2022 A/L Physics Past Paper · Question 25 (English Medium)

Answer: (2) $\sqrt{gR}$

Given information

What we know:

The track is frictionless. The circular part has radius $R$, and $B$ is the highest point of the circle.

At $B$ the ball is on the inner surface of the track, so the track pushes it towards the centre of the circle, which is downwards.

"Just touches" means the ball only touches the track. It does not press on it, so the force from the track is $N = 0$.

The question gives no numbers. So we keep symbols, and check with sample values $g = 10\ \text{m s}^{-2}$ and $R = 0.4\ \text{m}$.

What the question asks

We want: the speed $v$ of the ball at the point $B$.

What each letter means:

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

$R$ ($\text{m}$): the radius of the circular part.

$v$ ($\text{m s}^{-1}$): the speed of the ball at $B$.

$N$ ($\text{N}$): the force the track exerts on the ball at $B$.

$g$ ($\text{m s}^{-2}$): the acceleration due to gravity.

Relevant theory

Since a body is moving with an acceleration only under the action of an external force, a body performing circular motion is doing so under the action of an external force acting in the direction of the centripetal acceleration. This force is thus known as "centripetal force" and is given by. $F = ma$ $$F = \frac{mv^2}{r} = mr\omega^2$$

For example, swing a bucket of water round in a vertical circle. At the top the water stays in the bucket if the bucket moves fast enough. Swing it just slowly enough and the bucket stops pressing on the water at the top. The weight of the water alone is then exactly what keeps it on the circle. That is the slowest swing that still goes round.

At $B$ two forces point towards the centre: the weight $mg$ and the push $N$ of the track. Together they supply the centripetal force.

$$mg + N = \dfrac{mv^2}{R}$$

In words: the two forces towards the centre add up to $\dfrac{mv^2}{R}$.

The ball just touches, so $N = 0$.

$$mg = \dfrac{mv^2}{R}$$

In words: the weight alone supplies the centripetal force.

Required equations

$mg + N = \dfrac{mv^2}{R}$

$N = 0$

$v = \sqrt{gR}$

Applying it to this question

$\sqrt{\text{m s}^{-2} \times \text{m}} = \text{m s}^{-1}$, which is a speed. The mass $m$ cancels, so the answer does not depend on it.

$10$

$a = \dfrac{v^2}{R} = g$

$4$

$v^2 = gR = 10 \times 0.4$

$2$

$v = \sqrt{gR} = \sqrt{4}$

Reasoning and calculation

The speed does not depend on the mass of the ball. It depends only on $g$ and $R$.

So the speed at $B$ is $v = \sqrt{gR}$. That is option (2).

Checks

$v * v / r$

Does it make sense?

At this speed the acceleration towards the centre equals $g$, exactly what the weight alone can give.

A smaller speed would need less than $g$. The ball would drop away from the track. A larger speed would need more than $g$, and the track would have to push on the ball.

Conclusion

$v = \sqrt{gR}$

At $B$ the weight alone supplies the centripetal force, so $mg = \dfrac{mv^2}{R}$ and $v = \sqrt{gR}$.

The answer is (2).

"Just touches" means the force from the track is zero. Only the weight is left to supply the centripetal force.

Put $mg = \dfrac{mv^2}{R}$ at the top of the circle.

Exam tip