Mechanics 3 Revision Notes

A2 Mechanics 3 C. Luke

1

Mechanics 3

1 Kinematics 1.1 Acceleration as a Function of Time x - displacement from fixed point O ------- metre (m). v - velocity ---------- metre per second (m/s). a - acceleration --------- metre per second squared (m/s

2).

xO

v

a

P

a = dv/dt v = dx/dt a = d2x/dt

2

The acceleration is caused by a force. What if the force is changing with time? Then the acceleration will depend on time. a = a(t) Integration is the inverse process of differentiation, so

( ) ( )v t a t dt C

where C is a constant of integration. Now clearly v depends on time too.

( ) ( ) 'x t v t dt C

where C a constant of integration.

Example Constant acceleration Suppose a particle is moving with constant acceleration a m/s

2. At time t = 0 its velocity is u m/s

and its displacement from a fixed point O is 0 m. Find it terms of time t expressions for displace and velocity, respectively x(t) and v(t). Solution Formula: a(t) = a Initial conditions: t = 0 v = u and x = 0 Integrate

( ) 1v t adt C a dt C at C


2

Substitute initial condition: t=0, v = u. This implies that C = u. So, v = at+u More normally written as v = u + at. Integrate again:

212( ) ( ) ' ( ) ' 1 ' 'x t v t dt C u at dt C u dt a tdt C ut at C

Substitute initial condition: t=0, x = 0. This implies that C = 0. So,

212( )x t ut at

Summary of method Question typically gives formula for a(t) and initial conditions (values for v(t) and x(t) when t =

0). Note the direction of a(t). To get v(t), integrate a(t) wrt t, including constant of integration (C). Use initial condition to find C. To get x(t), integrate v(t) wrt t, including constant of integration (C). Use initial condition to find C.

1.2 Acceleration as a Function of Position Consider a particle on a spring. As the spring stretches or compresses the force on, and hence the acceleration of, the particle changes with position but not with time itself. Similarly, imagine a meteor falling to earth. As the meteor approaches our planet the force on it increases. Again the acceleration of the meteor is not primarily a function of time; instead its a function of position. dv/dt = a = a(x) Here were thinking of a and v as functions of x not t. What should dv/dt be replaced by? dv/dt = dv/dx x dx/dt = dv/dx x v

When acceleration depends on position: a = v dv/dx

Example A particle attached to a spring has an acceleration a = -4x m/s2, where x is the

displacement from the fixed origin. Given that when x = 0, v = 2 m/s. Obtain an expression in terms of x for v. Solution a = -4x v dv/dx = -4x Split up dv/dx, bringing all vs to the left and all xs to the right (separating the variables). v dv = -4x dx


3

Put integration signs on both sides:

4vdv xdx C Integrate:

2 212 2v x C

Substitute boundary (initial) condition: v = 2 when x = 0. C = 2

2 2

2

4 4

2 1

v x

v x

Alternatively without using constant of integration:

2 0

2 212

2 0

2 212

2 2

2

4

2

2 2

4 4

2 1

v x

vdv xdx

v x

v x

v x

v x

An interpretation: v

x1-1

2

-2

Phase Space

The particle can be pictured in phase space as moving along an ellipse (v

2+4x

2 = 4). Maximum

speed occurs when x=0. Maximum displacement occurs when v=0. Obviously during the motion the quantity v

2+4x

2 is conserved. This is closely related to the total

energy (kinetic + potential) of the particle. So what we have got here is an expression of the Principle of Conservation of Energy.

Example The acceleration due to gravity on a rocket in the gravitational field of a planet is k/x

2 m/s

2, where k is a positive constant and x is the displacement from the centre of the planet

and is directed towards the centre of the planet. If the rockets speed at the surface of the planet is u m/s and the radius of the planet is R m, find an expression for v in terms of x, R, k and u.


4

x

a = k/x2

R

v =u

Solution a = -k/x

2

v dv/dx = -k/x

2

Split up dv/dx, bringing all vs to the left and all xs to the right (separating the variables). v dv = -k/x

2 dx


2

1vdv k dx C

x

Integrate:

212

kv C

x

Substitute boundary (initial) condition: v = u when x = R.

212

kC u

R

2 21 12 2

22 2

k kv u

x R

k kv u

x R

Incidentally, if -2k/R + u

2>0, then the expression inside the square root is always positive

(because 2k/x>0). Hence, the rocket always moves away from the planet and doesnt fall back

down. The minimum value for u for which this occurs is u = 2k/R. This known as the escape velocity. For Earth the figures are R = 6.37 x 10

6 m and k = 3.99 x 10

14 Nm

2. This gives as the

escape velocity of Earth u = 11 200 m/s.

Example A particle P moves along the positive x-axis and its acceleration is (e2x

) m/s2

when its displacement from O is x. Given that when t = 0, P is at O and its velocity is 1 m/s in the direction Ox, obtain (a) v as a function of x; (b) x as a function of t. Solution a = e

2x

v dv/dx = e

2x

Split up dv/dx, bringing all vs to the left and all xs to the right (separating the variables). v dv = e

2xdx


5

Put integration signs on both sides: 2xvdv e dx C

Integrate:

2 21 12 2

xv e C

Substitute boundary (initial) condition: v = 1 when x = 0.

0C

2 21 12 2

x

x

v e

v e

We chose the positive square root here because the question informs us that P initially as a velocity in the direction Ox--- that is to say, its positive. To obtain x(t), write dx/dt = v. Therefore,

xdx edt

Split up dx/dt and bring xs to left and ts to right (separate the variables). e

-x dx = dt


1 'xe dx dt C Integrate:

'xe t C Substitute initial condition: x = 0 when t = 0.

' 1C Therefore,

1

1

x

x

e t

e t

Taking natural logs of both sides:

1

ln 1 ln1

x tt

Summary of method


6

Question typically gives formula for a(x) and boundary condition (value for v(x) for a given value of x).

Note the direction of a(x). Write v dv/dx = a(x). Separate the variables v and x. Integrate both sides, not forgetting the constant of integration (C). Use boundary condition to get C. If x(t) is required, write dx/dt = v(t) and separate variables and integrate.


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2 Elastic Strings and Springs 2.1 Introduction The basic difference between strings and springs is this: a string can pull but cannot push (it can be stretched but not compressed); a spring can pull and push (it can both be stretched and compressed). The length of a spring or string when it is not under tension is called its natural length. If it is stretched or compressed (in the case of a spring) there will be a force (tension) in it which tends to restore it to its natural length. This is called a restorative force. Its important to realise that the direction of the restorative force is always opposite to the direction of the extension or compression of the spring or string which causes it.

Natural length = l Natural length = l x

Natural length = l

-x

restorative force

restorative force

Relaxed Spring Extended Spring

Compressed Spring

If the external forces (stresses) on the spring or string are small, then it will return to its natural length when the forces are removed. This kind of behaviour is called elastic. If the stresses on the spring or string are too large however, it may become plastic --- that is to say it doesnt return to its natural length---or it may even snap. The elastic limit is the point at which a spring or string doesnt return to its natural length.

2.2 Hookes Law Named after the 17

th Century astronomer and physicist Robert Hooke, Hookes Law states that

the restorative force in a string or spring is proportional to the extension or compression as long as the extension is not too large.


8

Hookes Law

xT

l

T is the restorative force (newton). x is the extension or compression +ve for extension, -ve for compression (metre). l is the natural length (metre).

is called the modulus of elasticity. It depends on the material the string or spring is made from

and its cross-sectional area. For stiff springs or strings, is large (newton). The minus sign in the equation for Hookes Law indicates that the direction of the restorative force opposes the extension or compression.

Example The modulus of a spring in 30N and its natural length is L. When it is compressed to a length of 75cm, the restorative thrust is 12N. Calculate L. Solution Always draw a good diagram.

L

0.75

x = L - 0.75

T=12N

30( 0.75)12

12 30 22.5

18 22.5

1.25

xT

L

L

L

L L

L

L m

2.3 Equilibrium of Particles and String or Spring Systems When a mass is suspended in equilibrium from a string or spring, the extension produced will be just enough to produce the tension required to balance the weight of the mass.

extension = x

natural length = l

tension T = x/l

weight W = mg

P


9

If the particle P is in equilibrium,

xmg

l

Example AB is a light elastic string of modulus 40N and natural length 1m. CD is a light elastic string of modulus 50N and natural length 0.5m. AB and CD are attached together at B and C, and A is attached to a fixed point. A particle of mass 2kg is attached to the end D and the whole assembly hangs vertically. Calculate the extensions in each of the strings.

P

natural length = 1.0

A

B

C

D

extension = x1

natural length = 0.5

extension = x2

Forces on P

W = 2g

T

Forces at B-C

T

T2 1

2

Resolving vertically the forces at B-C T1-T2 = 0 So, T1 = T2

1 21 2

40 502.5

1.0 0.5

x xx x

Resolving vertically the forces on P: T2-W = 0 So, T2 = 2g

22

1

502 19.6 0.196

0.5

2.5 0.196 0.49

xg x

x

Answer x1 = 0.49 m, x2 = 0.196 m

Example A particle of mass 4kg is attached to one end A of a light elastic string AB of modulus 40N and natural length 1.6m. The end B of the string is attached to a point on a smooth plane inclined to an angle arcsin 3/5 to the horizontal. The particle rests in equilibrium on the plane with AB along a line of greatest slope. Calculate the extension of the string. Solution Draw a diagram, including one of the forces on the particle. Write down the sine and cosine of the angle of the slope.

sin = 3/5 cos = 4/5


10

arcsin 3/5

B

A

Forces on particle

extension = x

natural length = l

TR

W=4g

Resolve the forces on the particle parallel to the plane.

Wsin - T = 0 3/5 (4g) = T So T = 12g/5 = 23.52.

4023.52

1.6

0.941

xT

L

x

x m

2.4 Energy in a Stretched String or Compressed Spring A string or spring which is stretched can obviously produce motion (think about how a catapult works). Similarly, a compressed spring can produce motion (think of a jack-in-the-box). So by the Principle of Conservation of Energy, a compressed spring or a stretched string or spring contains energy. How much work is done in stretching a spring from its natural length?

When its length is x, the tension in it is T=x/L. The force you must exert on it to move it must be

at least this, F = x/L.

Natural length = l x

restorative f orceexternal f orce

If the spring moves a small amount x, then the force will not change (very much), and the work

done in moving over this distance is Fx = x/L.x. Now let x0, the work done becomes the

infinitesimal dW=x/L.dx. These infinitesimals must be integrated to find the total work done:

2

02

xx x

W dxL L


11

The energy stored in string (or spring) = work done in stretching it. Note: stored energy is called potential energy.

2

Elastic potential energy (EPE)2

x

L

Example A spring of modulus 30N and natural length 1.2m is initially compressed to a length of 0.7 m. How much energy is released when the springs length increases by 0.3 m. Solution

l = 1.2 m

x = 0.5 m

x = 0.2 m

When compressed to a length of 0.8m, the energy stored in the spring is

22 30 0.5

3.125 J2 2 1.2

xW

L

When compressed to a length of 1.0 m, the energy stored in the spring is

22 30 0.2

0.5 J2 2 1.2

xW

L

The energy released is 3.125 - 0.5 = 2.625 J

Example A particle of mass 2kg is attached to one end of an elastic string of natural length 1.25 m and modulus 200 N. The other end of the string is attached to a fixed point B on a ceiling. Initially the particle is stationary at B. The particle is then allowed to drop. (a) Calculate the distance the particle falls before it instantaneously comes to rest. (b) Calculate the greatest speed of the particle. If the particle was initially thrown down at a speed of 5 ms

-1, calculate the distance the particle

falls before it instantaneously comes to rest.


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Solution (a)

B

v = 0

v = 0B

L = 1.25 m

x

Initially: GPE = 0 J Finally: GPE = mgh =

-29.8(1.25+x) J

EPE = 0 J EPE = x2/2L =

200x2/2.5 J

KE = 0 J KE = 0 J By Principle of Conservation of Energy,

2

2

0 2 9.8 1.25 200 / 2.5

80 19.6 24.5 0

0.689 m or 0.444 m

x x

x x

x x

Reject the negative answer because its unphysical. Total distance fallen = 1.25 + 0.69 = 1.94 m (b) The greatest speed occurs when the particle falls through its equilibrium position. Lets work out where this is.

Bv = 0

B

L = 1.25 m

xT

W

Forces on particle:

T=200x/1.25=160x

W = 2g = 19.6

v

Equilibrium implies T = W 160x=19.6 x = 0.1225 m

Initially GPE = 0 J Finally GPE = mgh = -29.8(1.25+0.1225) = -26.901 J

EPE = 0 J EPE = (0.1225)2/2L = 1.2005

J


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KE = 0 J KE = 1/2mv2 = v

2 J

By Principal of Conservation of Energy,

2

-1

0 26.901 1.2005

5.07 ms

v

v

_________________________________________________________________________

B

v = 0

v = 5B

L = 1.25 m

x

Initially: GPE = 0 J Finally: GPE = mgh =

-29.8(1.25+x) J

EPE = 0 J EPE = x2/2L =

200x2/2.5 J

KE = mv2 = 25 J KE = 0 J

By Principle of Conservation of Energy,

2

2

25 2 9.8 1.25 200 / 2.5

80 19.6 49.5 0

0.919 m or 0.674 m

x x

x x

x x

Reject the negative answer because its unphysical. Total distance fallen = 1.25 + 0.92 = 2.17 m


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3 Further Dynamics 3.1 Variable Force We have seen that the acceleration of a particle can depend on time or position. From Newtons second law of motion, which for constant mass is equivalent to F = ma, this can occur if the force applied to the particle varies with time or position.

3.2 Impulse Consider a tennis ball hitting a wall at high speed. Imagine seeing the ball in contact with the wall in slow motion. The ball becomes enormously distorted and the stresses (forces) on the ball are highly variable in time.

time

force

We will derive the impulse equation: Let p = mv be the momentum of a particle subject to a time-varying force F(t). Suppose that F(t) acts on the particle between the times t1 and t2. By Newtons second law,

2 2 2

1 1 1

2 1( ) ( ) ( )

t t t

t t t

dpF t F t dt dp F t dt p t p t

dt


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The term 2

1

( )

t

t

F t dt

is defined as the impulse of the force. So we have the impulse equation:

impulse = change in momentum

Frequently, we find the concept of impulse useful when we are considering very large time-variable forces acting over short times --- impulsive forces!

Example A ball of mass 200g is subject is initially moving at a speed of 40 m/s towards a wall. It first makes contact with the wall at time t1 = -0.05 s and is in contact for 0.1 s. Stress-meters in the ball show that the force acting on the ball is given by the formula

2

360( ) 60

1 2000F t

t

Find the velocity of the ball after it leaves the wall. Solution

2

1

0.05

2

0.05

360impulse = ( ) 60

1 2000

t

t

F t dt dtt

0.05

2

0.05

360

1 2000dt

t

Substitute s = 205 t dt = 5/100 ds

0.05 5

5

2 2 5

0.05 5

360 18 5 1 18 5 36 5arctan arctan 5 18.518

1 2000 5 1 5 5dt ds s

t s

0.050.05

0.05 0.05

60 60 6dt t

Impulse = -18.518+6 = -12.518 Ns impulse = change in momentum -12.518 = mv - mu = 0.2 v - 0.2x40

0.2 v = -4.518 v = -22.6 m/s The balls velocity is 22.6 m/s in the opposite direction to its initial velocity.

3.3 Work and Force When a force acts on a particle, in general the particles kinetic energy will change. Since energy cannot be created or destroyed, force must have done work on the particle.


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Imagine a force F(x) acting on a particle as it moves a small distance x. As the distance moved is small, we can suppose that the force is approximately constant over it. So the work done by the force is

( ).W F x x

Now we let x0 and integrate the infinitesimal contributions to obtain the work done: 2

1

( )

x

x

W F x dx

3.4 Newtons Law of Gravitation Sir Isaac Newton, perhaps the greatest ever applied mathematician, in order to account for Keplers laws of planetary motion, proposed the following law of universal gravitation. A law which was believed to be literally absolutely true for over two hundred years until Albert Einsteins general theory of relativity superseded it. It is however still incredibly accurate for most purposes; space probes have been programmed to navigate courses of billions of kilometres using this seventeenth century law. Newtons law of universal gravitation:

1 2

2( )

GM MF x

x

x

M MF F

G is called the constant of gravitation. Its value is experimentally found to be 6.67 x 10

-11 kg

-1 m

3

s-2

. The force of gravitational attraction between extended bodies (i.e. not particles) can be computed using the above formula if the distance between their centres of gravity is used for x. (Newton had to invent the integral calculus to prove this.)

Example Assume that the Earth is a perfect sphere with its centre of gravity at its geometrical centre. The mass of the Earth is 5.96 x 10

24 kg and its radius is 6.37 x 10

6 m. Find in

terms of m the force of gravity between a particle of mass m kg and the Earth when the particle is situated at the Earths surface. Solution

x

11 24

1

22 6

6.67 10 5.96 10( ) 9.797

6.37 10

mGM mF x m

x

Hence, the acceleration due to gravity at the surface of the Earth is 9.797 (or 9.8) m/s2.


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In general, the acceleration due to gravity on a planet g is related to G by the following formula:

g = GM/R2

where M is the mass of the planet (kilogram) and R is its radius (metre).

Example A m kg mass is carried from the surface of the Earth to infinity by a force which just equals the weight of the object at every point. Calculate the work done. Leave your answer in terms of G, R, the Earths radius, and M, its mass. Hence calculate the Earths escape velocity. The escape velocity at the surface of a black hole is 3 x 10

8 m/s (the speed of light). What would

the radius of a compressed Earth have to be for it to be a black hole? Solution

Rx

Weight F F = GMm/x2

2

RR

GMm GMm GMmW dx

x x R

This is the minimum energy needed to escape the gravitational field of the Earth. The particle can escape if its kinetic energy (mv

2) equals the energy needed to escape

(GMm/R).

mv2 = GMm/R v = (2GM/R) = 1.12 x 10

4 m/s

R = 2GM/v

2 = 8.83 x 10

-3 m (or 8.83 mm radius length).

3.5 Simple Harmonic Motion (SHM) Many important mechanical systems (e.g. springs, waves) exhibit a motion called simple harmonic motion (SHM). In SHM the acceleration, a, is proportional to the displacement, x, from a fixed point O and is always directed towards O.

Acceleration equation of SHM

O

x

a

a = -2x

(omega) is a constant related to the frequency of the SHM.


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Recall that a = vdv/dx. So

vdv/dx = -2x

Integrating: v2 = -

2x

2 + C

The maximum displacement of the particle is called the amplitude of the motion, A. Obviously, when x = A (or -A), v = 0. We can use this to obtain C.

C = 2A

2

Therefore,

Energy equation of SHM

v2 +

2x

2 =

2A

2

Lets try to solve the equations of SHM to find x, v and a as functions of time (t). From the energy equation:

dx/dt = v = A2 - x

2

Separate the variables and integrate:

2 2

1dx dt C t C

A x

Substitute: sin cosx A dx A d

2 2 2 2 2

1 cos cosarcsin

cossin

A xdx d d d

AA x A A

arcsin sinx

t C x A t CA

General solution to the equations of SHM

2

sin

cos

sin

x A t C

v A t C

a A t C

Amplitude, maximum speed and maximum acceleration

Amplitude = A Maximum speed = A Maximum acceleration = 2A


19

Two special cases

1) C = 0

sin

cos

x A t

v A t

Initial velocity = A Initial displacement = 0

2) C = /2

cos

sin

x A t

v A t

Initial velocity = 0 Initial displacement = A

Period of SHM

The function x = A sin(t + C) is periodic with period 2/.

Period of SHM

2T

Frequency of SHM

1

2f

T

Example A particle P of mass 2kg is moving in a straight line with SHM centre O. P passes through O with speed 6 m/s and has a frequency of 4 Hz. (a) Calculate the amplitude of the motion. (b) Calculate the maximum force on the particle. (c) When the particle is 10 cm from O, calculate the particles speed.

Solution (a) Frequency = 4Hz period T = 0.25 s

=2/0.25 = 8 s-1

v2 +

2x

2 =

2A

2 When x = 0 , v = 6

36 + 0 = 642 A

2

A = 3/4 = 0.239 m

(b) Acceleration = - 2 x displacement

maximum acceleration = 2 x amplitude

= 642 (3/4)

= 48 m/s2

F = ma Maximum force = 2 x maximum acceleration = 96 N

(c) v2 +

2x

2 =

2A

2 When x = 0.1

v2 + 64

2 0.1

2 = 36 v = 5.45 m/s


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3.6 Springs, Strings and SHM

Example A particle of mass 2kg rests on a smooth horizontal table and one end is attached to a light elastic string of natural length 1m and modulus 32N. The other end of the string is attached to a fixed point O on the table. The particle is pulled back to a point A which is 1.5 m from O and released from rest. (a) Prove that while the string is taut the particle undergoes SHM. (b) How long does it take for the string to become slack for the first time? (c) What is the speed of the particle when the string becomes slack? (d) How long does it take for the particle to return to A for the first time? Solution

O 1m 0.5m

A

v=0

string goes slack here

BCD

1m0.5m

(a) Draw forces on P while string is taut.

O A1m

x

T

Resolve forces horizontally -T = ma

-1

322

1

16 4 s

xma

l

xa

a x

(b) The particle begins with zero initial velocity and initial displacement from equilibrium position = 0.5m.

cos

1.5cos 4

x A t

t

String becomes slack when x= 0. 4t = /2 t = /8 = 0.393 s

(c) v2 +

2x

2 =

2A

2 When x = 0 , A = 1.5, = 4

v2 + 0 = (16)(2.25) v = -6 m/s

Speed = 6 m/s


21

(d) Time taken from A to B = /8 ( cycle) Time taken from B to C = 2/6 = 1/3 (constant speed)

Time taken from C to D and back to C = /4 ( cycle) Time taken from C to B = 2/6 = 1/3 (constant speed)

Time taken from B to A = /8 ( cycle)

Total time from A to A = /2 + 2/3 = 2.24 s

Example A particle P of mass m is attached to two identical light springs of natural length l and modulus 5mg. The other ends of the springs are attached to fixed points A and B which are 4l apart on a smooth horizontal table. P is released from rest from a point C where AC = l. A, B and C are collinear. Show that the subsequent motion is SHM. Find the amplitude and period of the motion. Find the greatest kinetic energy of P, and find the speed of P when AP = 3l/2. Solution

A BC

l 2l

A BC

l lx l

v=0

P

P

M

M

l-x

l

Forces on P:

PT T1 2

Resolve forces on P horizontally

1 2

10

10 10

T T ma

l x l xma

l l

mgxma

l

g ga x

l l

_______________________________________

Particle starts with velocity = 0. So x = A cos t t=0 A = l

_______________________________________


22

Period

22

10

lT

g

_______________________________________

2 2 2 2 2

2 2 21 1 12 2 2

10 105

v x A

g gmv m x m l mgl

l l

Maximum K.E. = 5mgl

_______________________________________

2 2 2 2 2

22 2

2

2

10 10

4

30

4

30

2

lv x A x

g l gv l

l l

glv

glv

Example A particle P of mass 1kg is hanging in equilibrium attached to a light elastic string of natural length 1m and modulus 100N. The other end of the string is attached to a fixed point on the ceiling. P is pulled down a further 20cm and is then released from rest. (a) Show that P undergoes SHM in its subsequent motion while the string remains taut. (b) What is the closest distance between P and the ceiling in its subsequent motion. (c) How long after it is released is the particle first at its closest distance to the ceiling. Solution

1m

e

1m

x

string goes slack here

Calculate equilibrium extension:


23

9.8 100

0.098m

emg

l

x

e

_______________________________

Forces on P when extension is e+x:

T

x

W=mg Resolve forces on P vertically

1 0 0 0 . 0 9 8 9 . 8

1 0 0 1 0

T mg ma

x emg ma

l

x a

a x

_______________________________

Calculate the speed of P when the string goes slack: x = -e = -0.098 P is initially at x = 0.2-0.098 = 0.102 m when v = 0. So A = 0.102m.

2 2 2 2 2

2 22

2

100 0.098 100 0.102

2

25

2

5

v x A

v

v

v

When the string is slack the only force acting on the particle is gravity. So use

2 2 2

20 19.6

25

0.00408 m

v u as

s

s

Closest distance to ceiling = 1 - 0.00408 = 0.996 m

_______________________________

x = A cos t = 0.102 cos 10t


24

Time for the string to go slack: x = -e = -0.098

-0.098 = 0.102 cos 10t t = 0.286 s Time to reach top of parabolic path: v = u + at

0 = 2/5 - 9.8t t = 0.029 s Total time = 0.286 + 0.029 = 0.315 s


25

4 Circular Motion 4.1 Centripetal Force Angular speed (rad/s).

1

lr

dd dl v

dt dt r dt r

v r

r

l

v

r

i

j

cos sin sin cosd

r t t r t tdt

r

r i j v i j

From this we can see that r and v are perpendicular. Furthermore,

2 2

2 2 2

2 2 2 2

sin cos

cos sin

sin cos

r t t r

dr t t

dt

r t t r

v

va i j r

a

For a particle of mass m moving along a circular path of radius r the velocity is perpendicular to the displacement from the centre; the acceleration is towards the centre; the magnitude of velocity is v = r; the magnitude of acceleration is a = r2; there must be a resultant force of magnitude F = mr2 acting on it --- this is called the

centripetal force.

Centripetal force

22

mvF m r F

r


26

Example A particle of mass 0.1 kg rests on a rough horizontal disc which is rotating about its centre which is rotating at 2 revolutions per second. If the coefficient of friction between the particle and the discs surface is 0.4, calculate how far from the centre of the disc the particle can be before it will slip. Solution

OF

R

W = 0.98 N

r

2 revs/s = 4 rad/s = 4 Resolve forces vertically on particle: R - W = 0 R = 0.98 N

Resolve forces horizontally on particle: F = m2r = 0.1 (4)

2r = 1.6

2r

F R 1.62r (0.4)(0.98) r 0.0248 m (2.48 cm)

Example A car of mass m is moving at constant speed around a circular roundabout of diameter 70m. If the coefficient of friction between the cars tyres and the road is 0.3, by modelling the car as a particle calculate the maximum speed of the car if it is to maintain its circular course.

F

R

W = mg

35m

Solution Resolve forces vertically: R - mg = 0 R = mg Resolve forces horizontally: F = mv

2/r = mv

2/35


27

F R mv2/35 (0.3)mg v

2 (0.3)(35)g = 102.9

v 10.1 m/s (36.5 km/h 23 mph)

Example Keplers Third Law of Planetary Motion states that the square of the period of a planets orbit around the Sun is proportional to the cube of its orbit radius. By modelling the planet as a particle and its orbit as a circle centred on the Sun and by assuming that the circular motion is uniform and the only force acting on the planet is gravity given by Newtons law, and by neglecting the attraction of the planet on the Sun itself, prove Keplers law. Solution

Sun planetF

r

Force on planet F = GMm/r

2

M - mass of Sun m - mass of planet r - radius of orbit This is the only force, hence the resultant force on the planet.

Therefore, GMm/r2 = m

2r

2 = GM/r

3

Period of orbit T = 2/ T2 = 4

2/

2 = 4

2r3/GM

4.2 The Conical Pendulum

Example A light inextensible string AB of length 3l has its ends fixed to two points A and B which are in a vertical line with A a distance l above B. A smooth ring of mass m is threaded on the string and is made to move in a horizontal circle with centre B and at constant speed. By modelling the ring as a particle calculate its speed and the tension in the string. Solution

B

A

l

r

PT

T

W = mg


28

The string is threaded through the ring and not knotted so the tension in all parts of the string must be the same. Calculate r:

2 2

2 2 2 2

3

9 6

4

3

l r l r

l lr r l r

lr

34 45 5 3

sin cos tan =

Resolve forces on P vertically:

53

3cos

5

TT mg mg T mg

_________________________________

2

2 45

5 4 9sin 1 . . 4

3 3 5

2

mv Tr mg lT T v gl

r m m

v gl

4.3 Banked Curves

Example A car of mass m is moving at constant speed around a circular roundabout of diameter 70m. The road is banked at an angle of 20

o to the horizontal. If the coefficient of friction

between the cars tyres and the road is 0.3, by modelling the car as a particle calculate the maximum and minimum speeds of the car if it is to maintain its circular course.

35m

20o

Solution


29

R

W = mg

F

20a

Resolve forces vertically: R cos 200 - W + F sin 20

0 = 0 R cos 20

0 + F sin 20

0 =

mg Resolve forces horizontally: F cos 20

0 + R sin 20

0 = mv

2/r = mv

2/35

Therefore, R(cos

2 20

o - sin

2 20

o ) = mg cos 20

0 - mv

2/35 sin 20

0

F(cos

2 20

o - sin

2 20

o ) = mv

2/35 cos 20

0 - mg sin 20

0

Since -R F R ,

2 2 235 35 350.3 cos 20 sin 20 cos 20 sin 20 0.3 cos 20 sin 20o o o o o omv mv mvmg mg mg

Therefore,

2

2

cos 20 0.3sin 20 sin 20 0.3cos 20 4.96 m/s ( 11 mph)35

cos 20 0.3sin 20 sin 20 0.3cos 20 14.3 m/s ( 32 mph)35

o o o o

o o o o

vg v

vg v

4.3 Motion in a Vertical Circle When a particle moves in a vertical circle it gains and loses kinetic energy, and so the motion is not uniform. But we can still apply the equations of circular motion (it just means that the centripetal acceleration is not constant). What work does the centripetal force do on the particle? Nothing because the force is always at right-angles to the direction of motion. Therefore, the only work done on the particle is by gravity and we can say that

gain (resp. loss) in K.E. = loss (resp. gain) in GPE

Example A particle of mass m is attached to one end of a light inextensible string of length l. The other end of the string is attached to a fixed point O. The particle is hanging at the point A, which is vertically below O. It is projected horizontally with speed u. When the particle is at the

point P, AOP = as shown. The string oscillates through an angle on either side of OA where

cos = 2/3.


30

(a) Find u in terms of g and l.

When AOP = , the tension in the string is T.

(b) Show that T = mg/3 (9cos - 4). (c) Find the range of values of T.

Example A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is fixed at a point O. The particle is held with the string taut and OP horizontal. It is then projected vertically downwards with speed u, where u

2 = 3/2ga.

When OP has turned through an angle and the string is still taut, the speed of P is v and the tension in the string is T, as shown.

P

O

v (a) Find an expression for v

2 in terms of a, g and .

(b) Find an expression for T in terms of m, g and .

(c) Prove that the string becomes slack when = 210o.

(d) State, with a reason, whether P would complete a full vertical circle if the string were replaced by a light rod. After the string becomes slack, P moves freely under gravity and is at the same level as O when it is at the point A.

(e) Explain briefly why the speed of P at A is (3/2ga).

The direction of motion of P at A makes an angle with the horizontal.

(f) Find .


31

Solution (a)

P

O

v

P

O

u

Before After

GPE = 0 KE = mu

2 = mga GPE = -mga sin KE = mv

2

By Principle of Conservation of Energy,

mga = -mga sin + mv2 v

2= ga/2 (3 + 4 sin )

(b)

T

mg

Resolve forces radially: T - mg sin = mv

2/a

T = mg sin + mg/2 (3 + 4 sin )

= 3mg/2 (2 sin + 1)

(c) The string goes slack when T=0. Therefore, 2 sin + 1 = 0

sin = -

First value of = 210o

(d) No, the particle would not complete a vertical circle because when = 90o, v=0 when

= arcsin (-). The particle then falls back. (e) By the Principle of Conservation of Energy the sum of the kinetic energy and the gravitational potential energy is constant. Since A and O at the same height, the GPE is the same there. The KE and, hence, the velocity at A and O must be the same.

(f) Speed at = 210o = (ga/2).


32

A

O

30

60

phii

j

Velocity vector = (ga/2).cos 60

oi + (ga/2).sin 60

oj = (ga/2)i + (3ga/2)j

Horizontal component of velocity doesnt change in freefall. Velocity at A:

phi

= arccos (1/23)

Example A smooth solid sphere, with centre O and radius a, is fixed to the upper surface of a horizontal table. A particle P is placed on the surface of the sphere at a point A, where OA

makes an angle with the upward vertical, and 0 < < /2. The particle is released from rest.

When OP makes an angle with the upward vertical, and P is still on the surface of the sphere, the speed of P is v.

(a) Show that v2 = 2ga(cos - cos ).

Given that cos = ,find

(b) the value of when P loses contact with the sphere; (c) the speed of P as it hits the table. Solution (a)

a

BEFORE

a

AFTER

h

GPE = 0 GPE = -mgh = -mga(cos - cos ) KE = 0 KE = mv

2

By Principle of Conservation of Energy, 0 = -mga(cos - cos ) + mv2


33

Therefore, v2 = 2ga(cos - cos )

(b) Forces on P:

mg

R

Resolve forces radially: mg cos - R = mv2/a

Therefore, R = mg cos - mv2/a = mg cos - 2mg(cos - cos ) = mg(3cos - 2cos )

P loses contact with the sphere when R = 0, therefore cos = 2/3 cos = (2/3)(3/4) =

= /3

(c) Height fallen = a(1+cos ) = 7a/4 GPE lost = 7mga/4 GPE lost = KE gained mv

2 = 7mga/4

v = (7ga/2)


34

5 Statics of Rigid Bodies 5.1 Centre of Mass of a Uniform Plane Lamina Consider a system of particles of masses m1, m2, m3,, mn having coordinates (x1,y1), (x2,y2), (x3,y3),,(xn,yn). The centre of mass of this system has coordinates (X,Y).

i i i i

i i

m x m yx y

m m

Consider now a plane laminar bounded by the curve y=f(x), the x-axis and the parallel lines x=a and y=b. Split the laminar into a series of vertical strips. The laminar is uniform, and so its mass

per unit area is the same everywhere.

x=a x=b

y = f(x)

The length of each strip is y, and the width y.

The mass of each strip = area = .y.x The coordinates of the centre of mass of each strip are (x, y).

0

b bb

i i x a a a

b b bx

i

x a a a

xydx xydxy x xm x

xm

y x ydx ydx

2 2122

0

2=

b bby

i i x a a a

b b bx

i

x a a a

y dx y dxy xm y

ym

y x ydx ydx


35

Example (a)

x

y

R

0

1

A uniform lamina occupies the region R bounded by the x-axis and the curve

y = sin x 0 x Find the y coordinate of the centre of mass. (b) A uniform lamina occupies the region bounded by the curves

y = sin x and y = - sin x 0 x Find the y coordinate of the centre of mass.

x

y

0

1

-1/2

(c) A uniform lamina occupies the region bounded by the curves

y = sin x and y = sin x 0 x Find the y coordinate of the centre of mass.

x

y

0

1

1/2


36

Solution (a)

212

212

0

0

14

0

0

1 14 2 0

0

1 1 1 14 2 4 2

sin

sin

1 2cos 2

sin

sin 2

cos

sin 2 0 sin 0

cos cos 0

8

b

a

b

a

y dx

y

ydx

xdx

xdx

x dx

xdx

x x

x

(b) Mass of upper part = Mass of lower part =

0

sin 2xdx

12

0

sin 1xdx

Total mass = 2 + 1 = 3

Centre of mass of upper part = /8 above x-axis.

Centre of mass of lower part = /16 below x-axis. Centre of mass of combined laminar =

8 162 1

3 16

above x-axis. (c) We regard this as the combination of a laminar of positive mass and one of negative mass. Mass of positive part = 2 Mass of negative part = -1 Total mass = 2 - 1 = 0

Centre of mass of positive part = = /8 above x-axis.

Centre of mass of negative part = /16 above x-axis.


37

Centre of mass of combined laminar =

8 162 1 31 16

above x-axis.

Example Show that the centre of mass of a uniform wire in the form of an arc of a circle of

radius r, subtending an angle of 2 at the centre is at a distance r.sin / from the centre. Solution Arrange a pair of axes so that the centre of the arc is the origin, and the x-axis is the wires axis of symmetry. The centre of mass will therefore lie on the x-axis.

Suppose that the mass per unit length of the wire is .

x

y

x

y

r

r

Split the wire up into small elements each of which subtends an angle of from the centre.

Clearly, the mass of each element is r.. Since, x = r.cos , we have

2

2

0

cos2 sin sin

= 2

i i

i

r dr xm x r r

xm r

r r d

5.2 Centre of Mass of a Uniform Solid Body Let R be the region bounded by the curve y = f(x), the x-axis and the lines x=a and y=b. Suppose that R is rotated about the x-axis to obtain a solid of revolution. The centre of mass of this solid is obviously by symmetry located on the x-axis.


38

y=f(x)

x

y

We split the solid up into vertical cylinders of radius y and width x. If the mass per unit volume

(density) of the solid is , then the mass of each disk is y2.x.

2 22

02 2 2

b bb

i i x a a a

b b bx

i

x a a a

xy dx xy dxy x xm x

xm

y x y dx y dx

Example A childs toy consists of a uniform solid hemisphere, of mass M and base radius r, joined to a solid right circular cone of mass m. The cone has vertex O, base radius r and height 3r. Its plane face, with diameter AB, coincides with the plane face of the hemisphere, as shown.

A B

3r

O

2r Show that the distance of the centre of mass of the toy from AB is

3 2

8

M m r

M m

Solution The cone is generated by rotating the line y = r - 1/3 x about the x-axis.


39

r

y

x

y = r - x/3

3r

2

2

3

2

0

3

2

0

3

2 2 32 13 9

0

3

2 22 13 9

0

32 2 3 41 2 1

2 9 36 0

32 2 31 1

3 27 0

9 92 4

34

( / 3)

( / 3)

6

3 3 1

b

a

b

a

r

r

r

r

r

r

xy dx

x

y dx

x r x dx

r x dx

r x rx x dx

r rx x dx

r x rx x

r x rx x

r

r

The hemisphere is generated by rotating the region bounded by the curve

2 2y r x

the x-axis and the y-axis about the x-axis.


40

2

2

2 2

0

2 2

0

2 3

0

2 2

0

2 2 41 12 4 0

2 313 0

1 12 4

13

38

( )

( )

1

b

a

b

a

r

r

r

r

r

r

xy dx

x

y dx

x r x dx

r x dx

r x x dx

r x dx

r x x

r x x

r

r

x

y

3r/8 3r/4

Total mass of toy = M+m Centre of mass =

3 34 8

3 2

8

r r m Mm Mr

M m M m

5.3 Equilibrium of Rigid Bodies

A rigid body is in equilibrium if

the resultant force acting on it is zero; the algebraic sum of the moments about any point of the forces acting on it is

zero.


41

Suspended bodies: A suspended rigid body in the absence of external forces hangs in equilibrium iff (if and only if) its centre of mass is directly below the point of suspension.

Bodies resting on a plane: A body resting on a plane in the absence of external forces is in

equilibrium iff its centre of mass is directly above the base.

Example A closed container C consists of a thin uniform hollow hemisphere of radius a, together with a lid. The lid is a thin, uniform disc, also of radius a. The centre O of the disc coincides with the centre of the hemispherical bowl. The bowl and the lid are made from the same material. You may assume that the centre of mass of the hemispherical bowl is at a distance of a/2 from O on its axis of symmetry. (a) Show that the centre of mass of C is at a distance a/3 from the centre of the bowl. The container C has mass M. A particle of mass M is attached to the container at a point P in the circumference of the lid. The container is then placed with a point of its curved surface in contact with a horizontal plane. The container rests in equilibrium with O, P and the point of contact in the same vertical plane. (b) Find, to the nearest degree, the angle made by the line PO to the horizontal.

Solution (a) Mass of hemisphere = 2a2

Mass of disc = a2

Total mass of C = 3 a2

Distance of centre of mass of C from O=

2 2

2

2

2 0

3 3

a a a a

a

x

y

a/2O

(b) Find centre of mass of C and the particle. Total mass = 3/2 M


42

x

y

a/3

M/2

M

O

P

Distance of c.m. from y-axis =

13 2 2

932

0aM Ma

M

Distance of c.m. from x-axis =

12 1

332

0M M aa

M

x

y

2a/9

a/3

point of contact

O

P

vertical

13

29

3tan

2

a

a

Therefore, 32arctan

Example A childs toy consists of a uniform solid hemisphere, of mass M and base radius r, joined to a solid right circular cone of mass m. The cone has vertex O, base radius r and height


43

3r. Its plane face, with diameter AB, coincides with the plane face of the hemisphere, as shown.

A B

3r

O

2r The distance of the centre of mass of the toy from AB is

3 2

8

M m r

M m

Show that if the toy is not in equilibrium when it is resting on the surface of its cone, then M>26m. Solution

x

y

c.m.

horizontal

3tan 3rr

The toy is not in equilibrium when resting on the cone surface when

0 13

90

tan tan 90 cot

o

Therefore,

3 2 1

8 3

9 18 8 8

26

M m

M m

M m M m

M m


44

5.4 Toppling and Sliding on Inclined Planes

To decide whether equilibrium is to be broken by sliding or toppling, examine the following two situations:

when the body is on the point of sliding so that F=R; when the body is on the point of toppling, so that the reaction is acting at the

point about which the body is about to turn.

Example

x

y

R

0

1

A uniform prism has the cross-section R bounded by the x-axis and the curve

y = sin x 0 x

The height of the centre of gravity is /8 about the x-axis. Suppose that the prism lies on a rough plane and that the coefficient between it and the plane is

. The plane is inclined at an angle of to the horizontal.

(a) If is large enough to prevent sliding and if the prism is on the point of toppling, show that

tan = 4.

(b) If is small enough so that the prism slides before it topples, find an expression for tan when it is on the point of sliding.

(c) If the angle is gradually increased from 0o, write down the range of values of for which

the prism slides before it topples. Solution (a)

c.m.

R

F

+ = 90

o


45

1tan

8 2 4

tan tan 90 cot 4o

(b)

c.m.

R

F

W

F = R Resolve forces perpendicular to the plane: R - W cos = 0.

R = W cos

Resolve forces parallel to the plane: W sin - F = 0.

R = W sin

Therefore, tan = .

(c) If the prism slides before it topples then, < 4.

0 < < 4

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Mechanics 3 Revision Notes