MHT-CET Physics · Rotational Dynamics

Kinematics of Circular Motion

Motion on a circle is described by an angle: angular velocity ω = 2πn links it to the linear speed v = ωr, and with a constant angular acceleration the angle obeys the same equations as distance in a straight line.

All Rotational Dynamics notes MHT-CET Physics strategy

Why this matters

17 PYQs, three HARD. Three shapes: comparing speeds or accelerations of bodies that share a period, the angle turned in a given second under constant angular acceleration, and the moment the tangential and centripetal accelerations become equal.

Concept 1 of 3: Angular Velocity, Linear Speed and Centripetal Acceleration

Every point of a turning body sweeps the same angle in the same time, so the angular velocity is shared; the linear speed grows with the distance from the axis. Two bodies with the same period therefore have the same ω, and their speeds and centripetal accelerations are in the ratio of their radii.

Definition

$\omega = \dfrac{2\pi}{T} = 2\pi n$ ; $v = \omega r$ ; $a_c = \dfrac{v^2}{r} = \omega^2 r$ , directed to the centre.
Same period ⇒ same $\omega$ ⇒ $v \propto r$ and $a_c \propto r$ , whatever the masses.
$\dfrac{a_c}{r} = 4\pi^2n^2 \propto n^2$ .
A point at latitude $\lambda$ on the spinning Earth circles at radius $R\cos\lambda$ : speed $V\cos\lambda$ .
Uniform circular motion: velocity tangent to the circle, acceleration toward the centre, the two perpendicular. The acceleration is NOT tangent to the circle.

Circular motion

\omega = 2\pi n, \qquad v = \omega r, \qquad a_c = \omega^2 r = \frac{v^2}{r}

Worked example

A wheel of radius 0.5 m turns at 120 rpm. Find ω, the rim speed and the rim's centripetal acceleration.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q17Easy

Example 1 · Rotational Dynamics · Kinematics of Circular Motion

Two objects of masses

m_1

and

m_2

are moving in circles of radii

r_1

and

r_2

respectively. Their respective angular speeds

\omega_1

and

\omega_2

are such that they both complete one revolution in the same time

t

. The ratio of linear speed of

m_2

to that of

m_1

Letting the masses matter

Same period means same ω; speed ratio is the radius ratio. The masses in the stem are there to be picked by mistake, and 'm₁ : m₂' is always an option.

Drill 7 more on angular velocity, linear speed and centripetal acceleration

Concept 2 of 3: Constant Angular Acceleration

Swap distance for angle, speed for angular velocity and acceleration for angular acceleration, and the straight-line equations of motion work unchanged. The angle turned in the nth second from rest grows as the odd numbers 1, 3, 5, …

Definition

$\omega = \omega_0 + \alpha t$ , $\theta = \omega_0 t + \dfrac{1}{2}\alpha t^2$ , $\omega^2 = \omega_0^2 + 2\alpha\theta$ .
From rest, angle in the $n$ th second: $\theta_n = \dfrac{\alpha}{2}(2n - 1)$ ; consecutive seconds go 1 : 3 : 5 …, and equal intervals from rest 1 : 3.
A fan slowing uniformly: $\omega^2 = \omega_0^2 - 2\alpha\theta$ tells how many more turns it makes.
Non-uniform: $\omega(t) = \alpha - \beta t$ stops at $t = \dfrac{\alpha}{\beta}$ after turning $\displaystyle\int\omega\,dt = \dfrac{\alpha^2}{2\beta}$ .

Angle in the nth second, from rest

\theta_n = \frac{\alpha}{2}(2n - 1)

Worked example

A wheel starts from rest with α = 2 rad/s². Angle turned in the 3rd second, and in the first 4 seconds?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift II · Q3Moderate

Example 2 · Rotational Dynamics · Kinematics of Circular Motion

A wheel initially at rest, begins to rotate about its axis with constant angular acceleration. If it rotates through an angle

\theta_{1}

in first 2 s and a further angle

\theta_{2}

in the next 2 s , the ratio

\theta_{1}:\theta_{2}

Angle in the nth second versus angle in n seconds

'In the 3rd second' is one second's worth,

\frac{\alpha}{2}(2n - 1)

; 'in 3 seconds' is

\frac{1}{2}\alpha(9)

. Both land on options.

Drill 5 more on constant angular acceleration

Concept 3 of 3: Tangential and Centripetal Acceleration Together

A speeding-up particle on a circle has two accelerations at right angles: tangential, which changes the speed, and centripetal, which turns it. The tangential one is fixed at rα; the centripetal one grows as ω² — so from rest they become equal when ω² = α.

Definition

$a_t = r\alpha$ (along the path); $a_c = \omega^2 r$ (to the centre); net $a = \sqrt{a_t^2 + a_c^2}$ .
From rest, $\omega = \alpha t$ : $a_c = a_t$ when $\alpha^2t^2 = \alpha$ , i.e. $t = \dfrac{1}{\sqrt{\alpha}}$ .
Net force $= ma$ , using both components.

Net acceleration on a circle

a = \sqrt{(r\alpha)^2 + (\omega^2 r)^2}

Worked example

From rest with α = 9 rad/s². When are the tangential and centripetal accelerations equal?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q2Moderate

Example 3 · Rotational Dynamics · Kinematics of Circular Motion

A particle at rest starts moving with a constant angular acceleration of

4\,\text{rad/s}^2

in a circular path. The time at which magnitudes of its centripetal acceleration and tangential acceleration will be equal, is (in second)

Using only the centripetal part for the net force

When the speed changes, the net force needs both components. 5 kg at

a_t = 4

a_c = 20

m/s² feels

5\sqrt{416} = 20\sqrt{26}

N, not 100 N.

Drill 2 more on tangential and centripetal acceleration together

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Angular Velocity, Linear Speed and Centripetal Acceleration
Circular motion
$\omega = 2\pi n, \qquad v = \omega r, \qquad a_c = \omega^2 r = \frac{v^2}{r}$
Constant Angular Acceleration
Angle in the nth second, from rest
$\theta_n = \frac{\alpha}{2}(2n - 1)$
Tangential and Centripetal Acceleration Together
Net acceleration on a circle
$a = \sqrt{(r\alpha)^2 + (\omega^2 r)^2}$

Watch out for (3)

Letting the masses matter
→ Angular Velocity, Linear Speed and Centripetal Acceleration
Angle in the nth second versus angle in n seconds
→ Constant Angular Acceleration
Using only the centripetal part for the net force
→ Tangential and Centripetal Acceleration Together

Test yourself on Rotational Dynamics

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Rotational Dynamics chapter test Drill all 17 questions

Related notes

Angular Momentum and Torque →