Solution 21: Circular and Linear Motion Constraints
Given:
- Particle A: Moves on circle, radius $R=4$ m, speed $v_A = 2$ m/s.
- Particle B: Moves on horizontal line through center.
- Constraint: Distance $AB$ is always 4 m.
Angular velocity of A: $\omega = \frac{v}{R} = \frac{2}{4} = 0.5$ rad/s.
Position Analysis
Let the center of the circle be the origin $(0,0)$.
Position of A: $\vec{r}_A = (4 \cos \omega t) \hat{i} + (4 \sin \omega t) \hat{j}$.
Position of B: $\vec{r}_B = x_B \hat{i}$.
Using the distance constraint $|\vec{r}_A – \vec{r}_B|^2 = 4^2$: $$ (x_B – 4 \cos \omega t)^2 + (0 – 4 \sin \omega t)^2 = 16 $$ $$ x_B^2 – 8 x_B \cos \omega t + 16 \cos^2 \omega t + 16 \sin^2 \omega t = 16 $$ $$ x_B^2 – 8 x_B \cos \omega t + 16 = 16 $$ $$ x_B (x_B – 8 \cos \omega t) = 0 $$ Since B moves, $x_B = 8 \cos \omega t$.
Kinematics of B
Substituting $\omega = 0.5$: $x_B = 8 \cos(0.5t)$. This is Simple Harmonic Motion.
- Velocity: $v_B = \frac{dx}{dt} = -4 \sin(0.5t)$. Max speed = 4 m/s. (Option a True).
- Acceleration: $a_B = \frac{dv}{dt} = -2 \cos(0.5t)$. Max acc = 2 m/s$^2$. (Option b True).
Distance in One Revolution
One revolution of A corresponds to $t = \frac{2\pi}{0.5} = 4\pi$.
In this time, the argument of cosine ($0.5t$) goes from $0$ to $2\pi$.
B goes from $8 \to -8 \to 8$.
Total distance = $16 + 16 = 32$ m. (Option c True).
Relative Velocity
$\vec{v}_A = 2(-\sin \omega t \hat{i} + \cos \omega t \hat{j})$
$\vec{v}_B = -4 \sin \omega t \hat{i}$
$\vec{v}_{AB} = \vec{v}_A – \vec{v}_B = (4\sin \omega t – 2\sin \omega t)\hat{i} + 2\cos \omega t \hat{j}$
$\vec{v}_{AB} = 2\sin \omega t \hat{i} + 2\cos \omega t \hat{j}$
Modulus $|\vec{v}_{AB}| = \sqrt{4\sin^2 + 4\cos^2} = 2$ m/s. It is constant. (Option d True).
Correct Options: (a), (b), (c), and (d)