Question

The average acceleration vector for a particle having a uniform circular motion in one complete revolution.
A. A constant vector of magnitude $\dfrac{{{v^2}}}{r}$
B. $\dfrac{{{v^2}}}{r}$in magnitude and perpendicular to the plane of the circle.
C. Equal to the instantaneous acceleration vector at the start of the motion
D. A null vector.

Answer 1

Hint: Concept of circular motion and acceleration is used. Also average acceleration depends upon average velocity which in turn depends upon net displacement.

b>Complete step by step answer:
Uniform circular motion: If a particle covers equal distance along the circumference of the circle is equal intervals of time i.e with constant speed then the motion is said to be uniform circular motion.
Examples: Motion of tip of second hand of a clock
Angular velocity: The time rate of change of angular displacement of a particle is angular velocity. It is denoted by \[\overrightarrow w \,\,and\,\,\overrightarrow w = \dfrac{{\Delta \theta }}{{\Delta t}}\]where $\Delta \theta $is change in angle $\Delta t$is time interval
Linear velocity: Linear velocity is the rate of change of displacement of a particle with time. It is denoted by $v = \dfrac{{\Delta s}}{{\Delta t}}$.
$\Delta s$is displacement.
Both linear velocity and angular velocity are related as $v = rw$ where r is the radius of the circle.
Angular acceleration is the time rate of change of velocity. It is denoted by $\alpha $.
$\vec \alpha = \dfrac{{\Delta w}}{{\Delta t}}$
Linear acceleration is the time of change of linear velocity. It is denoted by a both linear and angular acceleration are related as $a = \alpha r$
Average acceleration is the total displacement to total time.
Now, the average acceleration vector during our complete rotation will be a null vector because in one complete rotation, net displacement is zero vector which implies that average velocity will also be a null vector which further implies that the average acceleration vector during one complete rotation will also be a null vector.

So, the correct answer is “Option D”.

Additional Information:
Angular frequency, $w = \dfrac{\theta }{t}$
Angular displacement, $\theta = \dfrac{s}{r}$
Alos, $w = \dfrac{{2\pi v}}{{}} = \dfrac{{2\pi }}{T}$where v is frequency and T is time period.

Note:
Remember that \[a = \dfrac{{{v^2}}}{r} = {w^2}r\]is the acceleration at a particular interval of time not for a complete rotation. Hence option A, B and C are incorrect.