Circular motion

WPH14Topic 4.23 questions

Radians, angular velocity, centripetal acceleration and centripetal force.

Specification points

Be able to express angular displacement in radians and in degrees, and convert between these units
Understand what is meant by angular velocity and be able to use the equations $v = \omega r$ and $T = \frac{2\pi}{\omega}$

ω2π

Be able to use vector diagrams to derive the equations for centripetal acceleration $a = \frac{v^2}{r} = r\omega^2$ , and understand how to use these equations

Understand that a resultant force (centripetal force) is required to produce and maintain circular motion

Be able to use the equations for centripetal force $F = ma = \frac{mv^2}{r} = mr\omega^2$

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Sample questions

Original practice questions written in the style of the exam.

Question 13 marks
A car travels around a level roundabout at a constant speed.

Explain why the car is accelerating, and identify the force that causes this acceleration.
Mark scheme
- The direction of the velocity is continuously changing, so the velocity changes (1)
- A changing velocity means the car is accelerating, towards the centre of the circle (1)
- The centripetal force is provided by friction between the tyres and the road, acting towards the centre (1)
Question 24 marks
A ball of mass $0.30\,\text{kg}$ is whirled in a vertical circle of radius $0.60\,\text{m}$ on the end of a light string.

(a) At the lowest point the ball's speed is $4.0\,\text{m s}^{-1}$ . Calculate the tension in the string at this point.

(b) Calculate the minimum speed the ball must have at the top of the circle for the string to remain taut.
Mark scheme
- At the bottom: $T - mg = \dfrac{mv^2}{r}$ (1)
- (10.9 N) (1)

T = 0.30 \times 9.81 + \dfrac{0.30 \times 4.0^2}{0.60} = 2.94 + 8.0 = 11\,\text{N}

T = 0.30 \times 9.81 + \frac{0.30 \times 4. 0 ^{2}}{0.60} = 2.94 + 8.0 = 11 N

At the top with minimum speed, tension is zero so weight alone provides the centripetal force:

mg = \dfrac{mv^2}{r}

(1)

v = \sqrt{gr} = \sqrt{9.81 \times 0.60} = 2.4\,\text{m s}^{-1}

(1)