Mechanics Circular Motions

Mechanics Circular Motions are handled by resolving forces horizontally and vertically in conjunction with the tension of the string, normal reactions, circular motion, and the particle’s mass.

Worked Examples of Mechanics Circular Motions

A particle of mass $m$ is attached to one end of a string of length $R$. The other end of the string is fixed at height $2h$ above the centre of a sphere of radius $R$. The particle moves in a circle of radius $r$ on the surface of the sphere and has constant angular velocity $\omega > 0$. The string makes an angle of $\theta$ with the vertical.
Three forces act on the particle: the tension force $F$ of the string, the normal reaction force $N$ to the surface of the sphere, and the gravitational force $mg$.

(a)    By resolving the forces horizontally and vertically show that

$F \sin \theta – N \sin \theta = m \omega ^2 r \\ F \cos \theta + N \cos \theta = mg$

Resolve forces for Mechanics Circular Motions.
\begin{aligned} \displaystyle \require{color} F \sin \theta &= N \cos \theta + m \omega r^2 r &\color{red} \text{horizontally} \\ \therefore F \sin \theta – N \sin \theta &= m \omega ^2 r &\color{red} (1) \\ F \sin \theta + N \sin \theta &= mg &\color{red} \text{vertically} \\ \therefore F \cos \theta + N \cos \theta &= mg &\color{red} (2) \\ \end{aligned} \\

(b)    Show that $\displaystyle N = \frac{1}{2} mg \sec \theta – \frac{1}{2} m \omega r \ \text{cosec} \ \theta$.

\begin{aligned} \displaystyle \require{color} \frac{F \sin \theta}{\sin \theta} – \frac{N \sin \theta}{\sin \theta} &= \frac{m \omega ^2 r}{\sin \theta} &\color{red} (1) \div \sin \theta \\ F – N &= m \omega ^2 r \ \text{cosec} \ \theta &\color{red} (3) \\ \frac{F \cos \theta}{\cos \theta} + \frac{N \cos \theta}{\cos \theta} &= \frac{mg}{\cos \theta} &\color{red} (2) \div \cos \theta \\ F + N &= mg \sec \theta &\color{red} (4) \\ 2N &= mg \sec \theta – m \omega^2 r \ \text{cosec} \ \theta &\color{red} (4) – (3) \\ \therefore N &= \frac{1}{2} mg \sec \theta – \frac{1}{2} m \omega^2 r \ \text{cosec} \ \theta \\ \end{aligned} \\

(c)    Show that the particle remains in contact with the sphere if $\displaystyle \omega \le \sqrt{\frac{g}{h}}$.

$\newcommand\ddfrac[2]{\frac{\displaystyle #1}{\displaystyle #2}}$
The particle stays in contact with the sphere $N \ge 0$.
\begin{aligned} \displaystyle \require{color} \frac{1}{2} mg \sec \theta – \frac{1}{2} m \omega^2 r \ \text{cosec} \ \theta &\ge 0 \\ mg \sec \theta – m \omega^2 r \ \text{cosec} \ \theta &\ge 0 \\ – m \omega^2 r \ \text{cosec} \ \theta &\ge – mg \sec \theta \\ m \omega^2 r \ \text{cosec} \ \theta &\le mg \sec \theta \\ \omega^2 &\le \ddfrac{mg \sec \theta}{mr \ \text{cosec} \ \theta} \\ \omega^2 &\le \frac{mg \times \ddfrac{R}{h}}{mr \times \ddfrac{R}{r}} &\color{red} \sec\theta = \frac{R}{h}, \text{cosec} \ \theta = \frac{R}{r}\\ \omega^2 &\le \frac{g}{h} \\ \therefore \omega &\le \sqrt{\frac{g}{h}} \\ \end{aligned} \\