Horizontal Circular Motion

Topic

AS June 2025 Q4

EdexcelCurrent spec12 marksHorizontal Circular Motion

4. A car is travelling round a circular track. The car moves with constant speed in a horizontal circle of radius \(r\).

In an initial model,

the car and driver are modelled as a single particle
the track is modelled as being rough, so that there is sideways friction between the tyres of the car and the track with coefficient of friction \(\mu\)
the track is modelled as being horizontal

Using this model, the maximum speed at which the car can move round the circle of radius \(r\) without slipping sideways is \(\dfrac{1}{2}\sqrt{gr}\).

(a) Show that \(\mu = \dfrac{1}{4}\) (5)

In a refined model,

the car and driver are modelled as a single particle
the track is modelled as being rough, so that there is sideways friction between the tyres of the car and the track with coefficient of friction \(\dfrac{1}{4}\)
the track is modelled as being banked at an angle \(\theta\) to the horizontal

Using this model, the minimum speed at which the car can move round the circle of radius \(r\) without slipping sideways is \(\sqrt{\dfrac{4rg}{35}}\)

(b) Find the value of \(\tan\theta\) (7)

Mark scheme (a)

Scheme	Marks	AO
\(R = mg\)	B1	1.1b
Equation of motion horizontally	M1	3.4
\(F = m\dfrac{\left(\frac{gr}{4}\right)}{r}\ \left(= \dfrac{1}{4}mg\right)\)	A1	1.1b
Use \(F = \mu R\)	M1	1.2
\(\mu = \dfrac{1}{4}\) *	A1*	2.2a
	(5)

Notes

B1: cao

M1: Correct no. of terms and use of given velocity seen

A1: Correct equation

M1: Use of \(F = \mu R\)

A1*: Correct answer correctly obtained

force diagram: particle on a slope inclined at theta to the horizontal, with R perpendicular to the slope, F up the slope, mg vertically down and the acceleration mv squared over r horizontally towards the centre — Mark scheme (b)

Scheme	Marks	AO

Resolve vertically or perpendicular to slope	M1	3.4
\(R\cos\theta + F\sin\theta = mg\) or \(R - mg\cos\theta = \dfrac{mv^2}{r}\sin\theta\)	A1	1.1b
Equation of motion horizontally or parallel to slope	M1	3.4
\(R\sin\theta - F\cos\theta = m\dfrac{\left(\frac{4gr}{35}\right)}{r}\) or \(mg\sin\theta - F = \dfrac{mv^2}{r}\cos\theta\)	A1	1.1b
Use of \(F = \dfrac{1}{4}R\)	M1	1.2
Eliminate \(F\) and \(R\) and obtain equation in \(\tan\theta\): \(\dfrac{\tan\theta - \frac{1}{4}}{1 + \frac{1}{4}\tan\theta} = \dfrac{4}{35}\)	M1	3.1b
\(\tan\theta = \dfrac{3}{8}\) \(\left(= \dfrac{51}{136} = 0.375\right)\)	A1	1.1b
	(7)
	(12 marks)

A2 June 2025 Q3

EdexcelCurrent spec7 marksHorizontal Circular Motion

Figure 4: particle P attached by two strings of length 0.6 m to fixed points A and B, with B vertically below A and AB = 0.8 m; P rotates about the vertical AB — Figure 4

A particle \(P\) of mass 0.2 kg is attached to one end of a light inextensible string of length 0.6 m. The other end of the string is attached to the fixed point \(A\).
A second light inextensible string of length 0.6 m has one end attached to \(P\) and the other end attached to the fixed point \(B\).
The point \(B\) is vertically below \(A\) such that \(AB = 0.8\) m, as shown in Figure 4.

The particle moves in a horizontal circle with constant angular speed \(\omega\) radians per second, with both strings taut.

Given that the tension in the string from \(P\) to \(A\) is twice the tension in the string from \(P\) to \(B\), find the value of \(\omega\). (7)

Diagram: P with tension 2T towards A, tension T towards B, weight 0.2g downwards, and angle theta between each string and the horizontal — Mark scheme

Scheme	Marks	AO
\(\left(\sin\theta = \dfrac{2}{3},\ \ \cos\theta = \dfrac{\sqrt{5}}{3},\ \ r = \dfrac{\sqrt{5}}{5}\right)\)
Resolve vertically	M1	3.4
\(\updownarrow\ 2T\sin\theta = 0.2g + T\sin\theta \qquad (T\sin\theta = 0.2g)\)	A1	1.1b
Equation of motion	M1	3.4
\(0.2r\omega^2 = 2T\cos\theta + T\cos\theta \qquad (0.12\omega^2 = 3T)\)	A1 A1	1.1b 1.1b
Complete strategy to find \(\omega\)	DM1	3.1a
\(\omega^2 = 25T = \dfrac{5g}{\sin\theta} = \dfrac{15g}{2} \qquad \omega = 8.6\ \ (8.57)\)	A1	1.1b
	(7 marks)

A2 June 2024 Q3

EdexcelCurrent spec9 marksHorizontal Circular Motion

Figure 2: hemispherical bowl with rim centre O and radius 10d; P moves in a horizontal circle of radius 8d with centre C vertically below O — Figure 2

Figure 2 shows a hemispherical bowl of internal radius \(10d\) that is fixed with its circular rim horizontal.

The centre of the circular rim is at the point \(O\).

A particle \(P\) moves with constant angular speed on the smooth inner surface of the bowl.

The particle \(P\) moves in a horizontal circle with radius \(8d\) and centre \(C\).

(a) Find, in terms of \(g\), the exact magnitude of the acceleration of \(P\). (6)

The time for \(P\) to complete one revolution is \(T\).

(b) Find \(T\) in terms of \(d\) and \(g\). (3)

Diagram: OC = 6d vertical, CP = 8d horizontal, OP = 10d; reaction R along PO at angle theta to the horizontal and weight mg downwards at P — Mark scheme (a)

Scheme	Marks	AO

Resolve vertically	M1	3.3
\(R\sin\theta = mg\)	A1	1.1b
Horizontal equation of motion	M1	3.3
\(R\cos\theta = ma\ \left(= mr\omega^2\right)\)	A1	1.1b
Solve for \(a\)	DM1	2.1
\(\dfrac{g}{a} = \tan\theta \Rightarrow a = \dfrac{4}{3}g\)	A1	1.1b
	(6)

Mark scheme (b)

Scheme	Marks	AO
\(8d\omega^2 = \dfrac{4}{3}g \Rightarrow \omega = \sqrt{\dfrac{g}{6d}}\)	M1	3.4
Use of \(T = \dfrac{2\pi}{\omega}\)	M1	1.1b
\(T = 2\pi\sqrt{\dfrac{6d}{g}}\) oe OR \(15\sqrt{\dfrac{d}{g}}\) or better	A1	1.1b
	(3)
	(9 marks)

Notes

M1: Use \(a = r\omega^2\) or \(a = \dfrac{v^2}{r}\) to obtain \(\omega\) or \(v\) \(\left(v = \sqrt{\dfrac{32dg}{3}}\right)\)

M1: Complete method to find \(T\)

A1: \(15\sqrt{\dfrac{d}{g}}\) or better - must be in terms of \(d\) and \(g\). Accept \(T = \dfrac{2\pi}{\sqrt{\dfrac{g}{6d}}}\)

AS June 2024 Q2

EdexcelCurrent spec10 marksHorizontal Circular Motion

Figure 2: hollow hemisphere with centre O and radius a, axis vertical; ball B moves in a horizontal circle of centre C and radius r on the inner surface, with C a distance h vertically below O — Figure 2

A thin hollow hemisphere, with centre \(O\) and radius \(a\), is fixed with its axis vertical, as shown in Figure 2.

A small ball \(B\) of mass \(m\) moves in a horizontal circle on the inner surface of the hemisphere. The circle has centre \(C\) and radius \(r\). The point \(C\) is vertically below \(O\) such that \(OC = h\).

The ball moves with constant angular speed \(\omega\)

The inner surface of the hemisphere is modelled as being smooth and \(B\) is modelled as a particle. Air resistance is modelled as being negligible.

(a) Show that \(\omega^2 = \dfrac{g}{h}\) (6)

Given that the magnitude of the normal reaction between \(B\) and the surface of the hemisphere is \(3mg\)

(b) find \(\omega\) in terms of \(g\) and \(a\). (3)

force diagram: ball on the inner surface with mg vertically down and R along the radius towards the centre, at angle theta above the horizontal radius r; depth h and radius a marked — Mark scheme (a)

Scheme	Marks	AO

Resolve horizontally	M1	3.1b
\(mr\omega^2 = R\cos\theta\)	A1	1.1b
Resolve vertically	M1	3.1b
\(mg = R\sin\theta\)	A1	1.1b
\(\dfrac{g}{r\omega^2} = \tan\theta = \dfrac{h}{r}\)	DM1	1.1b
\(\Rightarrow \omega^2 = \dfrac{g}{h}\) *	A1*	2.2a
	(6)

Mark scheme (b)

Scheme	Marks	AO
Resolve horizontally or vertically	M1	2.1
\(ma\cos\theta\,\omega^2 = 3mg\cos\theta \Rightarrow a\omega^2 = 3g\) or \(mg = 3mg\sin\theta \Rightarrow g = 3g \times \dfrac{g}{a\omega^2}\)	A1	1.1b
\(\omega = \sqrt{\dfrac{3g}{a}}\)	A1	1.1b
	(3)

Notes

M1: Resolve and substitute for \(R\) and \(r\). Dimensionally correct. Condone sine/cosine confusion

A1: Correct equation in \(a\), \(g\) and \(\omega\)

A1: Correct only

Mark scheme (c)

Scheme	Marks	AO
Have ignored the dimensions of \(B\)	B1	3.5b
	(1)
	(10 marks)

Notes

B1: Or equivalent or acceptable alternatives e.g. have ignored spin of \(B\)
B0 for weight acts through a point
B0 if any incorrect extras

A2 June 2023 Q6

EdexcelCurrent spec9 marksHorizontal Circular Motion

Figure 4: hollow cone with vertex V pointing downwards, height 0.8 m and base radius 0.6 m; particle P on the inner surface moving in a horizontal circle of radius 0.5 m — Figure 4

A hollow right circular cone, of internal base radius 0.6 m and height 0.8 m, is fixed with its axis vertical and its vertex \(V\) pointing downwards, as shown in Figure 4.

A particle \(P\) of mass \(m\) kg moves in a horizontal circle of radius 0.5 m on the rough inner surface of the cone.

The particle \(P\) moves with constant angular speed \(\omega\ \text{rad s}^{-1}\)

The coefficient of friction between the particle \(P\) and the inner surface of the cone is 0.25

Find the greatest possible value of \(\omega\) (9)

Diagram: P on the cone surface with normal reaction R perpendicular to the surface, friction F down the slope and weight mg; theta is the angle between the surface and the vertical axis — Mark scheme

Scheme	Marks	AO

Resolve vertically	M1	3.4
\(F\cos\theta + mg = R\sin\theta\) \((4F + 5mg = 3R)\)	A1 A1	1.1b 1.1b
Equation for motion towards centre	M1	3.4
\(F\sin\theta + R\cos\theta = m \times 0.5\omega^2\) \(\left(3F + 4R = m \times 2.5\omega^2\right)\)	A1 A1	1.1b 1.1b
At max \(\omega\) \(R + 5mg = 3R \quad \left(R = \dfrac{5mg}{2}\right)\) \(\dfrac{3}{4}R + 4R = m \times 2.5\omega^2 \quad \left(19R = 10m\omega^2\right)\)	M1	1.2
Solve for \(\omega\): \(19 \times \dfrac{5mg}{2} = 10m\omega^2\)	DM1	3.1a
\(\omega^2 = \dfrac{19g}{4} \Rightarrow \max\omega = 6.8\ \ (6.82)\)	A1	2.2a
	(9)
	(9 marks)

AS June 2023 Q3

EdexcelCurrent spec9 marksHorizontal Circular Motion

3. A girl is cycling round a circular track.
The girl and her bicycle have a combined mass of 55 kg.
The coefficient of friction between the track surface and the tyres of the bicycle is \(\mu\).

The track is banked at an angle of \(15^\circ\) to the horizontal.

The girl and her bicycle are modelled as a particle moving in a horizontal circle of radius 50 m
The minimum speed at which the girl can cycle round this circle without slipping is \(4.5\ \text{m s}^{-1}\)

Using the model, find the value of \(\mu\). (9)

force diagram: particle on a slope at 15 degrees to the horizontal, 50 m from the centre of the circle, with R perpendicular to the slope, F up the slope and 55g vertically down — Mark scheme

Scheme	Marks	AO

They need to form two equations, they could be in either order. Mark in the order seen.
Equation of motion horizontally	M1	3.4
\(\dfrac{55 \times 4.5^2}{50} = R\sin 15^\circ - F\cos 15^\circ\)	A1 A1	1.1b 1.1b
Resolve vertically	M1	3.4
\(55g = R\cos 15^\circ + F\sin 15^\circ\)	A1ft A1ft	1.1b 1.1b
Uses \(F = \mu R\)	B1	1.2
Solve for \(\mu\)	dM1	3.1b
\(\mu = 0.22\) or \(\mu = 0.224\)	A1	2.2a
	(9)
	(9 marks)

A2 June 2022 Q4

EdexcelCurrent spec8 marksHorizontal Circular Motion

Figure 2: B vertically above A with AB = 6a; the string runs from B down to the ring R, at angle theta to the vertical, and from R horizontally back to A; R moves with angular speed omega on a horizontal circle with centre A — Figure 2

A small smooth ring \(R\) of mass \(m\) is threaded onto a light inextensible string. One end of the string is attached to a fixed point \(A\) and the other end of the string is attached to the fixed point \(B\) such that \(B\) is vertically above \(A\) and \(AB = 6a\)

The ring moves with constant angular speed \(\omega\) in a horizontal circle with centre \(A\). The string is taut and \(BR\) makes a constant angle \(\theta\) with the downward vertical, as shown in Figure 2.

The ring is modelled as a particle.

Given that \(\tan\theta = \dfrac{8}{15}\)

(a) find, in terms of \(m\) and \(g\), the magnitude of the tension in the string, (3)

(b) find \(\omega\) in terms of \(a\) and \(g\) (5)

force diagram: B above A with AB = 6a and angle theta at B; BR = 10a - r with tension T towards B; AR = r horizontal with tension T towards A; weight mg at R — Mark scheme (a)

Scheme	Marks	AO

Resolve vertically	M1	3.4
\(T\cos\theta = mg\)	A1	1.1b
\(T = \left(\dfrac{mg}{\cos\theta} = \dfrac{6.8mg}{6}\right) = \dfrac{17mg}{15}\)	A1	1.1b
	(3)

Mark scheme (b)

Scheme	Marks	AO
Equation of motion	M1	3.1b
\(mr\omega^2 = T + T\sin\theta \qquad \left(m\times 3.2a\omega^2 = \text{their } T\left(1 + \dfrac{8}{17}\right)\right)\)	A1 A1	1.1b 1.1b
Solves for \(\omega\) or \(\omega^2\)	M1	1.1b
\(\left(\dfrac{r\omega^2}{g} = \dfrac{1 + \sin\theta}{\cos\theta} = \dfrac{6.8 + 3.2}{6},\quad \omega^2 = \dfrac{10g}{6\times 3.2a}\right) \qquad \omega = \sqrt{\dfrac{25g}{48a}} = \dfrac{5}{4}\sqrt{\dfrac{g}{3a}}\)	A1	1.1b
	(5)
	(8 marks)

Notes

M1: Equation for circular motion. Need all terms and dimensionally correct. Condone sin/cos confusion and sign errors.
Any correct form for acceleration

A1: Unsimplified equation with at most one error

A1: Correct unsimplified equation
Allow M1A1A0 for \(mr\omega^2 = T' + (\text{their (a)})\sin\theta\)

M1: Clear attempt to substitute for trig and tension or divide their two equations to solve for \(\omega\) or \(\omega^2\) in terms of \(a\) and \(g\)
Independent M mark but requires an equation using tension and trig.

A1: Any equivalent form
\(0.72\sqrt{\dfrac{g}{a}}\) or better \((0.7216\ldots)\)

AS June 2022 Q3

EdexcelCurrent spec11 marksHorizontal Circular Motion

3. A cyclist is travelling around a circular track which is banked at an angle \(\alpha\) to the horizontal, where \(\tan\alpha = \dfrac{3}{4}\)

The cyclist moves with constant speed in a horizontal circle of radius \(r\).

In an initial model,

the cyclist and her cycle are modelled as a particle
the track is modelled as being rough so that there is sideways friction between the tyres of the cycle and the track, with coefficient of friction \(\mu\),
where \(\mu \lt \dfrac{4}{3}\)

Using this model, the maximum speed that the cyclist can travel around the track in a horizontal circle of radius \(r\), without slipping sideways, is \(V\).

(a) Show that \(V = \sqrt{\dfrac{(3+4\mu)rg}{4-3\mu}}\) (7)

In a new simplified model,

the cyclist and her cycle are modelled as a particle
the motion is now modelled so that there is no sideways friction between the tyres of the cycle and the track

Using this new model, the speed that the cyclist can travel around the track in a horizontal circle of radius \(r\), without slipping sideways, is \(U\).

(b) Find \(U\) in terms of \(r\) and \(g\). (2)

Mark scheme (a)

Scheme	Marks	AO
Resolving vertically	M1	3.4
\(R\cos\alpha - F\sin\alpha = mg\)	A1	1.1b
Equation of motion horizontally	M1	3.4
\(R\sin\alpha + F\cos\alpha = \dfrac{mV^2}{r}\)	A1	1.1b
Use of \(F = \mu R\)	M1	3.4
Solve for \(V\)	M1	3.1b
\(V = \sqrt{\dfrac{(3+4\mu)rg}{4-3\mu}}\) *	A1*	1.1b
	(7)

Notes

M1: Correct no. of terms, dim correct, condone sin/cos confusion and sign errors

A1: Correct equation

M1: Correct no. of terms, dim correct, condone sin/cos confusion and sign errors

A1: Correct equation

M1: Independent but must be used in an equation

M1: Substitute for trig and solve for \(V\). Dependent on preceding M marks.

A1*: Correct given answer correctly obtained

Mark scheme (b)

Scheme	Marks	AO
Use of \(\mu = 0\) oe	M1	2.1
\(U = \sqrt{\dfrac{3rg}{4}}\)	A1	1.1b
	(2)

Notes

M1: If they don’t use \(\mu = 0\), we need to see the first 6 marks from (a), without friction

A1: cao

Mark scheme (c)

Scheme	Marks	AO
Since \(3 + 4\mu \gt 3\) and \(4 - 3\mu \lt 4\) oe	M1	2.1
\(\dfrac{3}{4} \lt \dfrac{3+4\mu}{4-3\mu}\) and hence \(U \lt V\) *	A1*	2.2a
	(2)
	(11 marks)

Notes

M1: Any convincing argument

A1*: Given answer correctly obtained

SC: Allow M1A0 if they work in reverse to show that if \(U \lt V\) then \(\mu \gt 0\) and make an appropriate comment

A2 October 2021 Q4

EdexcelCurrent spec10 marksHorizontal Circular Motion

Figure 4: A vertically above B with AB = l; the string runs from A down to the ring R and from R horizontally to B; R moves on a dashed horizontal circle with centre B — Figure 4

One end of a light inextensible string of length \(2l\) is attached to a fixed point \(A\). A small smooth ring \(R\) of mass \(m\) is threaded on the string and the other end of the string is attached to a fixed point \(B\). The point \(B\) is vertically below \(A\), with \(AB = l\). The ring is then made to move with constant speed \(V\) in a horizontal circle with centre \(B\). The string is taut and \(BR\) is horizontal, as shown in Figure 4.

(a) Show that \(BR = \dfrac{3l}{4}\) (2)

Given that air resistance is negligible,

(b) find, in terms of \(m\) and \(g\), the tension in the string, (4)

Scheme	Marks	AO
\(l^2 + r^2 = (2l - r)^2\), using Pythagoras	M1	1.1b
\(BR = \dfrac{3l}{4}\) *	A1*	1.1b
	(2)

Mark scheme (b)

Scheme	Marks	AO
Resolve vertically	M1	2.1
\(T\cos\alpha = mg\)	A1	1.1b
Overall strategy to solve problem: substitute for \(\cos\alpha\) and solve for \(T\)	M1	3.1b
\(T = \dfrac{5mg}{4}\)	A1	1.1b
	(4)

Notes

M1: Allow sin/cos confusion

A1: Correct equation

M1: Substituting for their trig ratio and solving for \(T\)

A1: cao

Mark scheme (c)

Scheme	Marks	AO
Equation of motion horizontally	M1	2.1
\(T + T\sin\alpha = \dfrac{mV^2}{r}\)	A1	1.1b
Overall strategy to solve problem: substitute for \(T\), \(\sin\alpha\) and \(r\) and solve for \(V\)	M1	3.1b
\(V = \sqrt{\dfrac{3gl}{2}}\)	A1	1.1b
	(4)
	(10 marks)

Notes

M1: Correct no. of terms, dimensionally correct

A1: Correct equation

M1: Substitute for \(T\), \(\sin\alpha\) and \(r\) and solve for \(V\)

A1: cao. Accept other equivalent forms

A2 October 2020 Q5

EdexcelCurrent spec11 marksHorizontal Circular Motion

Figure 4: A vertically above O on a horizontal table with OA = 40 cm; the string AP has length 60 cm and P is on the table — Figure 4

A particle \(P\) of mass 0.75 kg is attached to one end of a light inextensible string of length 60 cm. The other end of the string is attached to a fixed point \(A\) that is vertically above the point \(O\) on a smooth horizontal table, such that \(OA = 40\) cm. The particle remains in contact with the table, with the string taut, and moves in a horizontal circle with centre \(O\), as shown in Figure 4.

The particle is moving with a constant angular speed of 3 radians per second.

(a) Find

(i) the tension in the string,

(ii) the normal reaction between \(P\) and the table. (7)

The angular speed of \(P\) is now gradually increased.

(b) Find the angular speed of \(P\) at the instant \(P\) loses contact with the table. (4)

force diagram: A above O with OA = 0.4 and angle theta at A; AP = 0.6 with tension T; OP = root 20 over 10; reaction R up and weight 0.75g down at P — Mark scheme (a)

Scheme	Marks	AO

Resolve vertically	M1	3.4
\(\updownarrow 0.75g = T\cos\theta + R\) \(\left(\dfrac{3g}{4} = \dfrac{2}{3}T + R\right)\)	A1	1.1b
Equation of motion	M1	3.4
\(\leftrightarrow 0.75\times 0.6\sin\theta\times 9 = T\sin\theta\) \(\left(0.75\times\dfrac{\sqrt{20}}{10}\times 9 = T\times\dfrac{\sqrt{20}}{6}\right)\)	A1	1.1b
Complete strategy to find \(T\) and \(R\)	M1	3.1a
\(T = \dfrac{6\times 0.75\times 9}{10} = 4.05\) (N)	A1	1.1b
\(R = 0.75g - \dfrac{2}{3}T = 4.65\) (N) or 4.7 (N)	A1	1.1b
	(7)

Mark scheme (b)

Scheme	Marks	AO
Use \(R = 0\) to form revised equations	M1	3.4
\(T\cos\theta = 0.75g,\quad T\sin\theta = 0.75\times\dfrac{10\sqrt{20}}{100}\omega^2\) \(\big(\text{or } T\sin\theta = 0.75\times 0.6\sin\theta\times\omega^2\big)\)	A1	1.1b
Complete strategy to find \(\omega\) e.g. \(\Rightarrow \tan\theta = \dfrac{\sqrt{20}\omega^2}{10g} = \dfrac{\sqrt{20}}{4}\)	M1	1.1b
\(\omega = \sqrt{\dfrac{5g}{2}} = 4.95\) (rad/s)	A1	1.1b
	(4)
	(11 marks)

Notes

M1: Correct interpretation of loss of contact

A1: Revised equations

M1: Solve for \(\omega\)

A1: Exact, 4.9 or 4.95 (non-exact answer requires substitution for \(g\)).

AS October 2020 Q2

EdexcelCurrent spec13 marksHorizontal Circular Motion

Figure 2: point A vertically above B with AB = a root 3 shown dashed; the string runs from A to the bead P and from P back to B along the table; P moves in a horizontal circle with centre B — Figure 2

One end of a string of length \(3a\) is attached to a point \(A\) and the other end is attached to a point \(B\) on a smooth horizontal table. The point \(B\) is vertically below \(A\) with \(AB = a\sqrt{3}\)
A small smooth bead, \(P\), of mass \(m\) is threaded on to the string. The bead \(P\) moves on the table in a horizontal circle, with centre \(B\), with constant speed \(U\). Both portions, \(AP\) and \(BP\), of the string are taut, as shown in Figure 2.

The string is modelled as being light and inextensible and the bead is modelled as a particle.

(a) Show that \(AP = 2a\) (2)

(b) Find, in terms of \(m\), \(U\) and \(a\), the tension in the string. (4)

(d) Describe what would happen if \(U^2 \gt ag\sqrt{3}\) (1)

(e) State briefly how the tension in the string would be affected if the string were not modelled as being light. (1)

Mark scheme (a)

Scheme	Marks	AO
\((a\sqrt{3})^2 + (3a - AP)^2 = AP^2\)	M1	1.1b
\(AP = 2a\) *	A1*	1.1b
	(2)

Notes

M1: Use of Pythagoras \(3a^2 + 9a^2 - 6a \times AP + AP^2 = AP^2 \Rightarrow 6a \times AP = 12a^2\)

A1: \(AP = 2a\). GIVEN ANSWER

Mark scheme (b)

Scheme	Marks	AO
Equation of motion horizontally	M1	3.1b
\(T + T \times \dfrac{1}{2} = \dfrac{mU^2}{a}\)	A1 A1	1.1b 1.1b
\(T = \dfrac{2mU^2}{3a}\)	A1	2.2a
	(4)

Notes

M1: Use of horizontal equation to solve the problem, with correct no. of terms etc

A1: Equation with at most one error

A1: Correct equation

A1: Correct answer

Mark scheme (c)

Scheme	Marks	AO
Resolving vertically	M1	3.1b
\(R + T \times \dfrac{\sqrt{3}}{2} = mg\)	A1	1.1b
On the table \(\Rightarrow R \gt 0\)	M1	2.1
\(mg - \dfrac{2mU^2\sqrt{3}}{3a \times 2} \gt 0\)	A1	1.1b
\(U^2 \lt ag\sqrt{3}\) *	A1*	2.2a
	(5)

Notes

M1: Use of vertical resolution to solve the problem, with correct no. of terms etc

A1: Correct equation

M1: Use of \(R \gt 0\)

A1: Correct inequality

A1*: Correctly obtained given answer

Scheme	Marks	AO
Bead would lift off the table	B1	2.4
	(1)

Scheme	Marks	AO
Tension would vary along the string	B1	3.5b
	(1)
	(13 marks)

AS June 2019 Q3

EdexcelCurrent spec9 marksHorizontal Circular Motion

3. A light inextensible string has length \(8a\). One end of the string is attached to a fixed point \(A\) and the other end of the string is attached to a fixed point \(B\), with \(A\) vertically above \(B\) and \(AB = 4a\). A small ball of mass \(m\) is attached to a point \(P\) on the string, where \(AP = 5a\).

The ball moves in a horizontal circle with constant speed \(v\), with both \(AP\) and \(BP\) taut.

The string will break if the tension in it exceeds \(\dfrac{3mg}{2}\)
By modelling the ball as a particle and assuming the string does not break,

(a) show that \(\dfrac{9ag}{4} \lt v^2 \leqslant \dfrac{27ag}{4}\) (7)

(b) find the least possible time needed for the ball to make one complete revolution. (2)

force diagram: A vertically above B with AB = 4a; AP = 5a at angle theta to the vertical with tension T_A; BP = 3a horizontal with tension T_B; weight mg at P; right angle at B — Mark scheme (a)

Scheme	Marks	AO

No vertical motion: \(T_A\cos\theta = mg\)	M1	1.1b
\(T_A = \dfrac{5mg}{4}\)	A1	1.1b
Circular motion: \(T_B + T_A\sin\theta = m \times \dfrac{v^2}{r}\)	M1	3.1b
\(T_B + \dfrac{3}{5}T_A = m\dfrac{v^2}{3a}\)	A1	1.1b
\(T_B \gt 0\ \left(\Rightarrow v^2 \gt \dfrac{9ag}{4}\right)\)	DM1	2.1
\(T_B \leqslant \dfrac{3mg}{2} \Rightarrow m\dfrac{v^2}{3a} - \dfrac{3}{4}mg \leqslant \dfrac{3}{2}mg,\ \left(m\dfrac{v^2}{3a} \leqslant \dfrac{9mg}{4}\right)\)	DM1	2.1
\(\Rightarrow \dfrac{9ag}{4} \lt v^2 \leqslant \dfrac{27ag}{4}\) *	A1*	2.2a
	(7)

Mark scheme (b)

Scheme	Marks	AO
Use \(v^2 = \dfrac{27ag}{4}\) and \(T = \dfrac{2\pi r}{v}\) oe	M1	3.1b
\(T = 4\pi\sqrt{\dfrac{a}{3g}}\)	A1	1.1b
	(2)
	(9 marks)

Notes

M1: Correct method to find \(T\) in terms of \(a\) and (\(g\)) only.
They may sub 9.8 for \(g\) of course

A1: Any equivalent form but no fractions within fractions.
If they use 9.8 for \(g\), the numerical part needs to be to 2 sf or 3sf. i.e \(2.3\sqrt{a}\) or \(2.32\sqrt{a}\)

A2 June 2019 Q1

EdexcelCurrent spec6 marksHorizontal Circular Motion

Figure 1: hemispherical shell of radius a, rim horizontal and uppermost; bead B moves in a horizontal circle whose centre is a/4 below the level of the rim — Figure 1

A hemispherical shell of radius \(a\) is fixed with its rim uppermost and horizontal. A small bead, \(B\), is moving with constant angular speed, \(\omega\), in a horizontal circle on the smooth inner surface of the shell. The centre of the path of \(B\) is at a distance \(\dfrac{1}{4}a\) vertically below the level of the rim of the hemisphere, as shown in Figure 1.

Find the magnitude of \(\omega\), giving your answer in terms of \(a\) and \(g\). (6)

cross-section: reaction R along the radius from B to the centre of the rim at angle theta to the vertical; centre of the circle a/4 below the rim; radius r; weight mg — Mark scheme

Scheme	Marks	AO

\(\updownarrow R\cos\theta = mg\)	M1	3.1b
\(\leftrightarrow R\sin\theta = mr\omega^2\)	M1	3.3
	A1	1.1b
\(\tan\theta = \dfrac{r}{a/4} \qquad \left(\tan\theta = \sqrt{15}\right)\)	B1	1.1b
Complete strategy to find \(\omega\)	M1	3.1b
\(\tan\theta = \dfrac{mr\omega^2}{mg} = \dfrac{4r}{a},\quad \Rightarrow \omega^2 = \dfrac{4g}{a},\quad \omega = 2\sqrt{\dfrac{g}{a}}\)	A1	1.1b
	(6)
	(6 marks)

Scheme	Marks	AO
\(\updownarrow R\cos\theta = mg\)	M1	3.1b
\(\leftrightarrow R\sin\theta = ma\sin\theta\,\omega^2\)	M1	3.3
	A1	1.1b
\(\cos\theta = \tfrac{1}{4}\)	B1	1.1b
Complete strategy to find \(\omega\)	M1	3.1b
\(\Rightarrow R = 4mg,\quad 4mg = ma\omega^2 \quad\Rightarrow\quad \omega = 2\sqrt{\dfrac{g}{a}}\)	A1	1.1b

AS June 2018 Q2

EdexcelCurrent spec9 marksHorizontal Circular Motion

2. A car moves round a bend which is banked at a constant angle of \(\theta^\circ\) to the horizontal.

When the car is travelling at a constant speed of \(80\ \text{km h}^{-1}\) there is no sideways frictional force on the car. The car is modelled as a particle moving in a horizontal circle of radius \(500\ \text{m}\).

(a) Find the value of \(\theta\). (7)

(b) Identify one limitation of this model. (1)

The speed of the car is increased so that it is now travelling at a constant speed of \(90\ \text{km h}^{-1}\)
The car is still modelled as a particle moving in a horizontal circle of radius \(500\ \text{m}\).

force diagram: car on a slope inclined at theta degrees to the horizontal, with weight mg vertically down and normal reaction R perpendicular to the slope — Mark scheme (a)

Horizontal Circular Motion

Edexcel

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