Q.A cyclist starts from the centre O of a circular park of radius $1\ \text{km}$ and rides along the path O$\to$P$\to$R$\to$Q$\to$O. Here P is a point on the boundary circle reached by riding straight out along a radius (due east of O); the cyclist then rides along the circular boundary from P to R (R lies on the circle midway, in the north-east direction) and on to Q (the point on the circle due north of O); finally the cyclist returns straight along the radius QO back to the centre. Throughout the ride the speed is a constant $10\ \text{ms}^{-1}$. Find the magnitude and direction of the cyclist's acceleration at point R.

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

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

To convert rpm to rad/s:

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

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:

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

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

(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

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