M3.4 - Motion in a circle

Angular speed $\omega$ measures the rate at which angular position changes. Angles in circular-motion formulae are measured in radians, so one complete revolution is $2\pi$ radians.

For a point at radius $r$, arc length is $s=r\theta$. Differentiating gives tangential speed $v=r\omega$. All points on one rigid rotating body have the same angular speed, but a point farther from the axis has a larger linear speed.

A wheel completes $15$ revolutions in $6$ minutes. Its period is $360/15=24$ s and$$\omega=\frac{2\pi}{24}=\frac{\pi}{12}\,\mathrm{rad\,s^{-1}}.$$A point $8$ m from the axis therefore moves at $v=8(\pi/12)=2\pi/3\,\mathrm{m\,s^{-1}}$.

A particle moving in a circle accelerates even when its speed is constant, because its velocity direction changes. The acceleration points radially inward, towards the centre of the circle.

The inward acceleration is not an extra force. Choose inward as the positive radial direction and resolve the actual forces:$$\sum F_{\mathrm{inward}}=m\frac{v^2}{r}=mr\omega^2.$$Forces pointing away from the centre enter with a negative sign.

A $0.40$ kg particle moves at $3.0\,\mathrm{m\,s^{-1}}$ in a circle of radius $1.5$ m. Then$$a_r=\frac{3.0^2}{1.5}=6.0\,\mathrm{m\,s^{-2}},$$so the resultant inward force must be $0.40(6.0)=2.4$ N. Individual forces need not each equal $2.4$ N; their inward components must have that resultant.

For uniform motion in a horizontal circle, vertical acceleration is zero while the horizontal resultant towards the centre is $mr\omega^2$. Draw every real force, resolve vertically for equilibrium, and resolve horizontally inwards for circular motion.

Mark the circle's horizontal radius before resolving. Geometry determines $r$ and the angles of strings or reactions. Use limiting friction only when the speed is at a limiting value; otherwise the friction magnitude is an unknown no greater than $\mu R$. Check that every tension and reaction is non-negative.

A $0.50$ kg particle is a conical pendulum on a string of length $1.2$ m at $30^\circ$ to the vertical. Its circle has radius $r=1.2\sin30^\circ=0.60$ m. From$$T\cos30^\circ=mg,\qquad T\sin30^\circ=mr\omega^2,$$division gives $\tan30^\circ=r\omega^2/g$. Hence $\omega=3.07\,\mathrm{rad\,s^{-1}}$ and $T=mg/\cos30^\circ=5.66$ N using $g=9.8\,\mathrm{m\,s^{-2}}$.

In a vertical circle, speed changes with height while the required inward resultant remains $mv^2/r$. Use conservation of energy to find speed at a position, then use a separate radial equation to find tension or reaction.

Let $\theta$ be measured from the lowest point and let the bottom speed be $u$. The height above the bottom is $r(1-\cos\theta)$. With no non-conservative work,$$v^2=u^2-2gr(1-\cos\theta).$$For a particle on a taut string, resolving inward gives

A string must satisfy $T\ge0$; a smooth contact must satisfy reaction $R\ge0$. At the first loss of constraint, set the relevant force to zero. For a string to complete a full vertical circle, the limiting top condition is $v_{\mathrm{top}}^2=gr$, which with energy gives the minimum bottom speed $u=\sqrt{5gr}$. A rigid rod can push as well as pull, so it does not use the string-tension condition.

A particle on a string enters the bottom with $u^2=6gr$. At the top, energy gives $v^2=6gr-4gr=2gr$, so$$T+mg=2mg\quad\Rightarrow\quad T=mg>0.$$The string remains taut there. At the bottom, $T-mg=6mg$, so $T=7mg$; the larger bottom tension supplies both the inward acceleration and opposition to the downward weight.