2024 A/L Physics Past Paper · Question 15 (English Medium)

Answer: (4) $40\ \text{cm}$

Given information

This question is about a small coin riding on a spinning turntable.

To go round in a circle, the coin needs a force pulling it towards the centre. Here that force is friction from the turntable.

The further the coin is from the centre, the bigger the force it needs. Friction has a limit.

What we must find: the largest distance from the centre at which the coin stays without sliding.

How we find it: change $30$ rpm into angular velocity. Then set the force the coin needs equal to the largest friction.

The turntable is horizontal. The coefficient of static friction is $\mu=0.36$. It turns at $30$ revolutions per minute. Take $\pi=3$ and $g=10\ \text{m s}^{-2}$.

What the question asks

$m$ ($\text{kg}$): mass of the coin.

$r$ ($\text{m}$): distance of the coin from the centre.

$\omega$ ($\text{rad s}^{-1}$): angular velocity of the turntable.

$\mu$: coefficient of static friction.

$R$ ($\text{N}$): normal contact force on the coin.

$g$ ($\text{m s}^{-2}$): gravitational acceleration.

Relevant theory

This force is thus known as "centripetal force" and is given by. $F = ma$ $$F = \frac{mv^2}{r} = mr\omega^2$$

limiting friction between surfaces $\propto$ normal contact force between surfaces

$F \propto R$ $F = \mu R$

$\mu$ is called the coefficient of static friction and is roughly constant for a particular pair of surfaces.

For no slipping of the coin Centripeal force ≤ Limiting frictional force $mr\omega^2 \leq \mu R$

What is happening: the coin goes round with the turntable. Its velocity keeps changing direction, so it needs a force towards the centre, $mr\omega^2$.

Only friction can give that force. Friction can grow up to $\mu R$, and no more. The turntable is horizontal, so $R=mg$.

The needed force grows with $r$. At one distance it reaches the friction limit. Beyond that, the coin slides.

For example, put a small bag on a spinning playground roundabout. Near the centre, it stays where it is.

Move it out towards the edge, and spin at the same rate. It needs a bigger pull to keep going round.

At some point the grip is not enough, and the bag slides off.

In words: at a fixed spin, the needed force grows with distance. Friction sets the largest distance.

Required equations

$mr\omega^2\le\mu R$

$mr\omega^2=\mu mg$

$r=\dfrac{\mu g}{\omega^2}$

Applying it to this question

$0.5$

$f=\dfrac{30}{60}$

$\omega=2\pi f$

$\omega=2\times3\times0.5$

$3$

$\omega$

$mr\omega^2=\mu mg$

$r\omega^2=\mu g$

$r=\dfrac{\mu g}{\omega^2}$

$r=\dfrac{0.36\times10}{3^2}$

$r=\dfrac{3.6}{9}$

$0.4$

$r$

$40$

Reasoning and calculation

Inside $40\ \text{cm}$, friction can give the force the coin needs. Beyond it, it cannot. So $40\ \text{cm}$ is the largest distance. This is option (4).

$4\ \text{cm}$ comes from $\mu/\omega^2=0.36/9=0.04$, leaving out $g$. The friction limit is $\mu mg$, so we must also multiply by $g=10\ \text{m s}^{-2}$.

$12\ \text{cm}$ comes from $\mu g/30=3.6/30$, using the number $30$ from rpm in place of $\omega^2$. The rate must first be changed to $\omega=3\ \text{rad s}^{-1}$, and then squared.

$36\ \text{cm}$ comes from $\omega^2=\pi^2=10$. The question says $\pi=3$, so $\omega^2=9$. Dividing $3.6$ by $9$ gives $0.4\ \text{m}$.

$72\ \text{cm}$ is too far. There the needed force per kilogram is $r\omega^2=0.72\times9\approx6.5\ \text{N}$, but friction can give only $\mu g=3.6\ \text{N}$. The coin would slide.

Checks

$0.36*10/(2*3*30/60)^2$

$r\omega^2=0.4\times9=3.6$

$\mu g=0.36\times10=3.6$

$r\omega^2=\mu g$

In $r=\mu g/\omega^2$, the unit is $\text{m s}^{-2}$ divided by $\text{s}^{-2}$, which leaves $\text{m}$. The mass $m$ cancels, so any small coin gives the same distance.

A rougher surface (larger $\mu$) lets the coin sit further out. A faster turntable (larger $\omega$) brings the limit closer in. Both trends follow from $r=\mu g/\omega^2$.

Conclusion

$r=40\ \text{cm}$

The largest distance is $40\ \text{cm}$. The answer is (4).