P3 Dynamics

Mark Cannon

Hilary Term 2012

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Lecture 1

Introduction to Dynamics

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Introduction

Dynamics (d-namks)

from Greek: o powerful power, strength

Dynamics concerns the calculation of forces and motion for analysis & design

? For stationary objects, use statics & elasticity,e.g. bridges, buildings . . .

? For problems involving motion, use the laws of dynamics,

e.g. machines, vehicles, robots . . .

Dynamics is a component of Mechanics, which involves:

Kinematics motionDynamics forces and moments

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Introduction

Dynamics is essentially about Newtons 2nd law

even gyroscopic forces can be explained using Newtons 2nd law . . .

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Gyroscope_wheel_animation.mpgMedia File (video/mpeg)

Introduction. . . . . . despite how counter-intuitive they may appear to be

Henry W. Wallaces so-called anti-gravity kinemassic field generatorfrom U.S. Patent 3,626,605 (1971)

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Lectures & turorials

8 lectures covering:

Force and momentum as applied to particles

Work, power and energy

Circular motion

Gravity and satellite orbits

Rigid body dynamics

Two tutorial sheets:

1P3H dynamics of particles

1P3J dynamics of rigid bodies

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Reading

Favourite books:

Meriam and Kraige Engineering Mechanics Volume 2 Dynamics 5thedition, SI version, Wiley, 2003.

Meriam Dynamics 2nd edition, SI version, Wiley, 1975.

Other possibilities:

Hibbeler Engineering Mechanics - Dynamics SI edition, Prentice Hall,1997.

Bedford and Fowler Engineering Mechanics - Dynamics SI edition,Addison-Wesley, 1996.

...

Lecture notes & slides:

For all handouts (lecture notes & these slides), and the tutorial sheets, goto

http://www.eng.ox.ac.uk/conmrc/dcs

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Things you should know by the end of the course

How to:

use the definitions of velocity and acceleration to analyse straight-line andcurvilinear motion of particles

use Newtons second law to analyse the motion of particles under theaction of a steady or impulsive force

apply the principles of conservation of momentum and conservation ofenergy to the motion of a particle

describe planar motion of a particle in rectangular, normal-tangential, andpolar coordinates

calculate the moment of inertia of a planar rigid body from first principlesor from standard cases

apply the principles of conservation of angular momentum andconservation of energy to the motion of a planar rigid body

analyse the translation and rotation of a planar rigid body under theaction of a steady or impulsive force or moment.

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http://www.eng.ox.ac.uk/~conmrc/dcs

Force and Motion

A particle is a discrete mass concentrated at a point

treat an object as a particle when considering its translation

e.g. when analysing

? the trajectory of a golf ball

? or the motion of a spacecraft orbiting the earth

? or straight-line motion with variable acceleration

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Force and Motion

A rigid body is a system of particles rigidly connected to each other

treat any object with distributed mass as a rigid body when consideringits rotation

e.g. when analysing

? the motion of gears

? or forces and accelerations of pistons and crankshaft in a car engine

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piston_std4.mpgMedia File (video/mpeg)

Force and Motion

A particle moving in a straight line:

0 x Dx

P P

? instantaneous speed and acceleration:

V = limt0

x

t=

dx

dt= x a =

dV

dt= x

? integrate w.r.t. t:

V (t) = V0 +

t0

a(t) dt x(t) = x0 +

t0

V (t) dt

? If a = constant:

V (t) = V0 + at x(t) = x0 + V0 t +1

2at2

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Force and Motion

A particle moving in a straight line:

0 x Dx

P P

? Or use a =dx

dt=

dx

dt

dx

dx= x

dx

dxto get

a = VdV

dx

? integrate w.r.t. x using

VdV

dx=

d

dx

(12V

2)

= a = 12V2 12V

20 =

xx0

a(x) dx

? If a = constant: 12V2 12V

20 = a(x x0)

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Force and Motion

Newtons second law

for a particle:

force = rate of change of momentum

wheremomentum = mass velocity

Equivalent vector equation:

F =d

dtmV = m

dV

dt+ V

dm

dt

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Force and Motion

If mass is constant (dm/dt = 0), then

F = mdV

dtor F = m a

i.e.force = mass acceleration

If force and motion are in only one direction, then use the scalar form:

F =d

dtmV = m

dV

dt+ V

dm

dt

and

F = mdV

dtor F = ma

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Force and Motion

Some comments on weight . . .

A dropped object accelerates downwards at g = 9.81 m s2

(at earths surface)

The force of gravity causing this acceleration is the weight of the object:

weight = mass gravitational acceleration

The SI unit of force is the Newton: 1 N acting on 1 kg produces 1 m s2

acceleration, so

weight (Newtons) = mass (kg) 9.81 (m s2)

. . . and on speed

Velocity refers to the vector V

Speed is a scalar quantity V , equal to the magnitude of V

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Force and Motion

Example Motion in a straight line

A boat of mass 1500 kg is launched from a trolley on a sloping ramp.

The trolley is allowed to run down the ramp at 1 m s1 until theboat is just afloat. The trolley then stops and the boat continues tomove at 1 m s1.

Once afloat, a crew member of mass 70 kg stops the boat by pullingsteadily on a rope with a force equal to 30 % of his own weight.

How long will the boat take to stop, and what length of rope mustbe allowed to slip?

1 ms-1

force

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Force and MotionSolution: use the scalar form of Newtons 2nd law for constant mass:

F = ma gives 70 9.81 0.3 = 1500 a

so the acceleration is

a =dv

dt= 0.137 m s2

? To find the time to stop the boat:separate variables

dt =dv

a

and integrate

t =v2 v1

a=

0 10.137 = 7.28 s

? To find the distance: use

a =dv

dt=

dv

ds

ds

dt= v

dv

ds

separate variables

v dv = a ds

and integrate

1

2(v 22 v 21 ) = a(s2 s1)

s = s2 s1 =1

2

02 12

0.137 = 3.64 m

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Force and MotionExample terminal velocity in free fall

A free-fall parachutist has mass m = 75kg and frontal area A = 0.8m2.The air density at 2000 m is = 1.007 kg m3 (HLT p.68) and theaerodynamic drag is

D =1

2CD V

2A drag coefficient CD = 1.2 .

Find the terminal speed Vt. How far will the parachutist fall beforereaching 90 % of this terminal speed?

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Force and MotionExample terminal velocity in free fall

mg

D

Solution:

? Forces: weight mg downwards and aerodynamic drag D upwards.

? Hence Newtons second law: F = ma gives

mg 12

CD V2A = ma

? Terminal velocity Vt is reached when a = 0, so

V 2t =2mg

ACD=

2 75 9.811.007 0.8 1.2 = Vt = 39.01 m s

1

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Force and MotionExample terminal velocity in free fall

mg

D

Solution contd:

? To find the variation of V with s, use a = V dV /ds:

mVdV

ds= mg 1

2CDV

2A

? Simplify by rearranging and writing this in terms of V 2t :

VdV

ds=

g

V 2t

(V 2t V 2

)Separate variables and integrate between limits V = 0 and V = 0.9Vt:

s =V 2tg

0.9Vt0

V

V 2t V 2dV = V

2t

2gln

(V 2t (0.9Vt)2

V 2t 0

)= 77.58 ln

(1 0.81

1

)= 128.8 m

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Force and Motion

Highest parachute jump

Joseph Kittinger, Aug 16, 1960

Height: 31.3 kmTerminal velocity: 274 m s1

(988 km h1)

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Impulse and Momentum

What if F is time-varying and the magnitude F (t) is unknown?

This is often true in problems involving collisions

Use the concept of impulse the integral of F(t) over time

The impulse acting on a body is related to its momentum.

Given that

F = mdV

dtfor a body of mass m, then

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F dt = m (V2 V1)

In words:impulse = change in momentum

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Impulse and Momentum

The impulse-momentum equation requires no details of the time-variationof force

An impulse can describe an impact involving large forces over a short time

time

force

0

impulse = area under force-time graph

The area under graph of F(t) gives the magnitude of the impulse

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Impulse and Momentum

Example Impulse with unknown force variation

A batsman is struck by a cricket ball of mass 0.15 kg travelling at40 m s1. The ball is stopped by the impact. Can you estimate theforce exerted by the ball?

Solution:

? the force variation and duration of the impact is unknown

? but the impulse is mV = 0.15 40 = 6.0 kg m s1

so if the impact lasts t seconds, then the average force is

Fav =mV

t=

6.0

tN

e.g. t = 0.05 s gives an average force of 120 N

? shorter duration = greater force = more painful impact!

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Impulse and Momentum

Example Impulse with constant force

Use impulse & momentum to find the time needed to stop the boatin the first example.

Solution:The force F is constant, so 2

1

F dt = F

21

dt = F (t2 t1) = m (V2 V1)

giving

t = t2 t1 = mV2 V1

F= 1500 0 170 9.81 0.3 = 7.28 s

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Impulse and Momentum

Example Impulse and momentum as vectors

A cannon of mass M is free to roll without friction on horizontalground. An explosive charge projects a ball of mass m at speed vrelative to the barrel, which is inclined upward at angle .

u

v

q

At the instant after the ball leaves the muzzle find:

(a) the backward recoil speed u of the gun

(b) the absolute velocity components of the ball

(c) the magnitude and direction of any external impulse acting on the system.

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Impulse and Momentum

Solution:The components of the balls absolute velocityare:

vx = v cos u horizontallyvy = v sin vertically q

v ucos -

v sin

v

u

q

q

(a) Consider cannon plus ball together as a single system

theres no external horizontal impulse on the system, so the horizontalmomentum before and afterwards is zero

therefore0 = m (v cos u)Mu

giving

u =mv cos

M + m

[Note: the impulse between the cannon and ball is internal to the system]

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Impulse and Momentum

Solution contd:

(b) Substituting for u in vx = v cos u gives

vx = v cos mv cos

M + m= v cos

M

M + m

and the other component of absolute velocity is just vy = v sin

(c) The only external impulse is vertical: Qy (from the ground)

Qy is equal to the change of upward momentum of the entire system:

Qy = mv sin

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Impulse and Momentum

Example Rowing on a sliding seat

Between strokes, the crew of a boat slide a distance d towards thestern of the boat. If the crew has mass m and the boat has mass M,what happens to the boat?

Solution: Assume the crew slides distance d at constant speed for time t

theres no external impulse so the momentum of boat plus crew is unchanged

hence the velocity v of the centre of mass G is unchanged

x

G

d

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Impulse and Momentum

Solution contd:

Relative to G:

? the boat moves forward x at speedx/t

? the crew move backward d x at speed(d x)/t

Dx

x

t

Considering momentum before and after gives

(M + m)v + Mx

tm d x

t= (M + m)v

so the boat surges forward between strokes by an amount x =md

M + m

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Summary

Analyse translational motion of an objectby representing it as a particle concentrated at the centre of mass

Newtons second law

? general case:

force = rate of change of momentum F = ddt

mV = mdV

dt+V

dm

dt

? for objects with constant mass:

force = mass acceleration F = m dVdt

Impulse and momentum

impulse = change in momentum 2

1

F dt = m (V2 V1)

Accelerations are measured in an inertial frame of reference,i.e. non-accelerating, non-rotating

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