3.3 Circular motion

$$Angular speed is the rate of angular displacement:$$\omega=\frac{d\theta}{dt},\qquad \omega=\frac{\theta}{t}\text{ for constant }\omega.$$With angles in radians,$$v=r\omega,\qquad \omega=\frac{2\pi}{T}=2\pi f.Its unit is rad s$^{-1}$.

A disc turns at $120$ revolutions per minute. Then $f=2\text{ Hz}$ and\omega=2\pi(2)=4\pi\text{ rad s}^{-1}.A point $0.30\text{ m}$ from the axis hasv=r\omega=0.30(4\pi)=1.2\pi\text{ m s}^{-1}. $$$$

$$Its magnitude is$$a=r\omega^2=\frac{v^2}{r}.$$The resultant of the real force components in the inward radial direction must therefore be$$F_{\text{in,resultant}}=ma.$$Proof of the acceleration formulas is not required.$$

A $0.50\text{ kg}$ particle moves at $6.0\text{ m s}^{-1}$ in a circle of radius $2.0\text{ m}$. Thena=\frac{6.0^2}{2.0}=18\text{ m s}^{-2},\qquad F_{\text{in,resultant}}=0.50(18)=9.0\text{ N}. $$$$

For a conical pendulum of string length $l$, let the string make angle $\theta$ with the downward vertical. Its circular radius is $r=l\sin\theta$. If the tension is $T$,T\cos\theta=mg,\qquad T\sin\theta=m\omega^2r.$$$$

$$Dividing the equations gives$$\tan\theta=\frac{\omega^2r}{g}=\frac{\omega^2l\sin\theta}{g},$$so$$\omega^2=\frac{g}{l\cos\theta},\qquad T=\frac{mg}{\cos\theta}.A real taut-string model requires $T>0$ and $0<\cos\theta\le 1$.

For a mass on a light string of radius $r$, let $\theta$ be measured from the downward vertical and let the bottom speed be $u$. At angle $\theta$,v^2=u^2-2gr(1-\cos\theta),$$and inward radial balance gives$$T-mg\cos\theta=\frac{mv^2}{r},\qquad T=m\left(\frac{v^2}{r}+g\cos\theta\right).$$$$

At the top, $\theta=\pi$ andT+mg=\frac{mv_{\text{top}}^2}{r}.A string just remains taut when $T=0$, so $v_{\text{top}}^2=gr$. Energy from bottom to top gives $u^2=v_{\text{top}}^2+4gr$, hence the minimum bottom speed for a complete circle isu_{\min}=\sqrt{5gr}. $$$$