A-level Physics (Advancing Physics)/Circular Motion/Worked Solutions

1. A tennis ball of mass 10g is attached to the end of a 0.75m string and is swung in a circle around someone's head at a frequency of 1.5 Hz. What is the tension in the string?

$\omega =2\pi f=2\pi \times 1.5=3\pi {\text{ rad s}}^{-1}$

$F=T=m{\omega }^{2}r=0.01\times (3\pi {)}^2\times 0.75=0.0675{\pi }^2=0.666\text{ N}$

2. A planet orbits a star in a circle. Its year is 100 Earth years, and the distance from the star to the planet is 70 Gm from the star. What is the mass of the star?

100 years = 100 x 365.24 x 24 x 60 x 60 = 3155673600s

$f=\frac{1}{T}=\frac{1}{3155673600}=3.17\times 10^{-10}\text{ Hz}$

$\omega =2\pi f=2\pi \times 3.17\times 10^{-10}=1.99{\text{ nrad s}}^{-1}$

$\frac{GM}{r^2}={\omega }^{2}r$

$M=\frac{{\omega }^2{r}^3}{G}=\frac{(1.99\times 10^{-9}{)}^2\times (70\times 10^9{)}^3}{6.67\times 10^{-11}}=2.04\times 10^{25}\text{ kg}$

3. A 2000 kg car turns a corner, which is the arc of a circle, at 20kmh⁻¹. The centripetal force due to friction is 1.5 times the weight of the car. What is the radius of the corner?

20kmh⁻¹ = 20000 / 3600 = 5.56ms⁻¹

$W=2000\times 9.81=19620\text{ N}$

$F_r=1.5\times 19620=29430\text{ N}$

$29430=\frac{m{v}^2}{r}=\frac{2000\times 5.56^2}{r}=\frac{61728}{r}$

$r=\frac{61728}{29430}=2.1\text{ m}$

This is a bit unrealistic, I know...

4. Using the formulae for centripetal acceleration and gravitational field strength, and the definition of angular velocity, derive an equation linking the orbital period of a planet to the radius of its orbit.

${\omega }^{2}r=\frac{G{M}_{star}}{r^2}$

${\omega }^2{r}^3=G{M}_{star}$

$\omega =\frac{2\pi }{T}$

$\frac{4{\pi }^2{r}^3}{T^2}=G{M}_{star}$

$T^2=\frac{4{\pi }^2{r}^3}{G{M}_{star}}$

So, orbital period squared is proportional to radius of orbit cubed. Incidentally, this is Kepler's Third Law in the special case of a circular orbit (a circle is a type of ellipse).