3.6.1.1
Solving circular motion problems: equate centripetal force to the real force
Circular Motion — AQA A-Level Physics
- Identify which real force provides the centripetal forceThe resultant force directed towards the centre of a circular path that causes an object to move in a circle. It is not a separate force but the net force providing circular motion..
- Set that force equal to mv^2/r (or mr omega^2) and solve.
- For a mass orbiting a planet: gravitational $force = centripetal$ force.
- For a ball on a string: $tension = centripetal$ force.
Worked Example
A 300 g ball on a string of length 0.8 m. The string can withstand a maximum force of 60 N. Calculate the maximum speed of the ball.
Show Solution
1
List known values
- Mass: $m = 300 \text{ g} = 0.30 \text{ kg}$
- Radius: $r = 0.8 \text{ m}$
- Maximum force: $F = 60 \text{ N}$
2
Rearrange F = mv^2/r for v
$$v_{\max} = \sqrt{\frac{rF_{\max}}{m}}$$
3
Substitute values
$$v_{\max} = \sqrt{\frac{0.8 \times 60}{0.30}} = \sqrt{160} = 12.6 \text{ m s}^{-1}$$
Answer
$v_{\max} = 12.6$ m s$^{-1}$