(a) State what is meant by centripetal acceleration. (b) An unpowered toy car moves freely along a smooth track that is initially horizontal. The track contains a vertical circular loop around which the car travels, as shown in Fig. 1.1. The mass of the car is $230\text{ g}$ and the diameter of the loop is $62\text{ cm}$. Assume that the resistive forces acting on the car are negligible. (i) State what happens to the magnitude of the centripetal acceleration of the car as it…

(a) Acceleration directed towards the centre of the circular path. · (b)(i) It decreases. · (b)(ii) Because at the top $mv^2/r = mg + N$ and for contact $N>0$, so $v^2/r > g = 9.8\,\text{m s}^{-2}$. · (c) Not in contact at Y. · (d) No; the conclusion is unchanged because mass cancels in the energy and contact equations.

Physics 9702/42 — October/November 2021

Cambridge A-Level · A Level Structured Questions · worked solutions for every part, with the mark scheme

questions

100

marks

120

minutes

Topics Gravitational Fields · Motion in a Circle · Ideal Gases · Oscillations · Communication · Electric Fields · +7 more

Q1Motion in a CircleGravitational FieldsFree sample

(a)

State what is meant by centripetal acceleration.

DifficultyEasy

Worked solution

Answer

Centripetal acceleration is the acceleration of an object moving in a circle that is directed towards the centre of the circular path (responsible for changing the direction of the velocity).

Final answer

Acceleration directed towards the centre of the circular path.

Detailed explanation

Background Concept

For motion in a circle, the velocity is continually changing direction even if the speed is constant. Any change in velocity implies an acceleration.

The acceleration required to keep an object on a circular path is called centripetal acceleration. It always points towards the centre of the circle. Its magnitude is

a_c = \frac{v^2}{r} = r\omega^2

where $v$ is speed, $r$ is radius and $\omega$ is angular speed.

Understanding the Question

You are asked to state what centripetal acceleration means (a definition), not to calculate it.

Approach

Give the key defining feature: direction towards the centre of the circle (and optionally that it causes the direction of the velocity to change).

Step-by-Step Reasoning

In circular motion, the instantaneous direction of motion is tangential.
To bend the path into a circle, the acceleration must be perpendicular to the velocity and point radially inward.
That inward (centre-seeking) acceleration is called centripetal acceleration.

Key Takeaways

Centripetal acceleration is always towards the centre.
It is associated with changing direction of velocity, not necessarily changing speed.

Common Mistakes

Saying the acceleration is “along the direction of motion” (that would be tangential acceleration).
Giving only the formula $v^2/r$ without stating the direction.

Things to Be Careful About

The word “centripetal” refers to direction, not a new type of force; it is the net inward force that produces this acceleration.

Techniques used

state the direction of the acceleration for circular motionlink centripetal acceleration to change in direction of velocity

(b)

An unpowered toy car moves freely along a smooth track that is initially horizontal. The track contains a vertical circular loop around which the car travels, as shown in Fig. 1.1.

Fig. 1.1

The mass of the car is $230\text{ g}$ and the diameter of the loop is $62\text{ cm}$ . Assume that the resistive forces acting on the car are negligible.

(i)

State what happens to the magnitude of the centripetal acceleration of the car as it moves around the loop from X to Y.

DifficultyEasy

Worked solution

Answer

From X (bottom) to Y (top) the car’s speed decreases, so $a_c = v^2/r$ decreases in magnitude.

Final answer

It decreases.

Detailed explanation

Background Concept

The magnitude of centripetal acceleration for circular motion is

a_c = \frac{v^2}{r}

For a fixed loop, $r$ is constant, so $a_c$ depends only on $v^2$ .

Understanding the Question

The car moves freely around a vertical loop from the bottom point X to the top point Y, with negligible resistive forces. You must state how the magnitude of the centripetal acceleration changes from X to Y.

Approach

Decide how the speed changes as the car rises. Then use $a_c \propto v^2$ (since $r$ is constant) to infer how $a_c$ changes.

Step-by-Step Reasoning

From X to Y the car gains gravitational potential energy $mgh$ .
With negligible resistive forces, this gain comes from the car’s kinetic energy, so the speed $v$ decreases as it rises.
Since $r$ is constant for the loop, $a_c = v^2/r$ decreases as $v$ decreases.

Key Takeaways

For a given circle, changes in centripetal acceleration come from changes in speed.
In a vertical loop, speed is generally greatest at the bottom and smallest at the top.

Common Mistakes

Saying centripetal acceleration is constant because the radius is constant.
Confusing centripetal acceleration (radial) with tangential acceleration (along the track).

Things to Be Careful About

The question asks about magnitude (size), not direction. The direction is always towards the centre but rotates as the car goes round.

Techniques used

relate centripetal acceleration to speed using a = v^2/ruse energy/height reasoning qualitatively to compare speeds

(ii)

Explain, if the car remains in contact with the track, why the centripetal acceleration of the car at point Y must be greater than $9.8\text{ m s}^{-2}$ .

DifficultyMedium-Easy

Worked solution

Answer

At Y (top), the required centripetal force is towards the centre (downwards):

\frac{mv^2}{r} = mg + N

If the car remains in contact then $N > 0$ , so

\frac{mv^2}{r} > mg \;\Rightarrow\; \frac{v^2}{r} > g

Hence the centripetal acceleration at Y must be greater than $9.8\ \text{m s}^{-2}$ .

Final answer

Because at the top (mv^2/r = mg + N) and for contact (N>0), so (v^2/r > g = 9.8,\text{m s}^{-2}).

Detailed explanation

Background Concept

In circular motion, the resultant force towards the centre provides the centripetal acceleration:

F_{\text{towards centre}} = ma_c = m\frac{v^2}{r}

For an object constrained to a track, the track exerts a normal reaction force $N$ on the object, perpendicular to the surface. If the object is in contact, $N \ge 0$ (it cannot be negative because the track cannot "pull" the object).

Understanding the Question

At point Y (the top of the vertical loop), you must explain why, if the car stays in contact, its centripetal acceleration there must exceed $9.8\ \text{m s}^{-2}$ . This is a force-balance argument at the top of the loop.

Approach

Consider forces on the car at the top of the loop.
Resolve (take) forces towards the centre.
Use the condition for contact: $N > 0$ .

Step-by-Step Reasoning

At the top of the loop, the centre is below the car, so “towards the centre” is downwards.

diagram-bii-1

Forces on the car at Y:

Weight $mg$ acts downward (towards the centre).
Normal reaction $N$ from the track also acts downward if the car is on the inside of the loop (it pushes the car towards the centre).

So the resultant inward force is $mg + N$ , and Newton’s second law towards the centre gives

mg + N = m\frac{v^2}{r}

If the car remains in contact with the track, then $N > 0$ , hence

mg + N > mg

Therefore

m\frac{v^2}{r} > mg

Divide by $m$ :

\frac{v^2}{r} > g

But $a_c = v^2/r$ , so at Y

a_c > g = 9.8\ \text{m s}^{-2}

Key Takeaways

At the top of a loop, both $mg$ and (if in contact) $N$ act towards the centre.
Contact requires $N \ge 0$ ; the limiting case is $N = 0$ (just about to lose contact).
The condition for contact at the top is $a_c \ge g$ (strictly $> g$ if you insist on $N>0$ ).

Common Mistakes

Using $mv^2/r = mg - N$ with the wrong sign.
Saying $a_c = g$ always at the top; actually $a_c$ depends on speed, and $a_c = g$ is only the limiting case $N=0$ .
Forgetting that the centripetal acceleration is towards the centre (downwards at the top).

Things to Be Careful About

Be clear about the direction you choose as “towards the centre” at the top.
The mark is usually earned by writing $mv^2/r = mg + N$ and using $N>0$ (or $N\ge 0$ ) to conclude $a_c > g$ (or $\ge g$ ).

Techniques used

draw a free-body diagram at the top of a loopapply Newton's second law towards the centreuse the contact condition that the normal reaction is positive

(c)

The initial speed at which the car in (b) moves along the track is $3.8\text{ m s}^{-1}$ .

Determine whether the car is in contact with the track at point Y. Show your working.

DifficultyMedium

Worked solution

Working

Radius of loop:

r = \frac{0.62}{2} = 0.31\ \text{m}

Energy from bottom X to top Y (height gain $2r$ ):

\frac{1}{2}mv_X^2 = \frac{1}{2}mv_Y^2 + mg(2r)

v_Y^2 = v_X^2 - 4gr = (3.8)^2 - 4(9.8)(0.31) = 2.29

\Rightarrow v_Y = 1.51\ \text{m s}^{-1}

Centripetal acceleration at Y:

a_c = \frac{v_Y^2}{r} = \frac{2.29}{0.31} = 7.4\ \text{m s}^{-2}

For contact at Y, need $a_c \ge g = 9.8\ \text{m s}^{-2}$ . Since $7.4 < 9.8$ , the car is not in contact at Y.

Answer

Not in contact at Y.

Final answer

Not in contact at Y.

Detailed explanation

Background Concept

Two physics ideas combine here:

Energy conservation (negligible resistive forces):

\text{loss of KE} = \text{gain of GPE}

From bottom to top of a loop, the height increases by $2r$ .

Condition to maintain contact at the top (point Y): the required inward (downward) centripetal acceleration is

a_c = \frac{v^2}{r}

At the top, the limiting case for contact is when the normal reaction $N=0$ , giving

\frac{v^2}{r} = g

So contact requires $v^2/r \ge g$ .

Understanding the Question

Given:

initial speed on the track (at the bottom entry to the loop) $v_X = 3.8\ \text{m s}^{-1}$ ,
loop diameter $0.62\ \text{m}$ so radius $r = 0.31\ \text{m}$ ,
resistive forces negligible.

You must decide whether, at the top point Y, the car is still pressed against the track (contact) or has lost contact.

Approach

Use conservation of energy from X (bottom) to Y (top) to find the speed $v_Y$ .
Calculate $a_c = v_Y^2/r$ at the top.
Compare with $g$ : if $a_c < g$ then $N$ would have to be negative to supply enough inward force, which is impossible, so the car cannot be in contact.

Step-by-Step Reasoning

1) Find the radius

r = \frac{0.62}{2} = 0.31\ \text{m}

2) Use energy conservation from X to Y
Height gain from bottom to top is $2r$ , so GPE increase is $mg(2r)$ .

\frac{1}{2}mv_X^2 = \frac{1}{2}mv_Y^2 + mg(2r)

Mass cancels, giving

v_Y^2 = v_X^2 - 4gr

Substitute:

v_Y^2 = (3.8)^2 - 4(9.8)(0.31) = 14.44 - 12.15 = 2.29

v_Y = \sqrt{2.29} = 1.51\ \text{m s}^{-1}

3) Calculate centripetal acceleration at Y

a_c = \frac{v_Y^2}{r} = \frac{2.29}{0.31} = 7.4\ \text{m s}^{-2}

4) Check contact condition
For contact at the top, need $a_c \ge g$ .

7.4\ \text{m s}^{-2} < 9.8\ \text{m s}^{-2}

So the car does not have enough speed to provide the required centripetal acceleration using weight plus a non-negative normal reaction. Therefore it loses contact before or at Y.

Key Takeaways

In a vertical loop, speed at the top can be found using energy conservation if friction is negligible.
At the top, maintaining contact requires $v^2/r \ge g$ .
The limiting case $v^2/r = g$ corresponds to $N=0$ .

Common Mistakes

Using height change as $r$ instead of $2r$ (bottom to top).
Forgetting to convert diameter to radius.
Checking contact with $v \ge gr$ instead of $v^2/r \ge g$ .
Treating $a_c$ as constant around the loop.

Things to Be Careful About

Keep the bottom point and top point clear: the car rises by a full diameter.
The mass cancels in the energy equation and in the contact condition; do not let the given mass distract you.
Use consistent units (metres, seconds) throughout.

Techniques used

apply conservation of mechanical energy between two pointscalculate centripetal acceleration using a = v^2/ruse the top-of-loop contact condition v^2/r \ge g

(d)

Suggest, with a reason but without calculation, whether your conclusion in (c) would be different for a car of mass $460\text{ g}$ moving with the same initial speed.

DifficultyMedium-Easy

Worked solution

Answer

No. The speed at Y from energy and the contact condition $v^2/r \ge g$ are independent of mass (the factor $m$ cancels), so doubling the mass does not change whether the car stays in contact.

Final answer

No; the conclusion is unchanged because mass cancels in the energy and contact equations.

Detailed explanation

Background Concept

For this loop problem, two key relationships are used:

Energy conservation from bottom to top:

\frac{1}{2}mv_X^2 = \frac{1}{2}mv_Y^2 + mg(2r)

Contact condition at the top:

mg + N = m\frac{v_Y^2}{r},\quad N \ge 0

Both equations contain a factor of $m$ on every term, so mass often cancels.

Understanding the Question

You are asked (without calculation) whether changing the car mass from $230\ \text{g}$ to $460\ \text{g}$ , while keeping the same initial speed, would change the conclusion about contact at the top.

Approach

Decide whether the relevant conditions depend on $m$ :

Does $v_Y$ found from energy depend on $m$ ?
Does the top contact condition depend on $m$ ?

If neither depends on $m$ , then the conclusion will not change.

Step-by-Step Reasoning

1) Speed at the top from energy

\frac{1}{2}mv_X^2 = \frac{1}{2}mv_Y^2 + mg(2r)

Every term has $m$ , so dividing through by $m$ gives

\frac{1}{2}v_X^2 = \frac{1}{2}v_Y^2 + g(2r)

Thus $v_Y$ depends on $v_X$ , $g$ , and $r$ , but not on $m$ .

2) Contact condition at the top

mg + N = m\frac{v_Y^2}{r}

Again, the centripetal requirement involves $m$ on both sides. The limiting condition $N=0$ gives

mg = m\frac{v_Y^2}{r} \Rightarrow g = \frac{v_Y^2}{r}

So the requirement to remain in contact depends on $v_Y$ , $r$ , and $g$ only, not on $m$ .

Therefore doubling the mass does not change whether the car maintains contact at Y.

Key Takeaways

In many loop-the-loop problems with negligible resistive forces, the motion (speeds) is independent of mass.
The contact condition $v^2/r \ge g$ at the top is also independent of mass.

Common Mistakes

Assuming a heavier car “has more force” so it stays on: weight increases, but the required centripetal force $mv^2/r$ increases in the same proportion.
Thinking the given mass must be used because it is provided.

Things to Be Careful About

This independence holds under the stated assumption of negligible resistive forces. If friction or air resistance were significant and depended on mass differently, then mass could matter.

Techniques used

identify which quantities cancel in the governing equationsuse proportional reasoning to predict the effect of changing masslink contact condition to centripetal force balance

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