6.05a–6.05c - Uniform circular motion

Angular velocity and speed

Two particles pass the same point on a horizontal circle at the same speed, one clockwise and one anticlockwise: do they have the same velocity? No. Their tangent directions at that point are opposite. Speed measures how fast distance is covered; velocity also has a direction, which keeps changing around a circle.

Here is the question that will guide this lesson: how can a particle accelerate without getting faster? First separate the rate of turning from the rate of covering distance.

Take a point $0.4\,\mathrm{m}$ from a wheel's fixed axis. A quarter-turn covers angle $\pi/2$ but distance $0.4(\pi/2)=0.2\pi\,\mathrm{m}$. A point twice as far from the axis turns through the same angle and travels twice as far. All points on a rigid rotating wheel share an angular velocity; they do not share a speed.

The velocity vector is tangent to the circular path, in the direction of travel. It is not directed along the radius. For uniform circular motion, the speed and angular velocity are constant, but the velocity vector is not constant.

Speed is $v=|v_t|=r|\omega|$. If the positive sense is the direction of motion, write $v=r\omega$. One full revolution covers $2\pi$ radians, so for non-zero uniform rotation the period $P=2\pi/|\omega|$ and frequency $f=1/P$. The period and frequency relationships are not in the formula booklet: learn them. We use $P$ for period here so that $T$ can denote tension later.

Change only the sense of rotation. The distance travelled per second stays the same, but the signed angular velocity reverses.

The sign records a direction, rather than a different amount of turning.

Use a different turning rate with the same direction convention.

Acceleration towards the centre

Even at constant speed, a particle's velocity changes because its direction changes. Its acceleration points towards the centre, perpendicular to its instantaneous velocity.

You can see the direction and magnitude without imagining an extra force. At two nearby positions separated by angle $\Delta\theta$, the two equal-length velocity vectors also differ in direction by $\Delta\theta$. The triangle formed by the velocity vectors has change of velocity of magnitude $2v\sin(\Delta\theta/2)$. Hence

The change-of-velocity direction tends to the inward radius. This also follows by differentiating $\mathbf r=r\cos(\omega t)\mathbf i+r\sin(\omega t)\mathbf j$ twice: $\mathbf a=-\omega^2\mathbf r$. The minus sign means towards the centre. Both derivations assume a fixed centre and fixed radius, with constant angular velocity.

[DIAGRAM: asset_name: Velocity and acceleration in uniform circular motion; asset_slug: h245_6_05_inward_acceleration; caption: Velocity is tangent to the circle and acceleration points towards its centre; description: A blue circle with centre O and a point P in its upper right quadrant. A blue velocity arrow starts at P and runs anticlockwise along the tangent. A clay acceleration arrow starts at P and points towards O. A sage radius joins O to P. A separate velocity-vector triangle shows two equal-length vectors separated by a small angle delta theta and the arrow delta v joining their tips. Text states a = v squared over r = r omega squared, directed towards O]

The equivalent form $a_r=v|\omega|$ follows from $v=r|\omega|$; it is not printed separately in the formula booklet. “Centripetal” means centre-seeking: centripetal force is the inward resultant of the actual forces, not another force to add to a force diagram. If tension alone acts horizontally, it supplies that resultant.

Change only the speed, doubling it. Turning the velocity twice as fast while its length is also doubled makes the required acceleration four times as large.

For a particle moving uniformly on a rough horizontal plane without a string, friction can instead provide the inward force. Vertical balance gives $R=mg$; the condition $mv^2/r\le\mu R$ then gives $v^2\le\mu gr$. This limiting-speed result is derived from Newton's second law and the friction law; it is not in the formula booklet. Friction need not equal $\mu R$ below the limit.

Changing speed changes the required force even though the mass and path stay fixed.

Conical pendulums

A conical pendulum moves in a horizontal circle while its supporting string sweeps out a cone. Its vertical coordinate is constant, so vertical forces balance; its horizontal acceleration is towards the circle's centre.

Let the string length be $l$, its angle to the downward vertical be $\alpha$, and the tension be $T$. The horizontal circle has radius $r=l\sin\alpha$, not $l$. There are only two forces on the particle: tension along the string towards its support, and weight vertically downwards.

[DIAGRAM: asset_name: Conical pendulum geometry and forces; asset_slug: h245_6_05_conical_pendulum; caption: Vertical balance and horizontal acceleration use different components of the same tension; description: Side elevation of a conical pendulum with fixed support A above O, particle P to the right of O, AP of length l inclined at angle alpha to the downward vertical AO, and horizontal OP labelled r = l sin alpha. A tension arrow starts at P towards A and a weight arrow starts at P vertically down. A separate component panel shows T cos alpha upwards and T sin alpha horizontally towards O, with T cos alpha = mg and T sin alpha = mv squared over r]

The component equations are safer to rebuild than to recall a special formula. Divide the horizontal equation by the vertical one to eliminate both $T$ and $m$. Notice the surprising consequence: mass affects the tension but not the speed required for a given cone.

Change only the angle to the vertical. The radius and the tension must both be recalculated.

At $\alpha=90^\circ$, tension has no vertical component to balance weight, so a freely suspended conical pendulum cannot maintain a horizontal circle. This is a failure of vertical balance, not a claim that the tension itself is zero.

Keep the same string and particle, and change the cone once more.