Circular Motion

Angles in circular motion are measured in $\text{radians}$: an angle in radians is arc length divided by radius, so a full circle is $2\pi$ rad.

$\text{Angular velocity}$ $\omega$ is the angle swept per second, in $\mathrm{rad}{\mathrm{s}}^{-1}$. For one revolution in time $T$, $\omega =2\pi /T=2\pi f$.

Linear speed and angular velocity are linked by $v=r\omega$: points further from the centre move faster for the same rotation rate.

Velocity is a vector. Even if the speed is constant, a change of direction is a change of velocity — and a change of velocity over time is an acceleration.

For an object moving in a circle, the velocity at each instant is tangential, and the change in velocity over a short interval points towards the centre. So the $\text{centripetal acceleration}$ is directed towards the centre.

Its magnitude is $a=v^2/r=r{\omega }^2$. Faster motion or a tighter circle both demand greater acceleration.

By Newton’s second law, a centripetal acceleration needs a resultant force towards the centre: $F=m{v}^2/r=mr{\omega }^2$.

This is $\text{not}$ an additional force to draw on a free-body diagram. It is the resultant of the real forces. For a ball on a string it is the tension; for a car on a flat bend it is friction; for a satellite it is gravity.

If the centripetal force is removed — the string snaps — the object does not fly outwards. It continues in a straight line along the tangent, as Newton’s first law requires.

In a vertical circle, weight acts downwards throughout while the direction to the centre changes, so the forces must be analysed at specific points.

At the $\text{top}$ of a loop, both weight and the normal reaction (or tension) point towards the centre: $N+mg=m{v}^2/r$. The minimum speed to stay on the track is when $N=0$, giving $v_{\min }=\sqrt{gr}$.

At the $\text{bottom}$, the reaction points up towards the centre and weight points away: $N-mg=m{v}^2/r$. The reaction exceeds the weight, which is why riders feel heavier at the bottom of a dip.