3.5 Circular motion — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. In every problem, name the force that provides the centripetal force.

Warm-up: remember

A car of mass 1200 kg accelerates at $3.0\,\text{m}\,\text{s}^{-2}$ . Find the resultant force.
Write $90^\circ$ in radians.
A ball of mass 0.40 kg moves at $5.0\,\text{m}\,\text{s}^{-1}$ . Find its kinetic energy.

Answers

$F = ma = 1200 \times 3.0 = 3600\,\text{N}$ .
$90 \times \dfrac{\pi}{180} = \dfrac{\pi}{2} = 1.57\,\text{rad}$ .
$E_k = \tfrac{1}{2} \times 0.40 \times 5.0^2 = 5.0\,\text{J}$ .

If question 1 goes wrong, use $F = ma$ . If question 2 goes wrong, multiply the degrees by $\dfrac{\pi}{180}$ . If question 3 goes wrong, use $E_k = \tfrac{1}{2}mv^2$ .

Practise with help

1. Write $60^\circ$ and $200^\circ$ in radians.

Hint

Multiply the angle in degrees by $\dfrac{\pi}{180}$ .

Next step

For $60^\circ$ this gives $\dfrac{60\pi}{180}$ , which simplifies. For $200^\circ$ it gives $\dfrac{200\pi}{180}$ .

Full working

\begin{aligned} 60^\circ &= \frac{60\pi}{180} = \frac{\pi}{3} = 1.05\,\text{rad} \\ 200^\circ &= \frac{200\pi}{180} = \frac{10\pi}{9} = 3.49\,\text{rad} \end{aligned}

The radian is a pure number, so there is no unit to write after it.

2. A wheel of radius 0.30 m rotates at 4.0 revolutions per second. Find its angular velocity and the speed of a point on the rim.

Hint

The frequency is 4.0 Hz, and $\omega = 2\pi f$ . The rim speed is $v = \omega r$ .

Next step

Find $\omega = 2\pi \times 4.0$ first, then multiply by the radius.

Full working

\begin{aligned} \omega &= 2\pi f = 2\pi \times 4.0 = 25.1\,\text{rad}\,\text{s}^{-1} \\ v &= \omega r = 25.1 \times 0.30 = 7.5\,\text{m}\,\text{s}^{-1} \end{aligned}

A point on the rim moves at $7.5\,\text{m}\,\text{s}^{-1}$ , and a point nearer the centre moves more slowly.

3. A car of mass 1400 kg goes round a bend of radius 50 m at a constant $15\,\text{m}\,\text{s}^{-1}$ . Find the centripetal acceleration and the friction force on the car.

Hint

The friction force provides the centripetal force. Use $a = \dfrac{v^2}{r}$ , then $F = ma$ .

Next step

Find $a = \dfrac{15^2}{50}$ first, then multiply by the mass.

Full working

\begin{aligned} a &= \frac{v^2}{r} = \frac{15^2}{50} = 4.5\,\text{m}\,\text{s}^{-2} \\ F &= ma = 1400 \times 4.5 = 6300\,\text{N} \end{aligned}

The friction force is $6300\,\text{N}$ towards the centre of the bend. On ice, the friction could not reach this size, and the car would slide outwards.

4. A ball of mass 0.25 kg is whirled in a horizontal circle of radius 0.60 m on a string. The maximum tension the string can take is 6.0 N. Find the maximum speed of the ball.

Hint

At the maximum speed the tension equals the centripetal force, so $T = \dfrac{mv^2}{r}$ .

Next step

Rearrange to $v^2 = \dfrac{Tr}{m}$ and substitute the values.

Full working

\begin{aligned} v^2 &= \frac{Tr}{m} = \frac{6.0 \times 0.60}{0.25} = 14.4 \\ v &= 3.8\,\text{m}\,\text{s}^{-1} \end{aligned}

If the ball moves faster than $3.8\,\text{m}\,\text{s}^{-1}$ , the tension needed would pass 6.0 N and the string would break.

5. A bucket of water is swung in a vertical circle of radius 0.80 m. Find the minimum speed at the top for the water to stay in the bucket. Take $g = 9.81\,\text{m}\,\text{s}^{-2}$ .

Hint

At the minimum speed the water is held by its own weight alone, so $mg = \dfrac{mv^2}{r}$ .

Next step

Cancel the mass and rearrange to $v^2 = gr$ .

Full working

\begin{aligned} mg &= \frac{mv^2}{r} \\ v &= \sqrt{gr} = \sqrt{9.81 \times 0.80} \\ &= 2.8\,\text{m}\,\text{s}^{-1} \end{aligned}

At any speed lower than $2.8\,\text{m}\,\text{s}^{-1}$ , the water would leave the bucket at the top.

6. A passenger on a roundabout feels pushed outwards, towards the rim. The passenger says this shows there is an outward force. Explain what is really happening, and name the force that keeps the passenger moving in a circle.

Hint

The passenger is moving in a circle, so the acceleration points towards the centre. Think about what happens to the passenger's body when the roundabout turns.

Next step

Name the real force that points towards the centre, and say what the passenger would do if that force disappeared.

Full working

The passenger is moving in a circle, so the acceleration points towards the centre. The only real horizontal force on the passenger is the friction from the seat, which points towards the centre and provides the centripetal force.

The passenger feels pushed outwards because the body tends to keep moving in a straight line, by Newton's first law. When the roundabout turns, the body continues along the tangent for a moment, and the seat presses inwards on it. The outward feeling is the sensation of the inward push, not a real outward force. If the friction disappeared, the passenger would fly off along the tangent, not radially outwards.

There is no outward centrifugal force on the passenger in the ground frame.