity vector $\vec{v}$ (at angle $\phi$ with x-axis) is $\alpha = \phi - \theta$.

The transverse component of velocity $v_\theta$ is given by: $v_\theta = v \sin(\phi - \theta)$

In polar coordinates, the transverse component of velocity is also expressed as: $v_\theta = r \frac{d\theta}{dt}$

Equating these two expressions for $v_\theta$: $r \frac{d\theta}{dt} = v \sin(\phi - \theta)$

Therefore, the modulus of angular velocity is: $\frac{d\theta}{dt} = \frac{v \sin(\phi - \theta)}{r}$ Statement (a) is correct.

Analysis of Statement (b): Modulus of tangential component of acceleration is $r\frac{d^2\theta}{dt^2}$.

The tangential component of acceleration ($a_t$) is defined as the rate of change of the magnitude of velocity (speed): $a_t = \frac{dv}{dt}$ In polar coordinates, the speed $v$ is given by $v = \sqrt{\left(\frac{dr}{dt}\right)^2 + \left(r\frac{d\theta}{dt}\right)^2}$. So, $a_t = \frac{d}{dt} \left( \sqrt{\left(\frac{dr}{dt}\right)^2 + \left(r\frac{d\theta}{dt}\right)^2} \right)$.

The expression $r\frac{d^2\theta}{dt^2}$ is only part of the transverse component of acceleration in polar coordinates ($a_\theta = r\frac{d^2\theta}{dt^2} + 2\frac{dr}{dt}\frac{d\theta}{dt}$). This statement would be true only for circular motion centered at the origin (where $r$ is constant, so $\frac{dr}{dt}=0$, and $v = r\frac{d\theta}{dt}$, making $a_t = \frac{d}{dt}(r\frac{d\theta}{dt}) = r\frac{d^2\theta}{dt^2}$). However, the problem describes a general path AB. For a general path, $r$ can change, and $a_t$ is not simply $r\frac{d^2\theta}{dt^2}$.

Statement (b) is incorrect.

Analysis of Statement (c): Modulus of normal component of acceleration is $v\frac{d\theta}{dt}$.

The normal component of acceleration ($a_n$) is related to the rate of change of the direction of the velocity vector. From statement (a), we know $\frac{d\theta}{dt} = \frac{v \sin(\phi - \theta)}{r}$. So, $v\frac{d\theta}{dt} = v \left( \frac{v \sin(\phi - \theta)}{r} \right) = \frac{v^2 \sin(\phi - \theta)}{r}$.

The general expression for the normal component of acceleration is $a_n = \frac{v^2}{\rho}$, where $\rho$ is the radius of curvature. Another common expression for $a_n$ is $v \frac{d\phi}{dt}$, where $\frac{d\phi}{dt}$ is the rate of change of the direction of the velocity vector. The angular velocity of the position vector $\frac{d\theta}{dt}$ and the angular velocity of the velocity vector $\frac{d\phi}{dt}$ are generally not equal for a general curvilinear motion. They are equal only for circular motion centered at the origin. For example, in a straight line motion not passing through the origin, $\phi$ is constant, so $\frac{d\phi}{dt} = 0$, leading to $a_n = 0$. However, $\theta$ would be changing, so $\frac{d\theta}{dt} \neq 0$, which would incorrectly imply a non-zero normal acceleration.

Statement (c) is incorrect.

Analysis of Statement (d): Modulus of normal component of acceleration is $v\frac{d\phi}{dt}$.

The acceleration vector $\vec{a}$ can be decomposed into tangential and normal components: $\vec{a} = a_t \hat{e}_t + a_n \hat{e}_n$ where $\hat{e}_t$ is the unit tangent vector (in the direction of $\vec{v}$) and $\hat{e}_n$ is the unit normal vector (perpendicular to $\hat{e}_t$, pointing towards the center of curvature).

The velocity vector is $\vec{v} = v \hat{e}_t$. Differentiating with respect to time: $\vec{a} = \frac{d\vec{v}}{dt} = \frac{d(v \hat{e}_t)}{dt} = \frac{dv}{dt} \hat{e}_t + v \frac{d\hat{e}_t}{dt}$ The unit tangent vector $\hat{e}_t$ changes its direction as the particle moves along the path. If $\phi$ is the angle $\hat{e}_t$ makes with the x-axis, then: $\hat{e}_t = \cos\phi \hat{i} + \sin\phi \hat{j}$ $\frac{d\hat{e}_t}{dt} = (-\sin\phi \hat{i} + \cos\phi \hat{j}) \frac{d\phi}{dt}$ The vector $(-\sin\phi \hat{i} + \cos\phi \hat{j})$ is the unit normal vector $\hat{e}_n$. So, $\frac{d\hat{e}_t}{dt} = \frac{d\phi}{dt} \hat{e}_n$.

Substituting this back into the acceleration equation: $\vec{a} = \frac{dv}{dt} \hat{e}_t + v \frac{d\phi}{dt} \hat{e}_n$ Comparing this with $\vec{a} = a_t \hat{e}_t + a_n \hat{e}_n$, we identify: $a_t = \frac{dv}{dt}$ (tangential acceleration) $a_n = v \frac{d\phi}{dt}$ (normal acceleration)

Statement (d) is correct.

Conclusion: Statements (a) and (d) are correct.