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

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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 1.5Hz. 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 days, and the distance from the star to the planet is 70 Gm from the star. What is the mass of the star?

100 days = 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 2000kg 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).