Download Dynamics_Energy_Momentum - University of Manchester

Dynamics
Dynamics
READ the
Textbook!
Work/ Kinetic Energy
Potential Energy
Conservative forces
Conservation laws
Momentum
Part II – “We are
to admit no more
causes of natural
things than such as
are both true and
sufficient to explain
their appearances.”
Centre-of-mass
Impulse
http://www.hep.manchester.ac.uk/u/parkes/Chris_Parkes/Teaching.html
October 2013
Chris Parkes
Work & Energy
Work is the change in energy that results from applying a force
• Work = Force F times Distance s, units of Joules[J]
• More Precisely, W=F.x
– F,x Vectors so W=F x cos
s
F

x
F
– Units (kg m s-2)m = Nm = J (units of energy)
– Note 1: Work can be negative
• e.g. Friction Force opposite direction to movement x
– Note 2: Can be multiple forces, uses resultant force ΣF
– Note 3: work is done on a specific body by a specific force (or forces)
• The rate of doing work is the
P
dW
dt
So, for constant Force
Power
P=
[Js-1Watts]
d(F · x)
dx
= F·
= F ·v
dt
dt
Example
A particle is given a displacement
r  2iˆ  5 ˆj
in metres along a straight line. During the
displacement, a constant force
F  3iˆ  4 ˆj
in newtons acts on the particle.
Find (a) the work done by the force and (b) the
magnitude of the component of the force in the
direction of the displacement.
2
3
θ
F cos θ
F
-4
-5
r
Work-Energy Theorem
The work done by the resultant force (or the
total work done) on a particle is equal to the
change in the Kinetic Energy of the particle.
Wtot = K2 - K1 = DK
Meaning of K.E.
K.E. of particle is equal to the total work done
to accelerate from rest to present speed
Wtot = K
suggests K = 12 mv2
Work Done by Varying Force
W=F.x
becomes
W=
ò
x2
x1
F(x) ·d x
Energy, Work
• Energy can be converted into work
– Electrical, chemical, or letting a
weight fall (gravitational)
mgh of
water
• Hydro-electric power station
Potential Energy, U
In terms of the internal energy or potential energy
W  U  U    Fdx
Potential Energy - energy associated with the position or
configuration of objects within a system
Note: Negative sign
Gravitational Potential Energy
Choice of zero level is
arbitrary
Ug = mgh
mg h
Reference plane
Ug = 0
mg
-h
Ug = - mgh
mg
No such thing as a definitive
amount of PE
particle stays close to the Earth’s surface and so the
gravitational force remains constant.
This stored energy has the potential to do work Potential
We are dealing with changes in energy
• choose an arbitrary 0, and look at  p.e.
This was gravitational p.e., another example :
Energy
h
0
W  F  x  (mg)h
Stored energy in a Spring
Do work on a spring to compress it or expand it
Hooke’s law
BUT, Force depends on extension x
Work done by a variable force
Work done by a variable force
Consider small distance dx over which force is constant
F(x)
Work W=Fx dx
X
dx
So, total work is sum W   F  dx   F ( x)dx
0
X
0
Graph of F vs x,
F
integral is area under graph
work done = area
dx
X
Elastic Potential Energy
Unstretched position
For spring,F(x)=-kx:
x
F
X
Uel =
X
Uel =
1 2
kX
2
1 2
kX
2
-X
X
X
0
0
W   F ( x)dx    kxdx  [ 12 kx2 ]0X   12 kX 2
Stretched spring stores P.E. ½kX2
Potential Energy Function
k
1 2
U = -mgx + kx
2
Reference plane
x
Fs
mg
Conservation of Energy
DK + DU + DUint = 0
K.E., P.E., Internal Energy
Conservative & Dissipative Forces
• Conservative Forces
DK + DU = 0
– A system conserving K.E. + P.E. (“mechanical energy”)
• But if a system changes energy in some other way
(“dissipative forces”)
– e.g. Friction changes energy to heat, reducing mechanical energy
– the amount of work done will depend on the path taken against
the frictional force
– Or fluid resistance
– Or chemical energy of an explosion, adding mechanical energy
Conservative forces
frictionless surface
Example
A 2kg collar slides
without friction along a
vertical rod as shown.
If the spring is
unstretched when the
collar is in the dashed
position A, determine
the speed at which the
collar is moving when
y = 1m, if it is released
from rest at A.
Properties of conservative forces
•
The work done by them is reversible
•
Work done on moving round a closed path is zero
•
The work done by a conservative force is independent of
the path, and depends only on the starting and finishing
points
B
Work done by friction force is
greater for this path
A
Forces and Energy
W  U  U    Fdx
¶U(x)
Fx (x) = ¶x
e.g. spring
2
1
¶[
kx
]
U = 12 kx Þ F = - 2
= -kx
¶x
2
æ ¶U(x, y, z) ¶U(x, y, z) ¶U(x, y, z) ö
F(x, y, z) = - ç
i+
j+
k ÷ = -ÑU(x, y, z)
¶x
¶y
¶z
è
ø
• Partial Derivative – derivative wrt one variable, others held constant
• Gradient operator, said as grad(f)
Glider on a linear air track
Negligible
friction
Minimum on a
potential energy
curve is a position
of stable
equilibrium
- no Force
Maximum on a
potential energy
curve is a
position of
unstable
equilibrium
U
Linear Momentum Conservation
• Define momentum p=mv
d p d (mv)
nd
• Newton’s 2 law actually F  dt  dt  m ddtv  ma
• So, with no external forces, momentum is
dp
conserved. F  0, dt  0, p  const
Also true for net forces
on groups of particles
• e.g. two body collision on frictionless If F   F 0,
surface in 1D
then p   p const
i
i
i
i
before
m1
m2
0 ms-1
Initial momentum: m1 v0 = m1v1+ m2v2 : final momentum
after
v0
m1
m2
v2
v1
For 2D remember momentum is a VECTOR, must apply
conservation, separately for x and y velocity components
Energy Conservation
•Energy can neither be created nor destroyed
•Energy can be converted from one form to another
• Need to consider all possible forms of energy in a
system e.g:
–
–
–
–
–
Kinetic energy (1/2 mv2)
Potential energy (gravitational mgh, electrostatic)
Electromagnetic energy
Work done on the system
Heat (1st law of thermodynamics)
• Friction  Heat
Energy measured in Joules [J]
Collision revisited
m1
v1
m2
• We identify two types of collisions
– Elastic: momentum and kinetic energy conserved
Initial K.E.: ½m1 v02 = ½ m1v12+ ½ m2v22 : final K.E.
– Inelastic: momentum is conserved, kinetic energy is not
• Kinetic energy is transformed into other forms of energy
See lecture example for cases of elastic solution
Newton’s cradle
1. m1>m2
2. m1<m2
3. m1=m2
v2
Impulse
• Change in momentum from a force acting
for a short amount of time (dt)
Impulse J  p 2  p1  Fdt
Where, p1 initial momentum
p2 final momentum
• NB: Just Newton 2nd law rewritten
F
p 2  p1
dt

dp
dt
m
dv
 ma
dt
Approximating
derivative
Impulse is measured in Ns.
change in momentum is measured in kg m/s.
since a Newton is a kg m/s2 these are equivalent
Q) Estimate
the impulse
For Andy
Murray’s
serve
[135 mph]?
Centre-of-mass
• Average location for the total mass
xCM
m1 x1 + m2 x2 + m3 x3 +.... Smi xi
=
=
m1 + m2 + m3 +...
Smi
yCM
m1 y1 + m2 y2 + m3 y3 +.... Smi yi
=
=
m1 + m2 + m3 +...
Smi
Mass weighted average position
Centre of gravity – see textbook
• Position vector of centre-of-mass
rCM =
m1r1 + m2 r2 + m3r3 +.... Smi ri
=
m1 + m2 + m3 +...
Smi
Rigid Bodies –
Integral form
y
r
dm
x
z
r cm
1

r
dm

M
dm is mass of small element of body
r is position vector of each small element.
Momentum and centre-of-mass
• Differentiating position to velocity:
vCM
m1v1 + m2 v 2 + m3v 3 +.... Smi v i Smi v i
=
=
=
m1 + m2 + m3 +...
Smi
M
MvCM = Smi v i = P
• Hence momentum equivalent to
total mass × centre-of-mass velocity
Forces and centre-of-mass
• Differentiating velocity to acceleration:
MaCM = Smi ai = SF
• Centre-of-mass moves as acted on by the sum of the
Forces acting
Internal Forces
MaCM = Smi ai = SF
• Internal forces between elements of the body
• and external forces
– Internal forces are in action-reaction pairs and cancel
in the sum
– Hence only need to consider external forces on body
SF = SFext + SFint
SFint = 0
SFext = Ma CM
• In terms of momentum of centre-of-mass
dvCM d(MvCM ) dP
SFext = M
=
=
dt
dt
dt
Example
• A body moving to the right collides elastically
with a 2kg body moving in the same direction
at 3m/s . The collision is head-on. Determine
the final velocities of each body, using the
centre of mass frame.
3ms-1
6ms-1
4kg
C of M
2kg
Lab Frame before collision
6 ms-1
4kg
3 ms-1
5 ms-1
2kg
C of M
Centre of Mass Frame before collision
4kg
1
ms-1
C of M
2 ms-1
2kg
Centre of Mass Frame after collision
1 ms-1
4kg
C of M
2kg
2 ms-1