P3 Dynamics

P3 Dynamics Mark Cannon

Hilary Term 2012

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Lecture 1

Introduction to Dynamics

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Introduction Dynamics

(d¯ι-n˘ am0 ιks) from Greek:

δυναµικ´ o ς – powerful δ υ´ναµις – 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 Dynamics

−→ −→

motion forces and moments

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Introduction

Dynamics is essentially about Newton’s 2nd law

even gyroscopic forces can be explained using Newton’s 2nd law . . .

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

Henry W. Wallace’s so-called anti-gravity “kinemassic field” generator from U.S. Patent 3,626,605 (1971) 1-4

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 5th edition, 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, go to http://www.eng.ox.ac.uk/∼conmrc/dcs 1-6

Things you should know by the end of the course How to: use the definitions of velocity and acceleration to analyse straight-line and curvilinear motion of particles use Newton’s second law to analyse the motion of particles under the action of a steady or impulsive force apply the principles of conservation of momentum and conservation of energy to the motion of a particle describe planar motion of a particle in rectangular, normal-tangential, and polar coordinates calculate the moment of inertia of a planar rigid body from first principles or from standard cases apply the principles of conservation of angular momentum and conservation of energy to the motion of a planar rigid body analyse the translation and rotation of a planar rigid body under the action of a steady or impulsive force or moment.

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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 considering its 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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Force and Motion A particle moving in a straight line:

0

P

P

x

Dx

? instantaneous speed and acceleration: ∆x dx = = x˙ ∆t→0 ∆t dt

V = lim

? integrate w.r.t. t: Z t a(t) dt V (t) = V0 +

dV = x¨ dt

a=

Z x(t) = x0 +

0

t

V (t) dt 0

? If a = constant: V (t) = V0 + at

1 x(t) = x0 + V0 t + at 2 2

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Force and Motion A particle moving in a straight line:

0

? Or use a =

P

x

Dx

d x˙ dx d x˙ d x˙ = = x˙ to get dt dt dx dx a=V

? integrate w.r.t. x using dV d  1 2 V = V =a dx dx 2 ? If a = constant: 1 - 10

P

1 2 2V

dV dx

=⇒

1 2 2V

−

− 21 V02 = a(x − x0 )

1 2 2 V0

Z

x

=

a(x) dx x0

Force and Motion Newton’s second law for a particle: force = rate of change of momentum where momentum = mass × velocity

Equivalent vector equation: F=

dV dm d mV = m +V dt dt dt

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Force and Motion If mass is constant (dm/dt = 0), then F=m

dV dt

or

F = ma

i.e. force = mass × acceleration

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

d dV dm mV = m +V dt dt dt

and F =m

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dV dt

or

F = ma

Force and Motion Some comments on weight . . . A dropped object accelerates downwards at g = 9.81 m s−2 (at earth’s 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 s−2 acceleration, so weight (Newtons) = mass (kg) × 9.81 (m s−2 )

. . . 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 s−1 until the boat is just afloat. The trolley then stops and the boat continues to move at 1 m s−1 . Once afloat, a crew member of mass 70 kg stops the boat by pulling steadily 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 must be allowed to slip? 1 ms

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force

Force and Motion Solution: use the scalar form of Newton’s 2nd law for constant mass: F = ma

−70 × 9.81 × 0.3 = 1500 × a

gives

so the acceleration is a=

dv = −0.137 m s−2 dt ? To find the distance: use

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

a=

dv a

separate variables

and integrate t=

dv ds dv dv = =v dt ds dt ds

v dv = a ds

v2 − v1 0−1 = = 7.28 s a −0.137

and integrate 1 2 (v2 − v12 ) = a(s2 − s1 ) 2 1 02 − 12 = 3.64 m s = s2 − s1 = 2 −0.137

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Force and Motion Example – 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 ρ = 1.007 kg m−3 (HLT p.68) and the aerodynamic drag is D=

1 CD ρV 2 A 2

drag coefficient CD = 1.2 .

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

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Force and Motion Example – terminal velocity in free fall D

Solution:

mg

? Forces: weight mg downwards and aerodynamic drag D upwards. ? Hence Newton’s second law: F = ma gives mg −

1 CD ρV 2 A = ma 2

? Terminal velocity Vt is reached when a = 0, so Vt2 =

2mg 2 × 75 × 9.81 = ρACD 1.007 × 0.8 × 1.2

=⇒

Vt = 39.01 m s−1

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Force and Motion Example – terminal velocity in free fall D

Solution contd:

mg

? To find the variation of V with s, use a = V dV /ds: dV mV = mg − 12 CD ρV 2 A ds ? Simplify by rearranging and writing this  in terms ofVt2 : dV g V = 2 Vt2 − V 2 ds Vt Separate variables and integrate between limits V = 0 and V = 0.9Vt :  2  Z 0.9Vt Vt − (0.9Vt )2 Vt2 V Vt2 dV = − ln s= g 0 2g Vt2 − V 2 Vt2 − 0   1 − 0.81 = −77.58 ln = 128.8 m 1 1 - 16

Force and Motion

Highest parachute jump Joseph Kittinger, Aug 16, 1960 Height: Terminal velocity:

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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 dV F=m dt for a body of mass m, then 2

Z

F dt = m (V2 − V1 ) 1

In words: impulse = change in momentum 1 - 18

31.3 km 274 m s−1 (988 km h−1 )

Impulse and Momentum

The impulse-momentum equation requires no details of the time-variation of force An impulse can describe an impact involving large forces over a short time

force impulse = area under force-time graph

time 0

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 at 40 m s−1 . The ball is stopped by the impact. Can you estimate the force exerted by the ball?

Solution: ? the force variation and duration of the impact is unknown ? but the impulse is m∆V = 0.15 × 40 = 6.0 kg m s−1 so if the impact lasts t seconds, then the average force is Fav =

m∆V 6.0 = N t t

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 boat in the first example. Solution: The force F is constant, so Z 2 Z F dt = F 1

2

dt = F (t2 − t1 ) = m (V2 − V1 )

1

giving t = t2 − t1 = m

V2 − V1 0−1 = 1500 × = 7.28 s F −70 × 9.81 × 0.3

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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 horizontal ground. An explosive charge projects a ball of mass m at speed v relative to the barrel, which is inclined upward at angle θ. v

q u

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. 1 - 22

Impulse and Momentum Solution: The components of the ball’s absolute velocity are: vx = v cos θ − u

horizontally

vy = v sin θ

vertically

u v v sinq

q v cos q - u

(a) Consider cannon plus ball together as a single system there’s no external horizontal impulse on the system, so the horizontal momentum before and afterwards is zero therefore 0 = m (v cos θ − u) − Mu giving u=

mv cos θ M +m

[Note: the impulse between the cannon and ball is internal to the system] 1 - 23

Impulse and Momentum

Solution contd: (b) Substituting for u in vx = v cos θ − u gives vx = v cos θ −

mv cos θ M = v cos θ M +m M +m

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

(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 θ

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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 the stern 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 there’s 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 d

G

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Impulse and Momentum Solution contd:

x

Relative to G: ? the boat moves forward ∆x at speed ∆x/∆t

Dx

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

t

Considering momentum before and after gives (M + m)v + M

∆x d − ∆x −m = (M + m)v ∆t ∆t

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

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md M +m

Summary Analyse translational motion of an object by representing it as a particle concentrated at the centre of mass Newton’s second law ? general case: force = rate of change of momentum

dV dm d mV = m +V dt dt dt

⇐⇒

F=

⇐⇒

F=m

? for objects with constant mass: force = mass × acceleration

dV dt

Impulse and momentum 2

Z impulse = change in momentum

⇐⇒

F dt = m (V2 − V1 ) 1

Accelerations are measured in an inertial frame of reference, i.e. non-accelerating, non-rotating 1 - 27