Read each statement below carefully and state, with reasons, if it is true or false: (a) The net acceleration

Exercises · 3.16

Q. 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,

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Concept understanding — Uniform Circular Motion

Uniform Circular Motion

Uniform circular motion is motion along a circular path at constant speed. Although the speed stays the same, the velocity does not — its direction keeps changing at every instant — so the motion has an acceleration even though the speed never changes. This concept covers the kinematics of that motion; the force that causes it (centripetal force) belongs to the Laws of Motion unit.

1. Angular Quantities

As the body sweeps through an angle $\theta$ , its angular velocity is

\omega = \frac{d\theta}{dt}

For one full revolution in a period $T$ , at frequency $f$ :

\omega = \frac{2\pi}{T} = 2\pi f, \qquad T = \frac{1}{f}

To convert rpm to rad/s:

\omega = \frac{2\pi \cdot (\text{rpm})}{60} = \frac{\pi}{30}\,(\text{rpm})

The linear (rim) speed $v$ and the angular speed $\omega$ are related by

v = \omega r

2. Centripetal Acceleration

Even though the speed is constant, the velocity vector keeps turning — this produces an acceleration directed toward the centre of the circle:

a_c = \frac{v^2}{r} = \omega^2 r = \frac{4\pi^2 r}{T^2} = 4\pi^2 f^2 r

Use whichever form matches the data you are given. This acceleration is often expressed as a multiple of $g$ (as $a_c/g$ , taking $g = 10\ \text{m/s}^2$ ).

3. Directions

The velocity is always tangential (along the direction of motion); the acceleration is centripetal — radial, pointing inward, and perpendicular to the velocity.

Because the direction of the velocity keeps changing, the change in the velocity vector over an angle $\Delta\theta$ has magnitude

|\Delta v| = 2v\sin\left(\frac{\Delta\theta}{2}\right)

So a quarter turn gives $|\Delta v| = v\sqrt{2}$ , a half turn gives $2v$ , and a full turn gives $0$ .

The average acceleration over an arc is $|\Delta v|/\Delta t$ — this is smaller in magnitude than the instantaneous centripetal acceleration $v^2/r$ , and its direction lies along the perpendicular bisector of the chord joining the two points (which, for a circle, always passes through the centre) rather than being radially inward from either endpoint's own position.

4. Points on a Rotating Body

Every point on a rigid rotating body (a wheel, a disc, a clock hand, the Earth) shares the same $\omega$ , but the rim speed $v = \omega r$ grows with the radius.

Clock hands: the second hand turns at $\omega = \dfrac{2\pi}{60}$ rad/s, the minute hand at $\dfrac{2\pi}{3600}$ rad/s, the hour hand at $\dfrac{2\pi}{43200}$ rad/s.
The Earth spins with $\omega = \dfrac{2\pi}{86400} \approx 7.3\times10^{-5}$ rad/s. A point on the equator moves at $\omega R \approx 465$ m/s, and a point at latitude $\lambda$ moves at $\omega R\cos\lambda$ (since the circle of latitude has radius $R\cos\lambda$ ).

5. Connected Systems

Wheels joined by a belt (or two gears in mesh) share the same rim speed, so $\omega_1 r_1 = \omega_2 r_2$ — the larger wheel turns with the smaller $\omega$ .
Wheels on the same axle (concentric) share the same $\omega$ , so the outer rim moves faster ( $v \propto r$ ).

6. Non-Uniform Circular Motion

If the speed also changes, there is a tangential acceleration

a_t = \frac{dv}{dt} = r\alpha

(from the angular acceleration $\alpha = d\omega/dt$ ), in addition to the radial $a_c = v^2/r$ . These two are perpendicular, so the total acceleration is

$a = \sqrt{a_c^2 + a_t^2}, \qquad \tan\beta = \frac{a_t}{a_c}$ …

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False: Net acceleration in circular motion is not always purely radial; it has a tangential component if speed changes.
True: The velocity vector is always tangent to the path at any given point.
True: The average acceleration vector over one cycle in uniform circular motion is a null vector due to symmetry.

Understanding the nature of velocity and acceleration in circular motion is fundamental. Velocity describes how an object's position changes, while acceleration describes how its velocity changes. In circular motion, even if the speed is constant, the direction of velocity is continuously changing, which means there must always be an acceleration.

Let's analyze each statement:

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

Concept of Acceleration in Circular Motion:

In circular motion, a particle's velocity vector is constantly changing direction. This change in direction necessitates an acceleration component directed towards the center of the circle, known as centripetal or radial acceleration ( $a_r$ ). Its magnitude is given by $a_r = v^2/r$ , where $v$ is the speed and $r$ is the radius.

However, if the particle's speed is also changing (i.e., it's speeding up or slowing down), there is an additional component of acceleration called tangential acceleration ( $a_t$ ). This component is directed along the tangent to the circle, either in the direction of motion (speeding up) or opposite to it (slowing down). Its magnitude is $a_t = |dv/dt|$ .
Net Acceleration:

The net acceleration ( $\vec{a}_{net}$ ) is the vector sum of the radial and tangential accelerations: $\vec{a}_{net} = \vec{a}_r + \vec{a}_t$ .
- If the motion is uniform circular motion (constant speed), then $dv/dt = 0$ , so $a_t = 0$ . In this specific case, the net acceleration is purely radial, pointing towards the center.
- If the motion is non-uniform circular motion (changing speed), then $a_t \neq 0$ . The net acceleration will have both radial and tangential components. Its direction will not be purely along the radius towards the center; it will be at an angle to the radius.
Conclusion for Statement (a):

The statement uses the word "always." Since tangential acceleration exists in non-uniform circular motion, the net acceleration is not always purely radial.

Watch out

A common mistake is to assume all circular motion is uniform circular motion. Remember that "circular motion" encompasses both uniform (constant speed) and non-uniform (changing speed) cases.

Therefore, statement (a) is False.

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

Definition of Instantaneous Velocity:

The velocity vector at any instant represents the instantaneous rate of change of the particle's position. Its direction indicates the direction of motion at that precise moment.
Relation to Path:

By definition, the direction of instantaneous velocity is always tangent to the path (trajectory) of the particle at that point. If it were not tangent, the particle would immediately move off the path. This holds true for any type of motion, not just circular motion. For example, if you swing a stone on a string and release it, the stone flies off in the direction tangent to the circular path at the point of release.
Conclusion for Statement (b):

This is a fundamental definition in kinematics.

Important

The direction of the instantaneous velocity vector is always tangent to the path of motion.

Therefore, statement (b) is True.

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

Acceleration in Uniform Circular Motion (UCM):

In UCM, the speed ( $v$ ) is constant, but the direction of the velocity vector continuously changes. This change in direction is caused by the centripetal acceleration, which always points towards the center of the circle. Its magnitude is constant, $a = v^2/r$ .
Averaging a Vector over a Cycle:

To find the average of a vector quantity $\vec{A}(t)$ over a time period $T$ , we calculate:

\vec{A}_{avg} = \frac{1}{T} \int_0^T \vec{A}(t) dt

In UCM, let the particle move in a circle of radius $R$ in the $xy$-plane. If the particle starts at $(R, 0)$ and moves counter-clockwise with angular speed $\omega = v/R$, its position at time $t$ is $\vec{r}(t) = R \cos(\omega t) \hat{i} + R \sin(\omega t) \hat{j}$.
The velocity is $\vec{v}(t) = -R\omega \sin(\omega t) \hat{i} + R\omega \cos(\omega t) \hat{j}$.
The acceleration is $\vec{a}(t) = -R\omega^2 \cos(\omega t) \hat{i} - R\omega^2 \sin(\omega t) \hat{j}$. …

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Concept: A Two-Term Lookup Table for (a)/(b), a Hodograph-Symmetry Argument for (c)

Method: Radial + Tangential Decision Table, Then "The Acceleration Hodograph Is a Full Circle" for the Averaging Question

The existing LONG answer proves (c) with an explicit calculus integral over one period; the existing SHORT answer argues (c) via " $\Delta\vec v=0$ over one cycle." This method reaches (c) a third way — by picturing the set of all acceleration vectors over one cycle as tracing out a full circle of its own (the "hodograph"), and using a purely geometric symmetry argument on that picture, distinct from both an integral and a velocity-endpoint argument.

(a) and (b) — via a single lookup table

State the general rule once: in any circular motion (uniform or not), the net acceleration decomposes into two perpendicular pieces —

\vec a_{\text{net}} = \vec a_{\text{radial}} + \vec a_{\text{tangential}}, \qquad a_{\text{radial}}=\frac{v^2}{r}\ (\text{always present}), \qquad a_{\text{tangential}}=\left|\frac{dv}{dt}\right|\ (\text{present iff speed changes})

Statement	Always true?
(a) Net acceleration is always purely radial	Only if $a_{\text{tangential}}=0$ , i.e. only for uniform circular motion — not always, so False
(b) Velocity is always tangent to the path	This is the very definition of instantaneous velocity, for any curve, not just circles — True

(c) — via the acceleration hodograph

In uniform circular motion, the acceleration vector at every instant has the same constant magnitude $a=v^2/r$ and points radially inward, toward the centre. As the particle sweeps once around its circular path, the direction of $\vec a(t)$ sweeps once around a circle of its own, of radius $a$ , centred at the origin (this second circle, traced out by the tip of $\vec a(t)$ , is called the hodograph of the acceleration). …

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Q.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.

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