# 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 1.5Hz. What is the tension in the string?

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

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

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

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

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

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

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

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

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

${\displaystyle F_{r}=1.5\times 19620=29430{\mbox{ N}}}$

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

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

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.

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

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

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

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

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

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).