#### Transcript Lecture 24 - McMaster University

```Energy
•
•
•
Potential energy
Examples with rotation
Force and Potential Energy
Serway 8.4-8.6; 10.8
Physics 1D03 - Lecture 24
Mechanical Energy
E = K + U = K + Ugravity + Uspring + ...
Mechanical energy is conserved by “conservative”
forces; the total mechanical energy does not change if
only conservative forces do work.
Remember, for a conservative system:
ΔK+ΔU=0
Physics 1D03 - Lecture 24
Non-Conservative Forces
Non-conservative forces cause the mechanical energy of
the system to change. Divide work W into work by
conservative forces and work by “other” forces.:
DK = ΣW = Wc + Wother
Wc can be replaced by potential energy terms: Wc = − DU ,
so
and, since E=K+U,
DK = - DU + Wother
DE = Wother
The change in mechanical energy is equal to the work by “other”
forces.
“Other” means any force not represented by a term in the potential
energy. It includes non-conservative forces, but also externallyapplied forces, conservative or not, that transfer energy into or out
of the system.
Physics 1D03 - Lecture 24
Conservation of Energy
Other types of energy:
-electrostatic P.E.
-chemical P.E.
-nuclear P.E.
-etc., etc.,
and thermal energy (actually just kinetic and potential
energy at a microscopic scale)
Then:
The total energy of the universe is conserved.
Physics 1D03 - Lecture 24
Rotation about a fixed axis
Kinetic Energy:
Work:
Power:
K = ½ Iw 2
dW   d
(infinitesimal rotation)
W   D
(for a constant torque)
P  w
Physics 1D03 - Lecture 24
Example: Pole vault
Jumping animals, from a frog (10-2 kg) to a horse (103 kg) can all jump
about 1 or 2 metres vertically. Why can a pole-vaulter jump 6 m?
Guess: Initial kinetic energy built up during several seconds of effort
can be converted to elastic and then gravitational potential energy. The
pole-vaulter doesn’t need to get all his mechanical energy from a single
muscle contraction.
Kinetic energy
Elastic
potential energy
Gravitational
potential energy
Physics 1D03 - Lecture 24
To test our guess, see if the numbers fit:
If he runs at 9 m/s, and can convert all his kinetic energy into
gravitational potential energy, how high would he get?
Ki = (1/2) mv2 = Uf = mgy
so
y = v2/2g = 4.1 m for v = 9 m/s
( for v = 10 m/s, y = 5.1 m)
This is enough to explain a 6-m vault. His mechanical energy
actually increases as the pole straightens, as his muscles (e.g.,
arms) do work.
Physics 1D03 - Lecture 24
Quiz: Bungee cord
m
A mass attached to a 50-m elastic
cord is dropped. The mass falls 80
m (stretching the cord 30 m) before
the block stops momentarily.
How far has it fallen when it
reaches its maximum speed ?
50 m
a) 50 m
b) less than 50 m
c) more than 50 m
30 m
m
Physics 1D03 - Lecture 24
Quiz
A thin (uniform) pole of length L and mass
m, initially vertical and stationary, is
allowed to fall over. Assuming it falls as if
hinged at the base, what is its angular
velocity just before it hits the ground?
How much potential energy does it lose as it falls?
A) mgL
B) 2mgL
C) mgL/2
Physics 1D03 - Lecture 24
Gravitational potential energy depends on the height of the center
of mass. As the stick falls, gravitational potential energy is
converted to kinetic energy of rotation; the total mechanical energy
is constant.
Initial: yCM = L/2, U = mgL/2, K = 0
so E = mgL/2
CM
L/2
Final: yCM = 0, U =0,
so E = K = ½ I w 2
Initial energy = final energy: ½ I w 2 = mgL/2
For a uniform thin rod, I =
1/ mL2,
3
3g
so (after a little work), w 
L
Physics 1D03 - Lecture 24
Questions: What is the (linear) speed of the tip and of the centre
of mass? How do these compare with a particle that falls a
distance L/2?
Linear velocities: v = rw
At the tip: r = L,
g
vtip  wL  3 L  3gL
L
At the CM: r = L/2,
vCM
L 1
 w  2 3gL
2
-------------------------------------------------------------------------------For a particle falling from a height L/2, mgL/2 = ½ m v2, and
v  gL
(faster than the CM of the rotating stick).
Physics 1D03 - Lecture 24
Example
The end of the string is glued to the rim of the
pulley. After the mass reaches its lowest point,
the pulley continues to turn, and raises the
mass back to ¾ of its original height. What is
the frictional torque at the axle of the pulley?
R
m
Physics 1D03 - Lecture 24
Initial Energy = Final Energy + Work Done
There is no kinetic energy at
(1) and (2)
MgL
0  
 W
4
W  14 MgL
(1)
R
(2)
y  0
For the full problem. . .
frictional torque
W   f total
y  
1
4
L
total angle rotation
 total
f 
L  34 L
7L


R
4R
W
total

1
4
MgL
MgR

L
7
7
4
R
y  L
Physics 1D03 - Lecture 24
Forces and Potential Energy
One dimension, motion along the x axis: Given U(x), find F.
For a small displacement Dx, the work done is W = Fx Dx
If the force is conservative, then W  DU as well.
So,
DU
Fx  
Dx
,
or
dU
Fx  
dx
(Extra) Similarly, dU = -dW = -Fx dx, so
x2
U ( x2 )  U ( x1 )    Fx ( x)dx
x1
Physics 1D03 - Lecture 24
Quiz
U
The potential energy of a particle,
acted on by conservative forces
only, is shown as a function of
position.
At which of the points on the
graph is the net force on the
particle:
i)
A
E
B
x
D
C
zero? – QUIZ (A,B,C,D,E)
ii) in the positive x direction?
iii) a maximum?
Physics 1D03 - Lecture 24
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