Further Mechanics — Circular and rotational motion: A-level Further Maths OCR revision notes

6.01

Dimensional analysis

Dimensions and dimensional consistency 6.01

Definitions

Dimensions: the expression of a quantity in terms of the base dimensions mass M, length L and time T, written in square brackets, for example $[force] = M L T^{- 2}$ .

Key results

A valid equation is dimensionally consistent: every term has the same dimensions.
Pure numbers, angles and the arguments of functions such as $sin$ , $e^{x}$ and $ln$ are dimensionless.
If a quantity depends on others as a product of powers, equating the powers of M, L and T gives simultaneous equations for the unknown indices.

Method

Write each quantity's dimensions from a defining formula: velocity from distance ÷ time, force from $F = ma$ , energy from force × distance, power from energy ÷ time.
To find a formula, assume $Q = k a^{α} b^{β} c^{γ}$ , substitute dimensions, and equate the indices of M, L and T.

In practiceAssume the period of a pendulum is $T = k m^{α} l^{β} g^{γ}$ . Find $α$ , $β$ and $γ$ .

$T = M^{α} L^{β} (L T^{- 2})^{γ}$ substitute the dimensions
$M : α = 0; T : - 2 γ = 1; L : β + γ = 0$ equate the indices
$α = 0, γ = - \frac{1}{2}, β = \frac{1}{2} ⟹ T = k \frac{l}{g}$ the constant ( $2 π$ ) is not found this way

Notes

Dimensional analysis cannot find dimensionless constants such as the $2 π$ in a pendulum's period; these come from theory or experiment.
A dimensionally consistent equation can still be wrong, but an inconsistent one is certainly wrong: a quick check of any formula you derive.

Worked example

The speed $v$ of a wave on a stretched string depends on the tension $T$ and the mass per unit length $μ$ .

Assuming $v = k T^{a} μ^{b}$ , with $k$ dimensionless, find $a$ and $b$ .

Show worked solution

$[v] = L T^{- 1}$ , $[T] = M L T^{- 2}$ and $[μ] = M L^{- 1}$ .

So:

L T^{- 1} = M^{a + b} L^{a - b} T^{- 2 a}

T: $- 2 a = - 1$ , so:

a = \frac{1}{2}

M: $a + b = 0$ , so:

b = - \frac{1}{2}

Check L: $a - b = 1$ . ✓ So $v = k T / μ$ .

6.02

Work, energy and power

Work, energy and power 6.02

Definitions

Work done by a constant force $F$ moving its point of application a distance $s$ in its own direction: $W = F s$ . For a variable force, $W = \int F d x$ .
Power: the rate of doing work, $P = \frac{d W}{d t} = F v$ for a force $F$ at speed $v$ ; measured in watts.

Key results

Work–energy principle: work done by external forces other than weight $=$ change in kinetic energy $+$ change in gravitational potential energy (+ change in elastic potential energy).
A vehicle working at constant power $P$ has driving force $\frac{P}{v}$ . At maximum speed the acceleration is zero, so $\frac{P}{v} =$ total resistance.

Method

For a vehicle, write $\frac{P}{v} - R - m g sin α = ma$ along the slope, with $R$ the resistance.
Use energy for questions about distance and speed; use $F = ma$ with $P = F v$ for questions about acceleration at an instant.

In practiceA car of mass 1200 kg climbs a slope with $sin α = \frac{1}{20}$ against a resistance of 400 N, its engine working at 30 kW. Find its acceleration at 15 m s $^{- 1}$ .

$F = \frac{P}{v} = \frac{30 000}{15} = 2000 N$ kW to W first
$2000 - 400 - 1200 (9.8) (\frac{1}{20}) = 1200 a$ Newton's second law along the slope
$1012 = 1200 a ⟹ a = 0.843 m s^{- 2}$

Notes

Convert kW to W before substituting.
Work done against friction is $F_{r} s$ only if the friction is constant; otherwise integrate.

Elastic strings and springs 6.02

Definitions

Hooke's law: $T = \frac{λ x}{l}$ , where $λ$ is the modulus of elasticity, $l$ the natural length and $x$ the extension (or compression, for a spring).
Elastic potential energy: $EPE = \frac{λ x ^{2}}{2 l}$ , the work done in stretching.

Key results

EPE is the area under the tension–extension graph, $\int_{0}^{x} \frac{λ s}{l} d s$ .
A string is slack, with zero tension and zero EPE, when its length is less than $l$ ; a spring can be compressed.

Method

Choose a reference level for gravitational PE, then write total energy at two positions and equate (no other forces doing work), or account for the work done against resistance.
At the lowest point of a bounce, the speed is zero, giving a quadratic in the extension.

In practiceA particle of mass 0.5 kg is attached to an elastic string of natural length 1 m and modulus 20 N, and released from rest at the fixed end O. Find its greatest extension ( $g = 9.8$ ).

At the lowest point $v = 0$ ; the particle has fallen $1 + x$ and the extension is $x$ .
$0.5 (9.8) (1 + x) = \frac{20 x ^{2}}{2 ( 1 )}$ GPE lost = EPE gained
$10 x^{2} - 4.9 x - 4.9 = 0 ⟹ x = \frac{4.9 + 220.01}{20} = 0.987 m$ the positive root

Notes

Use the extension $x$ , not the total length, in both $T$ and EPE.
Maximum speed occurs where the acceleration is zero, at the equilibrium position: $\frac{λ x}{l} = m g$ .

Worked examples

Worked example

A car of mass 1200 kg has engine power 24 kW and resistance $15 v$ N.

Find its acceleration at 20 m s⁻¹ on level ground, and its maximum speed up a slope with:

sin α = \frac{1}{20}

Show worked solution

On the level at 20 m s⁻¹: driving force:

= \frac{24000}{20} = 1200

and resistance 300 N, so:

a = \frac{900}{1200} = 0.75

s⁻².

Uphill at maximum speed:

\frac{24000}{v} = 15 v + 1200 (9.8) (\frac{1}{20})

so:

15 v^{2} + 588 v - 24000 = 0

and $v = 24.9$ m s⁻¹.

Worked example

A particle of mass 0.4 kg is attached to one end of a light elastic string of natural length 0.5 m and modulus 20 N, whose other end is fixed at O.

It is released from rest at O.

Find the greatest extension of the string.

Show worked solution

At the lowest point the speed is zero.

Loss of GPE $=$ gain in EPE:

0.4 (9.8) (0.5 + x) = \frac{20 x ^{2}}{2 ( 0.5 )} = 20 x^{2}

So:

20 x^{2} - 3.92 x - 1.96 = 0

giving $x = 0.426$ m (the negative root is rejected).

6.04

Centre of mass

Centres of mass of composite bodies 6.04

Definitions

Centre of mass: the point at which the total weight can be taken to act. For a uniform lamina it is the centroid of its shape.

Key results

Composite body: $\overset{x}{ˉ} = \frac{\sum m _{i} x _{i}}{\sum m _{i}}$ , and similarly for $\overset{y}{ˉ}$ (and $\overset{z}{ˉ}$ ).
Standard results: triangle $\frac{2}{3}$ along each median from the vertex; semicircular lamina $\frac{4 r}{3 π}$ from the diameter; solid hemisphere $\frac{3}{8} r$ and solid cone $\frac{1}{4} h$ from the plane face.
A hole or a removed piece is treated as a negative mass.

Method

Split the body into standard shapes, tabulate mass (or area, or volume, for a uniform body) and the coordinates of each centre.
Use symmetry first: a uniform body's centre of mass lies on every axis of symmetry.

In practiceA uniform L-shaped lamina is the rectangle $0 \leq x \leq 6$ , $0 \leq y \leq 2$ together with $0 \leq x \leq 2$ , $2 \leq y \leq 6$ . Find its centre of mass.

$A : area 12, centre (3, 1); B : area 8, centre (1, 4)$ uniform, so areas stand in for masses
$\overset{x}{ˉ} = \frac{12 ( 3 ) + 8 ( 1 )}{20} = 2.2, \overset{y}{ˉ} = \frac{12 ( 1 ) + 8 ( 4 )}{20} = 2.2$
Check: the L is symmetric in the line $y = x$ , so $\overset{x}{ˉ} = \overset{y}{ˉ}$ , as found.

Notes

Measure every coordinate from the same origin and axes. Mixing references is the most common error.
For a lamina made of different materials, use masses, not areas.

Centres of mass by integration 6.04

Key results

Uniform lamina under $y = f (x)$ , $a \leq x \leq b$ : $\overset{x}{ˉ} = \frac{\int _{a}^{b} x y d x}{\int _{a}^{b} y d x}$ and $\overset{y}{ˉ} = \frac{\int _{a}^{b} \frac{1}{2} y ^{2} d x}{\int _{a}^{b} y d x}$ .
Uniform solid of revolution about the $x$ -axis: $\overset{x}{ˉ} = \frac{\int _{a}^{b} π y ^{2} x d x}{\int _{a}^{b} π y ^{2} d x}$ , with $\overset{y}{ˉ} = 0$ by symmetry.
Standard results: a solid cone is $\frac{1}{4} h$ above its base; a solid hemisphere is $\frac{3}{8} r$ from its plane face; a uniform semicircular lamina is $\frac{4 r}{3 π}$ from its diameter.

Method

Divide the shape into thin strips (or discs) of width $δ x$ , write the mass of each and the position of its own centre, then sum by integration.
Use symmetry to avoid unnecessary integrals.

In practiceFind the centre of mass of the uniform lamina under $y = x^{2}$ for $0 \leq x \leq 2$ .

$A = \int_{0}^{2} x^{2} d x = \frac{8}{3}$
$\overset{x}{ˉ} = \frac{1}{A} \int_{0}^{2} x \cdot x^{2} d x = \frac{4}{8/3} = 1.5$ a strip at $x$ of height $x^{2}$
$\overset{y}{ˉ} = \frac{1}{A} \int_{0}^{2} \frac{1}{2} (x^{2})^{2} d x = \frac{16/5}{8/3} = 1.2$ each strip's own centre is at height $\frac{1}{2} y$

Notes

A strip under the curve has its centre at $(x, \frac{1}{2} y)$ , which is where the $\frac{1}{2} y^{2}$ in $\overset{y}{ˉ}$ comes from.
For a lamina between two curves, use the difference of the two functions as the height of each strip.

Suspended bodies and toppling 6.04

Key results

A body hanging freely from a point P rests with its centre of mass G vertically below P.
A body on a plane is on the point of toppling when G is vertically above the edge it would turn about.
On an inclined plane, a body slides first if $μ < tan α$ at the angle where it would topple; otherwise it topples first.
Hanging from a corner $P$ at the origin, with $G$ at $(\overset{x}{ˉ}, \overset{y}{ˉ})$ , the edge along the $x$ -axis makes an angle $θ$ with the vertical where $tan θ = \frac{y ˉ}{x ˉ}$ ; for $G = (2.2, 2.2)$ , $θ = 4 5^{\circ}$ .
A uniform cuboid of width $w$ and height $h$ on a slope topples about its lower edge when $tan α = \frac{w}{h}$ , and slides first if $μ < \frac{w}{h}$ .

Method

Find G, then draw the line PG; the required angle is between PG and a named edge of the body.
For toppling, take moments about the edge: the weight's moment changes sign as G passes over the edge.

In practiceThe L-shaped lamina above hangs freely from the corner $P (6, 0)$ . Find the angle between its lower edge and the vertical.

$G = (2.2, 2.2), PG = (- 3.8, 2.2)$ PG hangs vertically
$tan θ = \frac{2.2}{3.8} ⟹ θ = 30. 1^{\circ}$ the lower edge runs along $PO = (- 6, 0)$

Notes

State which angle you have found; questions usually ask for the angle between a given edge and the vertical.

Worked examples

Worked example

The region under $y = x$ for $0 \leq x \leq 4$ is rotated through $2 π$ about the $x$ -axis to form a uniform solid.

Find the distance of its centre of mass from the origin.

Show worked solution

\overset{x}{ˉ} = \frac{\int _{0}^{4} π y ^{2} x d x}{\int _{0}^{4} π y ^{2} d x} = \frac{\int _{0}^{4} x ^{2} d x}{\int _{0}^{4} x d x} = \frac{64/3}{8} = \frac{8}{3}

By symmetry the centre of mass is on the $x$ -axis, $\frac{8}{3}$ units from the origin.

Worked example

A uniform solid consists of a cylinder of radius 3 cm and height 6 cm, with a hemisphere of radius 3 cm on top.

Find the height of its centre of mass above the base.

Show worked solution

Volumes: cylinder $54 π$ , hemisphere $18 π$ .

Centres above the base: cylinder 3; hemisphere:

6 + \frac{3}{8} (3) = 7.125

\overset{y}{ˉ} = \frac{54 ( 3 ) + 18 ( 7.125 )}{72} = 4.03

Worked example

The lamina on the sheet (a 6 × 4 rectangle ABCD, with A at (0, 4) and D at (6, 4), and a hole of radius 1 centred at (4, 2)) hangs freely from A.

Find the angle between AD and the vertical.

Show worked solution

\overset{x}{ˉ} = \frac{24 ( 3 ) - π ( 4 )}{24 - π} = 2.85

and $\overset{y}{ˉ} = 2$ , so G is 2.85 along and 2 below A.

AG is vertical, so the angle between AD and the vertical is:

tan^{- 1} \frac{2}{2.85} = 35. 1^{\circ}

6.05

Motion in a circle

Circular motion: the radial equation 6.05

Definitions

Angular speed $ω$ : the rate of change of angle, in rad s⁻¹, with $v = r ω$ .
Radial (centripetal) acceleration: $\frac{v ^{2}}{r} = r ω^{2}$ , directed towards the centre.

Key results

Newton's second law towards the centre: resultant inward force $= \frac{m v ^{2}}{r} = m r ω^{2}$ .
Conical pendulum, string at $θ$ to the vertical: $T cos θ = m g$ and $T sin θ = m r ω^{2}$ .
Banked track at angle $α$ with no friction: $tan α = \frac{v ^{2}}{r g}$ .

Method

Draw the forces, not 'the centripetal force': it is the resultant of the real forces, not an extra one.
Resolve vertically (no vertical acceleration) and horizontally towards the centre.

In practiceA conical pendulum: a particle of mass 0.3 kg on a string 0.8 m long moves in a horizontal circle with the string at $3 0^{\circ}$ to the vertical. Find the tension and the angular speed ( $g = 9.8$ ).

$T cos 3 0^{\circ} = 0.3 (9.8) ⟹ T = 3.39 N$ vertically: no acceleration
$r = 0.8 sin 3 0^{\circ} = 0.4 m$ the radius, not the string length
$T sin 3 0^{\circ} = 0.3 (0.4) ω^{2} ⟹ ω^{2} = 14.1, ω = 3.76 rad s^{- 1}$ horizontally, towards the centre

Notes

For a conical pendulum, $r = l sin θ$ : the radius is not the string length.
With friction on a banked track, find the speed range by taking friction up the slope (minimum speed) and down the slope (maximum speed).

Motion in a vertical circle 6.05

Key results

Energy, with $θ$ measured from the lowest point: $v^{2} = u^{2} - 2 g a (1 - cos θ)$ .
Radial equation on a string: $T - m g cos θ = \frac{m v ^{2}}{a}$ .
A string must stay taut, $T \geq 0$ ; to reach the top needs $v_{top}^{2} \geq g a$ , so $u^{2} \geq 5 g a$ .
A light rod can push as well as pull, so a particle on a rod completes the circle if $v_{top} \geq 0$ , that is $u^{2} \geq 4 g a$ .

Method

Use conservation of energy to find $v^{2}$ at a general position.
Substitute into the radial equation to find the tension or reaction there.
Set the tension (or reaction) to zero to find where the string goes slack or the particle leaves a surface.

In practiceA particle on a light string of length $a$ is projected from the lowest point with speed $u$ . Find the least $u$ for complete circles.

$v^{2} = u^{2} - 2 g a (1 - cos θ)$ energy, $θ$ from the lowest point
$top (θ = π) : T + m g = \frac{m ( u ^{2} - 4 g a )}{a}$ radial equation, with $v^{2}$ substituted
$T \geq 0 ⟺ u^{2} \geq 5 g a ⟹ u_{m i n} = 5 g a$ the string must stay taut at the top

Notes

Tangential acceleration $g sin θ$ changes the speed but never appears in the radial equation.
After a string goes slack, the particle moves as a projectile until the string becomes taut again.

Leaving a circular surface 6.05

Key results

On the outside of a smooth sphere of radius $a$ : $m g cos θ - R = \frac{m v ^{2}}{a}$ , with $θ$ from the upward vertical.
Starting from rest at the top, the particle leaves the surface where $cos θ = \frac{2}{3}$ , about $48. 2^{\circ}$ from the top, independent of $a$ , $m$ and $g$ .

Method

Write $v^{2}$ at a general angle $θ$ from the top by energy.
Substitute into the radial equation $m g cos θ - R = \frac{m v ^{2}}{a}$ , and set $R = 0$ for the point where the particle leaves the surface.

In practiceA particle is projected at 1 m s $^{- 1}$ along the top of a smooth sphere of radius 1 m. Where does it leave the surface? ( $g = 9.8$ )

$v^{2} = 1 + 2 (9.8) (1) (1 - cos θ)$ energy, from the top
$R = 0 : 9.8 cos θ = \frac{v ^{2}}{1} = 1 + 19.6 - 19.6 cos θ$ radial equation
$29.4 cos θ = 20.6 ⟹ cos θ = 0.701, θ = 45. 5^{\circ}$ sooner than $48. 2^{\circ}$ : it starts faster

Notes

The reaction from a surface cannot pull, so contact ends when $R$ would become negative.
On the inside of a circular track (a loop), the reaction acts towards the centre and the condition at the top becomes $R \geq 0$ , as for a string.

Worked examples

Worked example

A particle of mass 0.5 kg on a light string of length 0.8 m moves in a vertical circle, with speed 6 m s⁻¹ at the lowest point.

Show that it does not complete the circle, and find where the string goes slack. ( $g = 9.8$ m s⁻²)

Show worked solution

With $θ$ from the lowest point:

v^{2} = 36 - 2 (9.8) (0.8) (1 - cos θ) = 20.32 + 15.68 cos θ

Then:

T = \frac{0.5 v ^{2}}{0.8} + 0.5 (9.8) cos θ = 12.7 + 14.7 cos θ

At the top, $cos θ = - 1$ gives:

T = - 2.0 < 0

impossible, so the string goes slack first.

$T = 0$ when $cos θ = - 0.864$ , $θ = 149. 8^{\circ}$ , at a height:

0.8 (1 + 0.864) = 1.49

above the lowest point, with speed:

6.77 = 2.60

s⁻¹.

Worked example

A conical pendulum has a string 0.5 m long making 30° with the vertical.

Find its angular speed.

Show worked solution

r = 0.5 sin 3 0^{\circ} = 0.25

Dividing $T sin θ = m r ω^{2}$ by $T cos θ = m g$ :

ω^{2} = \frac{g tan 3 0 ^{\circ}}{r} = \frac{9.8 ( 0.577 )}{0.25} = 22.6

so $ω = 4.76$ rad s⁻¹.

6.06

Further dynamics and kinematics

Variable forces 6.06

Definitions

Variable force: a force that changes with time, position or velocity, so the acceleration is not constant and the suvat equations do not apply.

Key results

Newton's second law with $a = \frac{d v}{d t} = v \frac{d v}{d x} = \frac{d ^{2} x}{d t ^{2}}$ .
Force a function of $t$ : use $m \frac{d v}{d t} = F (t)$ . Force a function of $x$ : use $m v \frac{d v}{d x} = F (x)$ . Force a function of $v$ : separate the variables in whichever form gives the quantity asked for.
In two dimensions, with $r$ , $v$ and $a$ functions of $t$ , differentiate or integrate each component separately.

Method

Choose $\frac{d v}{d t}$ if the question involves time, and $v \frac{d v}{d x}$ if it involves distance.
Separate the variables, integrate, and use the initial conditions to find the constant.

In practiceA particle of mass 2 kg moves in a straight line against a resistance of $4 v$ N, starting at 6 m s $^{- 1}$ . Find $v$ in terms of $x$ , and how far it travels.

$2 v \frac{d v}{d x} = - 4 v$ $a = v \frac{d v}{d x}$ , because distance is involved
$\frac{d v}{d x} = - 2 ⟹ v = 6 - 2 x$ using $v = 6$ when $x = 0$
$v \to 0 as x \to 3 m$ but only as $t \to \infty$ , since $v = 6 e^{- 2 t}$

Notes

A resistance acts against the motion, so it appears with a negative sign when the positive direction is the direction of motion.
Check the answer for physical sense. Under a resistance $k v$ alone, $v = v_{0} - \frac{k}{m} x$ : the particle stops after a finite distance $\frac{m v _{0}}{k}$ , but only as $t \to \infty$ . Under $k v^{2}$ alone, $v = v_{0} e^{- k x / m}$ : it never stops, and the distance travelled grows without limit.

Worked example

A particle of mass 0.5 kg moves in a straight line against a resistance of $0.1 v^{2}$ N, with no other horizontal force.

Its initial speed is 20 m s⁻¹.

Find its speed after it has travelled 10 m.

Show worked solution

Distance is involved, so:

0.5 v \frac{d v}{d x} = - 0.1 v^{2}

giving:

\frac{1}{v} d v = - 0.2 d x

Integrating,

ln v = - 0.2 x + c

and $v = 20$ at $x = 0$ , so:

v = 20 e^{- 0.2 x}

At $x = 10$ :

v = 20 e^{- 2} = 2.71

s⁻¹.