Question

Each situation in column I gives graph of a particle moving in circular path. The variables ω, θ and t represent angular speed (at any time t), angular displacement (in time t) and time respectively. Column II gives certain resulting interpretation. Match the graphs in column I with statements in column II.

Column-I	Column-II
(A) ω - θ graph	(p) Angular acceleration of particle is uniform
(B) ω² - θ graph	(q) Angular acceleration of particle is non-uniform
(C) ω - t graph	(r) Angular acceleration of particle is directly proportional to t.
(D) ω - t² graph	(s) Angular acceleration of particle is directly proportional to θ.

• A. A-(q,s)
• B. B-(p)
• C. C-(p)
• D. D-(q,r)

Answer 1

Answer: (A), (B), (C), (D)

A-(q,s), B-(p), C-(p), D-(q,r)

Answer 2

The problem requires matching graphs of angular motion variables with interpretations of angular acceleration. We will use the definitions of angular acceleration and kinematic equations for rotational motion.

Key Concepts:

1. Angular acceleration $\alpha = \frac{d\omega}{dt}$

2. Angular acceleration $\alpha = \omega \frac{d\omega}{d\theta}$

3. For constant angular acceleration: $\omega = \omega_0 + \alpha t$ and $\omega^2 = \omega_0^2 + 2\alpha\theta$

Analysis of each graph:

(A) $\omega - \theta$ graph is a straight line passing through the origin.

This means $\omega \propto \theta$, so we can write $\omega = k\theta$, where $k$ is a constant. To find angular acceleration $\alpha$, we use the relation $\alpha = \omega \frac{d\omega}{d\theta}$. From $\omega = k\theta$, differentiating with respect to $\theta$ gives $\frac{d\omega}{d\theta} = k$. Substituting these into the formula for $\alpha$: $\alpha = (k\theta) \cdot k = k^2\theta$.

• Since $\alpha = k^2\theta$, the angular acceleration is directly proportional to the angular displacement $\theta$. This matches statement (s).

• As $\alpha$ depends on $\theta$ (and is not constant), the angular acceleration is non-uniform. This matches statement (q).

Therefore, (A) matches with (q) and (s).

(B) $\omega^2 - \theta$ graph is a straight line passing through the origin.

This means $\omega^2 \propto \theta$, so we can write $\omega^2 = k\theta$, where $k$ is a constant. We compare this with the kinematic equation for rotational motion: $\omega^2 = \omega_0^2 + 2\alpha\theta$. Assuming the particle starts from rest ($\omega_0 = 0$), the equation becomes $\omega^2 = 2\alpha\theta$. Comparing $\omega^2 = k\theta$ and $\omega^2 = 2\alpha\theta$, we find $k = 2\alpha$. This implies $\alpha = k/2$. Since $k$ is a constant, $\alpha$ is also a constant.

• Therefore, the angular acceleration of the particle is uniform. This matches statement (p).

Therefore, (B) matches with (p).

(C) $\omega - t$ graph is a straight line passing through the origin.

This means $\omega \propto t$, so we can write $\omega = kt$, where $k$ is a constant. To find angular acceleration $\alpha$, we use the definition $\alpha = \frac{d\omega}{dt}$. From $\omega = kt$, differentiating with respect to $t$ gives $\frac{d\omega}{dt} = k$. So, $\alpha = k$. Since $k$ is a constant, $\alpha$ is also a constant.

• Therefore, the angular acceleration of the particle is uniform. This matches statement (p).

Therefore, (C) matches with (p).

(D) $\omega - t^2$ graph is a straight line passing through the origin.

This means $\omega \propto t^2$, so we can write $\omega = kt^2$, where $k$ is a constant. To find angular acceleration $\alpha$, we use the definition $\alpha = \frac{d\omega}{dt}$. From $\omega = kt^2$, differentiating with respect to $t$ gives $\frac{d\omega}{dt} = 2kt$. So, $\alpha = 2kt$.

• Since $\alpha = 2kt$, the angular acceleration is directly proportional to time $t$. This matches statement (r).

• As $\alpha$ depends on $t$ (and is not constant), the angular acceleration is non-uniform. This matches statement (q).

Therefore, (D) matches with (q) and (r).

Final Matches:

• (A) matches with (q), (s)

• (B) matches with (p)

• (C) matches with (p)

• (D) matches with (q), (r)