MHT-CET Physics · Motion in a Plane

Uniform Circular Motion

A body going round a circle at constant speed v has constant kinetic energy but a velocity that keeps turning, so it accelerates toward the centre at v²/r = ω²r; a banked track or a vertical loop supplies that centripetal force from gravity.

All Motion in a Plane notes MHT-CET Physics strategy

Why this matters

11 PYQs, none HARD: what stays constant, the change in momentum over half a circle, the average acceleration over half a turn, the acceleration from a frequency, the tangential speed from revolutions, banking of a railway track, and the least speed at the top of a vertical circle. One card.

Concept 1 of 1: Centripetal Acceleration, Banking and the Vertical Circle

In uniform circular motion only the speed, and so the kinetic energy, stays constant; velocity, momentum and acceleration all keep changing direction. The velocity is tangential, the acceleration points to the centre: a = v²/r = ω²r = 4π²f²r. Half a turn reverses the velocity, a change of 2v in momentum 2mv, and over the time πr/v that is an average acceleration 2v²/πr. A track banked by raising the outer rail h over a gauge d has tan θ = h/d = v²/rg. At the top of a vertical circle gravity alone can supply the centripetal force, so the least speed is √(gr).

Definition

Constant: speed, KE. Changing: velocity, momentum, acceleration (direction).
$a = \dfrac{v^2}{r} = \omega^2 r$ , $\omega = 2\pi f$ (d = 1 m, f = 4 Hz ⇒ $32\pi^2$ m/s²).
Half a circle: $\Delta p = 2mv$ ; average acceleration $\dfrac{2v^2}{\pi r}$ .
Banking: $\tan\theta = \dfrac{h}{d} = \dfrac{v^2}{rg}$ ⇒ $r = \dfrac{v^2d}{gh}$ .
Vertical circle, top: $v_{\min} = \sqrt{gr}$ , $f_{\min} = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{r}}$ .

Centripetal acceleration

a = \frac{v^2}{r} = \omega^2 r, \qquad \tan\theta = \frac{v^2}{rg}

Worked example

A stone on a 0.8 m string is whirled at 5 m/s. Centripetal acceleration, and the least speed at the top if whirled vertically? (g = 10 m/s²)

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 1 · Q47Moderate

Example 1 · Motion in a Plane · Uniform Circular Motion — Centripetal Acceleration, Banking

A particle moves around a circular path of radius '

r

' with uniform speed '

V

'. After moving half the circle, the average acceleration of the particle is

Calling the acceleration constant

Its magnitude v²/r is constant but its direction turns with the body. Only the kinetic energy and speed are truly constant.

Confusing instantaneous and average acceleration

Over half a circle the average acceleration is 2v²/(πr), smaller than the instantaneous v²/r.

Drill 10 more on centripetal acceleration, banking and the vertical circle

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 (1)

Centripetal Acceleration, Banking and the Vertical Circle
Centripetal acceleration
$a = \frac{v^2}{r} = \omega^2 r, \qquad \tan\theta = \frac{v^2}{rg}$

Watch out for (2)

Calling the acceleration constant
→ Centripetal Acceleration, Banking and the Vertical Circle
Confusing instantaneous and average acceleration
→ Centripetal Acceleration, Banking and the Vertical Circle

Test yourself on Motion in a Plane

15 past MHT-CET questions from this chapter, timed at 14 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 Motion in a Plane chapter test Drill all 11 questions