3.5 Circular motion — Practise

Work through these on your own. For each question, name the force that provides the centripetal force before you calculate its size.

The questions and the answers below come from a set that is checked by computer every time the book is built.

Write $120^\circ$ in radians.
A wheel rotates at 2.5 revolutions per second. Find its angular velocity in radians per second.
An object moves in a circle of radius 0.40 m with an angular velocity of $20\,\text{rad}\,\text{s}^{-1}$ . Find its speed.
An object moves at $6.0\,\text{m}\,\text{s}^{-1}$ in a circle of radius 12 m. Find its centripetal acceleration.
An object of mass 0.60 kg moves at $4.0\,\text{m}\,\text{s}^{-1}$ in a circle of radius 0.80 m. Find the centripetal force on it.
A car of mass 1000 kg goes round a bend of radius 80 m at $20\,\text{m}\,\text{s}^{-1}$ . Find the centripetal acceleration and the friction force. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
A ball is whirled in a vertical circle of radius 1.2 m. Find the minimum speed at the top of the circle for the ball to keep moving in the circle. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
A bucket of mass 0.40 kg is swung in a vertical circle of radius 0.50 m. At the bottom of the circle its speed is $4.0\,\text{m}\,\text{s}^{-1}$ . Find the tension in the string. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .
A wheel rotates with an angular velocity of $12\,\text{rad}\,\text{s}^{-1}$ . Find its period of rotation.
The speed of an object in a circle is doubled and the radius is halved. By what factor does the centripetal acceleration change?
A passenger on a roundabout says there is an outward force pushing them outwards. Explain why there is no such force in the ground frame.
A stone is whirled in a horizontal circle on a string. Say how you would find the tension, and name the force that provides the centripetal force.

Answers and concise solutions

1. The angle in radians

Answer: $120 \times \dfrac{\pi}{180} = \dfrac{2\pi}{3} = 2.09\,\text{rad}$ .

2. The angular velocity

Answer: $\omega = 2\pi f = 2\pi \times 2.5 = 15.7\,\text{rad}\,\text{s}^{-1}$ .

3. The speed

Answer: $v = \omega r = 20 \times 0.40 = 8.0\,\text{m}\,\text{s}^{-1}$ .

4. The centripetal acceleration

Answer: $a = \dfrac{v^2}{r} = \dfrac{6.0^2}{12} = 3.0\,\text{m}\,\text{s}^{-2}$ , directed towards the centre of the circle.

5. The centripetal force

Answer: $F = \dfrac{mv^2}{r} = \dfrac{0.60 \times 4.0^2}{0.80} = 12\,\text{N}$ towards the centre.

6. The acceleration and the friction force

Answer: $a = \dfrac{v^2}{r} = \dfrac{20^2}{80} = 5.0\,\text{m}\,\text{s}^{-2}$ and $F = ma = 1000 \times 5.0 = 5000\,\text{N}$ . The friction from the road provides the centripetal force.

7. The minimum speed at the top

Answer: At the minimum speed the weight provides the whole centripetal force: $mg = \dfrac{mv^2}{r}$ , so $v = \sqrt{gr} = \sqrt{9.81 \times 1.2} = 3.4\,\text{m}\,\text{s}^{-1}$ .

8. The tension in the string

Answer: At the bottom the tension acts upwards and the weight acts downwards, so $T - mg = \dfrac{mv^2}{r}$ . Then $T = m\left(g + \dfrac{v^2}{r}\right) = 0.40(9.81 + 32) = 16.7\,\text{N}$ .

9. The period

Answer: $\omega = \dfrac{2\pi}{T}$ , so $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{12} = 0.52\,\text{s}$ .

10. The factor

Answer: The acceleration is $a = \dfrac{v^2}{r}$ . Doubling $v$ multiplies it by 4, and halving $r$ multiplies it by 2 again, so it increases by a factor of 8.

11. Why there is no outward force

The passenger is moving in a circle, so the acceleration points towards the centre and a resultant force in that direction is needed.
The real force is friction from the seat, pointing towards the centre. It provides the centripetal force.
The passenger feels pushed outwards because the body tends to keep moving in a straight line, by Newton first law, and the seat presses inwards on it.

Answer: The acceleration points towards the centre, and friction from the seat provides the centripetal force. No outward force acts. The sensation comes from the body tending to travel in a straight line.

12. The method and the force

Find the speed from the radius and the period or the angular velocity, using v = omega r.
Use F = m v squared over r to find the size of the centripetal force.
The tension in the string provides the centripetal force, and it points towards the centre of the circle.

Answer: Find the speed from $v = \omega r$ , then use $F = \dfrac{mv^2}{r}$ . The tension in the string provides the centripetal force.