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?

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

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

'''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

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

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

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

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

'''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

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

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

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

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

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.

$$\omega^2r = \frac{GM_{star}}{r^2}$$

$$\omega^2r^3 = GM_{star}$$

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

$$\frac{4\pi^2r^3}{T^2} = GM_{star}$$

$$T^2 = \frac{4\pi^2r^3}{GM_{star}}$$

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).