Circular Motion

We start from the definition of angle in radians and reach the two formulas that connect the linear quantities of a point on a wheel to the angular quantities of the wheel itself.

Step 1 — the definition of the radian. If a point moves a distance s along a circle of radius r, the angle turned in radians is \theta = s/r. Multiply both sides by r:

Step 2 — differentiate both sides with respect to time. The radius r is a fixed number for a given circle, so it is a constant multiplier and stays outside the derivative:

Step 3 — name the two derivatives. ds/dt is the linear speed v along the circle, and d\theta/dt is the angular velocity \omega:

Step 4 — differentiate once more with respect to time. Same reasoning: r stays outside:

Step 5 — name these derivatives. dv/dt is the rate at which the speed changes — the tangential acceleration a_t — and d\omega/dt is the angular acceleration \alpha:

Result.

In vector form \vec{v} = \vec{\omega}\times\vec{r} and \vec{a}_t = \vec{\alpha}\times\vec{r}.

What it means. Every point on a rigid rotating body shares the same \omega and \alpha, but a point twice as far from the axis moves twice as fast and gains speed twice as quickly.

Conditions used. \theta must be in radians — in degrees, s = r\theta is simply false. The radius r must be constant (the body is rigid, the point does not spiral in or out). Also, a_t is only the part of the acceleration along the tangent; there is always a second, radial part as well, which the next section works out.

We start from the fact that a body going round a circle at constant speed keeps changing the direction of its velocity, and we reach a formula for how big the resulting acceleration is and where it points.

Setting up. A particle moves on a circle of radius r at constant speed v. In a short time \Delta t it travels from point A to point B, and the radius to it turns through a small angle \Delta\theta.

Step 1 — the angle turned. By the definition of angular velocity, \omega = \Delta\theta/\Delta t, so

Step 2 — the velocity turns through the same angle. The velocity is always along the tangent, and the tangent turns exactly as fast as the radius does. So \vec{v}_A and \vec{v}_B have the same length v but differ in direction by \Delta\theta.

Step 3 — find the change in velocity. Redraw \vec{v}_A and \vec{v}_B from one common point (right-hand figure). The change is the vector from the tip of \vec{v}_A to the tip of \vec{v}_B:

Step 4 — measure its length. The two vectors and \Delta\vec{v} form an isosceles triangle: two equal sides of length v with \Delta\theta between them. Splitting that triangle down the middle gives two right-angled triangles, each with angle \Delta\theta/2 and hypotenuse v, so half the base is v\sin(\Delta\theta/2) and the whole base is twice that:

Step 5 — use the small-angle approximation. When an angle in radians is very small, \sin x \approx x (try x = 0.01: \sin x = 0.0099998). Since we are about to let \Delta t \to 0, the angle is as small as we like:

Step 6 — divide by the time taken. Acceleration is change of velocity divided by time:

Step 7 — take the limit as \Delta t \to 0 so that the answer is the instantaneous acceleration. The constant v comes outside the limit:

Step 8 — recognise the limit as \omega:

Step 9 — get the first standard form. Substitute \omega = v/r (from v = r\omega):

Step 10 — get the second standard form. Instead substitute v = r\omega into a = v\omega:

Step 11 — get the third form in terms of the period. Since \omega = 2\pi/T:

Step 12 — the direction. In Step 4's isosceles triangle, the two base angles are each (180^{\circ} - \Delta\theta)/2. As \Delta\theta \to 0 each base angle approaches 90^{\circ} — so \Delta\vec{v} becomes perpendicular to the velocity, pointing inwards along the radius, i.e. towards the centre.

Result.

That inward direction is why it is called the centripetal ("centre-seeking") acceleration.

What it means. A body can accelerate without speeding up. Here the size of the velocity never changes, only its direction — and changing direction is itself an acceleration, which needs a force pointing at the centre.

Conditions used. The speed v is constant (uniform circular motion) and the radius r is constant. Step 5 needs the angle in radians. If the speed also changes there is an extra tangential part a_t = r\alpha, and the total acceleration is \sqrt{a_c^{2} + a_t^{2}}.

We start from the two force equations for the bob — one vertical, one horizontal — and reach an expression for the time of one revolution.

Setting up. A bob of mass m hangs from a string of length L fixed at a point, and it moves in a horizontal circle so that the string sweeps out a cone at a constant angle \theta to the vertical. From the geometry of that cone,

Only two forces act on the bob: its weight mg straight down, and the tension T along the string towards the support.

Step 1 — resolve the tension. The string makes angle \theta with the vertical, so the tension has a vertical part T\cos\theta (upwards) and a horizontal part T\sin\theta (pointing at the centre of the circle).

Step 2 — vertical equation. The bob stays at the same height, so it has no vertical acceleration and the vertical forces balance:

Step 3 — horizontal equation. Horizontally the bob is accelerating — it is going round a circle — and the only horizontal force is T\sin\theta. So this component must supply the whole centripetal force:

Step 4 — divide Step 3 by Step 2. We divide because both T and m then cancel, which is exactly what we want:

Step 5 — cancel and use \sin/\cos = \tan:

(The same division with mv^{2}/r instead gives the equally useful \tan\theta = v^{2}/rg.)

Step 6 — replace r by L\sin\theta:

Step 7 — write \tan\theta as \sin\theta/\cos\theta:

Step 8 — cancel \sin\theta from both sides. This is allowed because the bob is actually circling, so \theta \ne 0 and \sin\theta \ne 0:

Step 9 — make \omega^{2} the subject. Multiply both sides by g and divide by L:

Step 10 — take the square root:

Step 11 — convert to a period. One revolution takes T_p = 2\pi/\omega:

Step 12 — turn the divided root upside down. Dividing by \sqrt{g/x} is the same as multiplying by \sqrt{x/g}:

Step 13 — recognise L\cos\theta as the depth h:

Step 14 — the tension. Go back to Step 2 and divide both sides by \cos\theta:

Result.

What it means. The mass cancelled, so a heavy bob and a light bob on the same string at the same angle take the same time to go round. The period depends only on the depth h of the circle below the support — so as the bob spins faster, \theta grows, h shrinks and the period gets shorter. Because \cos\theta < 1, the tension mg/\cos\theta is always bigger than the weight, and it would become infinite at \theta = 90^{\circ} — which is why the string can never be truly horizontal.

Conditions used. The string is light and inextensible, the angle \theta stays constant (steady motion, not a wobble), air resistance is neglected, and the circle is horizontal so there is no vertical acceleration.

We start from F = mv^{2}/r applied at the top and at the bottom of the circle, add conservation of energy between those two points, and reach three results: the least speed at the top, the least speed at the bottom, and the fact that the tension difference is always 6mg.

Setting up. A mass m on a string of length r is whirled in a vertical circle. Let v_1 be its speed at the highest point and v_2 at the lowest point. At both of those two points the string is vertical, so the tension lies exactly along the radius — which is why these two points are the easy ones to analyse.

At the highest point

Step 1 — which way does each force point? At the top, the centre of the circle is below the mass. Both the weight mg and the tension T_1 (pulling towards the support, which is below) therefore point downwards, i.e. both point towards the centre.

Step 2 — write the centripetal equation. The sum of the inward forces equals mv^{2}/r:

Step 3 — make T_1 the subject by subtracting mg from both sides:

Step 4 — apply the physical limit. A string can pull but cannot push, so the tension can never be negative:

Step 5 — substitute Step 3 into that inequality:

Step 6 — add mg to both sides:

Step 7 — cancel m and multiply both sides by r:

Step 8 — take the square root. The smallest allowed speed is the boundary case:

At exactly this speed T_1 = 0: the string goes slack and gravity alone supplies the whole centripetal force.

At the lowest point

Step 9 — which way does each force point? At the bottom the centre is above the mass. The tension T_2 points up (towards the centre) and the weight points down (away from the centre), so the net inward force is T_2 - mg:

Step 10 — make T_2 the subject:

Linking top and bottom by energy

Step 11 — apply conservation of mechanical energy. The tension does no work (it is always perpendicular to the motion), so only gravity matters. Take the lowest point as the zero of potential energy; the highest point is a height 2r above it:

Step 12 — multiply every term by 2:

Step 13 — cancel m throughout:

Step 14 — put in the minimum top speed v_1^{2} = gr:

Step 15 — take the square root:

The tension difference

Step 16 — subtract the top equation (Step 3) from the bottom one (Step 10):

Step 17 — remove the brackets carefully; the two mg terms now add:

Step 18 — combine the first two terms over the common denominator r:

Step 19 — use Step 13, which says v_2^{2} - v_1^{2} = 4gr:

Step 20 — cancel r:

Result.

What it means. To keep the string taut all the way round you need a speed at the bottom of at least \sqrt{5gr} — that is why you must swing a bucket of water briskly. The 6mg result is remarkable: the speed cancelled out, so the tension difference between bottom and top is the same no matter how fast you whirl it (as long as it does get round).

Conditions used. The string is light and inextensible, air resistance is neglected, and Step 11 assumed no energy is lost to friction. The 6mg result and v_2^{2} = v_1^{2} + 4gr hold for any speed; only v_1 = \sqrt{gr} and v_2 = \sqrt{5gr} are the special minimum case. For a rod instead of a string the rod can push, so T_1 may be negative and the minimum top speed is zero.

We start from the two force equations for a car on a frictionless banked road and reach a formula for the correct banking angle.

Setting up. The road surface is tilted at an angle \theta to the horizontal, with the outer edge raised. Take friction as negligible, so only two forces act on a car of mass m: its weight mg straight down, and the normal reaction N at right angles to the road surface. Because the surface is tilted by \theta, the normal N is tilted by \theta from the vertical.

Step 1 — resolve N into vertical and horizontal parts. Since N makes angle \theta with the vertical:

Step 2 — the vertical equation. The car stays on the road and does not rise or sink, so it has no vertical acceleration; the vertical forces must balance:

Step 3 — the horizontal equation. Horizontally the car is accelerating, because it is turning. The only horizontal force is N\sin\theta, so that component must be the whole centripetal force:

Step 4 — divide Step 3 by Step 2. We divide because both N and m then cancel:

Step 5 — cancel N on the left and m on the right:

Step 6 — use \sin\theta/\cos\theta = \tan\theta:

Step 7 — take the inverse tangent to get the angle:

Step 8 — or make v the subject instead. Multiply Step 6 by rg and take the square root:

Step 9 — the height of the outer edge. If the road is b wide, the raised edge is the "opposite" side of a right triangle whose hypotenuse is b:

Result.

What it means. The mass cancelled in Step 5, so one banking angle serves a motorcycle and a loaded truck equally — the engineer only has to choose the design speed v and the radius r. Sharper bends (small r) and faster roads (large v) need steeper banking. A cyclist leaning into a turn uses the identical relation.

Conditions used. Friction is neglected, so this gives the one speed at which no sideways friction is needed at all. The car is treated as a point mass (no toppling), the road is a circular arc of fixed radius, and \theta is the same across the whole width. At speeds away from the design speed, friction must make up the difference — which is where the v_{\max} formula below comes from.