Class 11 NCERT PHYSICS • Chapter 3: Motion in a Plane • Question 16

QUESTION 16

Read each statement below carefully and state, with reasons, if it is true or false :

(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre

(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point

(c) The acceleration vector of a particle in uniform circular motion averaged over one

cycle is a null vector

SOLUTION

We need to evaluate the validity of three kinematic statements regarding circular motion and velocity vector orientations.

Concept: Uniform Circular Motion

Uniform Circular Motion (UCM): Motion with constant speed where acceleration is purely centripetal.
Non-Uniform Circular Motion: Motion where speed varies, involving both centripetal acceleration and tangential acceleration.
Instantaneous Velocity: The rate of change of position, always tangent to the path.
Average Acceleration: The change in velocity over a time interval.

Given:

(a) Claim: Net acceleration is always radial.
(b) Claim: Velocity is always tangential.
(c) Claim: Average acceleration over one cycle in UCM is a null vector.

Solving:

Statement (a):
- In circular motion, the net acceleration vector $\vec{a}$ is the vector sum of two components:

\vec{a} = \vec{a}_c + \vec{a}_t

where $\vec{a}_c$ is the centripetal acceleration (radial) and $\vec{a}_t$ is the tangential acceleration.
If the motion is non-uniform, the speed $v$ changes, meaning $a_t = dv/dt \neq 0$ . In this case, the net acceleration is not purely radial.
Therefore, the statement is False.

Statement (b):
- The velocity vector is defined as the derivative of the position vector $\vec{r}$ with respect to time:

\vec{v} = \frac{d\vec{r}}{dt} = \lim_{\Delta t \to 0} \frac{\Delta \vec{r}}{\Delta t}

As the time interval $\Delta t$ approaches zero, the displacement vector $\Delta \vec{r}$ becomes a chord that eventually aligns with the tangent to the path at that point.
Therefore, the statement is True.

Statement (c):
- The average acceleration $\vec{a}_{\text{avg}}$ over a period $T$ is given by:

\vec{a}_{\text{avg}} = \frac{1}{T} \int_{0}^{T} \vec{a}(t)\, dt

Since $\vec{a}(t) = \frac{d\vec{v}}{dt}$ , the integral evaluates to the change in velocity:

\vec{a}_{\text{avg}} = \frac{\vec{v}(T) - \vec{v}(0)}{T}

In uniform circular motion, the velocity vector $\vec{v}$ has a constant magnitude but its direction changes. After one full cycle ( $T$ ), the particle returns to its starting point with its original velocity vector direction. Thus, $\vec{v}(T) = \vec{v}(0)$ .
Consequently, the numerator is zero, making $\vec{a}_{\text{avg}}$ a null vector.
Therefore, the statement is True.

Answer

(a) False. In non-uniform circular motion, there is a tangential acceleration component, so the net acceleration is not purely radial.
(b) True. By definition, the instantaneous velocity is the rate of change of position, which is always tangent to the trajectory.
(c) True. In uniform circular motion, the velocity vector returns to its initial state after one cycle, so the change in velocity (and thus average acceleration) is zero.