Question

A particle is performing uniform circular motion of radius 3 m with an angular velocity 2 radians per second in clockwise direction as shown. Find acceleration of particle at point Q.

• A. (-6$\hat{i}$ - 6√3 $\hat{j}$) m/s²
• B. (-6√3$\hat{i}$-6$\hat{j}$) m/s²
• C. (-6$\hat{i}$ + 6√3$\hat{j}$) m/s²
• D. (6$\hat{i}$ - 6√3$\hat{j}$) m/s²

Answer 1

Answer: (A)

(-6$\hat{i}$ - 6√3 $\hat{j}$) m/s²

Answer 2

The particle is performing uniform circular motion. In uniform circular motion, the acceleration is always centripetal, directed towards the center of the circle.

1. Identify the given parameters:

   • Radius of the circular path, $r = 3$ m.
   • Angular velocity, $\omega = 2$ radians per second.
   • The particle is at point Q, which makes an angle of $60^\circ$ with the positive x-axis.

2. Calculate the magnitude of the centripetal acceleration ($a_c$): The formula for centripetal acceleration in terms of angular velocity is: $a_c = r\omega^2$ Substitute the given values: $a_c = (3 \text{ m}) \times (2 \text{ rad/s})^2$ $a_c = 3 \times 4 = 12 \text{ m/s}^2$

3. Determine the direction of the acceleration vector: At any point in uniform circular motion, the centripetal acceleration vector points from the particle's position towards the center of the circle. In this case, the center is the origin (0,0). The position vector of point Q is $\vec{r}_Q$. Since Q is at an angle of $60^\circ$ with the positive x-axis, its position vector can be written as: $\vec{r}_Q = r \cos 60^\circ \hat{i} + r \sin 60^\circ \hat{j}$ The acceleration vector $\vec{a}$ is directed opposite to the position vector $\vec{r}_Q$. Therefore, the unit vector in the direction of acceleration is $-\hat{r}_Q$. So, $\vec{a} = -a_c \hat{r}_Q$

4. Express the acceleration vector in component form: $\vec{a} = -a_c (\cos 60^\circ \hat{i} + \sin 60^\circ \hat{j})$ Substitute the values $a_c = 12 \text{ m/s}^2$, $\cos 60^\circ = \frac{1}{2}$, and $\sin 60^\circ = \frac{\sqrt{3}}{2}$: $\vec{a} = -12 \left(\frac{1}{2} \hat{i} + \frac{\sqrt{3}}{2} \hat{j}\right)$ $\vec{a} = - \left(12 \times \frac{1}{2}\right) \hat{i} - \left(12 \times \frac{\sqrt{3}}{2}\right) \hat{j}$ $\vec{a} = -6 \hat{i} - 6\sqrt{3} \hat{j} \text{ m/s}^2$