Relation Between Angular and Linear Quantities: MDCAT Physics notes
Relation Between Angular and Linear Quantities notes for MDCAT: s = rθ, v = rω, a = rα, centripetal acceleration and force, vertical circles and traps.
Linking linear and angular quantities
For a point at distance $r$ from the axis of a rotating body (angles in radians):
| Linear | Angular | Relation |
|---|---|---|
| Arc length $s$ | Angular displacement $\theta$ | $s = r\theta$ |
| Tangential velocity $v$ | Angular velocity $\omega$ | $v = r\omega$ |
| Tangential acceleration $a_t$ | Angular acceleration $\alpha$ | $a_t = r\alpha$ |
For the same $\theta$ or $\omega$, points farther from the axis move farther and faster; reducing $r$ reduces the linear displacement.
Centripetal acceleration
In uniform circular motion the speed is constant but the direction of velocity changes, so there is an acceleration directed towards the centre:
$$a_c = \frac{v^2}{r} = r\omega^2$$
The velocity is tangential and the centripetal acceleration is radial, so in uniform circular motion velocity and acceleration are perpendicular.
Centripetal force
The net force needed to keep a body on a circular path, directed towards the centre:
$$F_c = \frac{mv^2}{r} = mr\omega^2$$
Centripetal force is not a new kind of force; some real force supplies it:
- Car turning on a flat road: friction between the tyres and the road.
- Stone on a string: tension. Planet or satellite: gravity. Electron in an atom: electric attraction.
- Rotating spaceship: the outer wall pushes objects towards the centre; occupants feel pressed against the wall, which acts as artificial gravity.
A car moving at constant speed around a bend has a resultant (centripetal) force on it; a car at rest, on a straight road at constant velocity, or going uphill at constant velocity has zero net force.
Scaling problems
Use $F \propto mv^2/r$:
- $v$ doubled: $F$ becomes $4F$.
- $v$ doubled and $r$ four times: $F' = 4F/4 = F$.
- $m$ halved and $r$ doubled: $F' = F/4$.
- Roller coaster, $v = 30\ \text{m s}^{-1}$, $r = 30\ \text{m}$: $F = m(900)/30 = 30m$.
Vertical circle
At the top of a vertical circle, weight and tension (or normal force) both point to the centre: $T + mg = mv^2/r$. The minimum speed at the top occurs when $T = 0$, so gravity alone provides the centripetal force:
$$v_{min} = \sqrt{gr}$$
This applies to a stone on a string, a bucket of water, or a fighter plane looping the loop. At this critical speed the tension at the top is zero.
Key formulas
- $s = r\theta$, $v = r\omega$, $a_t = r\alpha$
- $a_c = v^2/r = r\omega^2$
- $F_c = mv^2/r = mr\omega^2$
- Top of vertical circle: $v_{min} = \sqrt{gr}$
Common MDCAT traps
- $a_c = r\omega^2$, not $\omega^2/r$ or $r\omega$.
- Centripetal acceleration points towards the centre, never away (centrifugal is a pseudo force).
- Doubling speed quadruples centripetal force.
- On a flat road, friction (not weight or engine power) gives the centripetal force.
- Minimum speed at the top of a loop is $\sqrt{gr}$, not $\sqrt{2gr}$.
Quick revision
- $v = r\omega$ links linear and angular velocity.
- Uniform circular motion: $\vec v \perp \vec a$.
- Centripetal force does no work.
- Tension is zero at the top at critical speed.
- Circular motion at constant speed still needs a net force.