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Newton to Einstein Mid-Term Test, 12am Wed. 12th Nov. 2014
These revision summaries contain what you need to know in order just to pass the test
(40%). This is a bare competency across the subject, the minimum knowledge you
need to carry forward as a prerequisite for other Physics modules. It is tested in
Section A of the test paper. For Section B, you require to be able to use this
knowledge in more complicated problem-solving.
0. Vectors A vector is a quantity with magnitude and direction.


The Magnitude of the vector u is u  u  u x2  u 2y .
uy

The Angle (to the x-axis) of the vector u is   arctan
ux
Unit Vectors: vectors of unit length in the direction implied by the name, e.g.

u  uuˆ
Unit vectors along x, y and z directions are
iˆ  (1, 0, 0)
ˆj  (0, 1, 0)
kˆ  (0, 0, 1)

And u  u x iˆ  u y ˆj  u z kˆ

Dot Product or scalar product is a scalar quantity: u .v  uv cos  where  is the


angle between u and v .
iˆ.iˆ  1 iˆ. ˆj  0 etc

iˆ.u  u x etc

u  u x , u y , u z  u x iˆ  u y ˆj  u z kˆ


 
Cross Product or vector product is a vector: u  v  uv sin  where  is the angle


between u and v .
iˆ  ˆj  kˆ ˆj  kˆ  iˆ kˆ  iˆ  ˆj Note : iˆ  kˆ   ˆj


 
u  v  uv sin  nˆ


where n̂ is the unit vector normal to both u and v and in the direction conventionally


given by the right-hand rule (screw forwards when rotating from u to v
1. Dynamics
Newton’s Laws of Motion (learn them!), and what they mean:
First Law implies the Principle of Relativity.
Second Law defines Momentum (by force equals
rate of change of momentum).
Third Law implies the Conservation of Momentum.
Resolution of Forces. Forces are vectors, and you should be able to add their
components, resolved along any convenient pair of perpendicular directions (vertical
and horizontal, along a ramp and normal to it, etc).
The Moment of a force about a point is the force times the perpendicular distance to
the axis.
Static Equilibrium: Forces balance.
Rotational Equilibrium: Moments balance.
Friction: We distinguish sliding and static friction. The frictional force is parallel to
the surface and is
F friction  Fnormal
Sliding friction
F friction  Fnormal
Energy:
Static friction
Kinetic energy is ½mv 2
Potential energy is gained if a body is moved, doing work against a
force-field (e.g. against gravity, or against an electrostatic
 

repulsion. The change in potential energy is dE  F .dr if dr
is a small displacement, or dE  Fdr if dr is the component of
the displacement parallel to the force.
Total energy equals kinetic energy plus potential energy
TOTAL ENERGY IS CONSERVED
 
Work: The work done by a force if its point of application moves is dW  F.dr .
If the force is due to a change in potential energy E, then dW  dE
 dr  
dW
 F.
 F .v
Power: Power is rate of doing work, so P 
dt
dt
Potential Energy Curves: If potential energy E varies with position, then the force
dE
is F  
. This vanishes (F = 0) at maxima and minima of the curve.
dx
2. Curvilinear Motion Motion along general curved paths.
Velocity and Speed: The velocity vector v is always along the path, or tangential.
So the speed v, the magnitude of v, is also the tangential speed.
Radial and Tangential Acceleration: The acceleration can be in any direction. It
has therefore components which, resolving parallel and perpendicular to the
motion, are the tangential and radial accelerations:
dv
aT 
dt
v2
aR 
R
where R is the radius of curvature of the path at the point under consideration.
The radial component is also called the centripetal acceleration.
UNIFORM CIRCULAR MOTION: Motion on a circular path at constant speed v.
ds Rd 

 R
dt
dt
Angular Velocity is  in radians per second.
v
Tangential Acceleration is aT 
dv
d
R
 R
dt
dt
Angular Acceleration is  
d

dt
Radial Acceleration is a R 
v2
 R 2 , also called centripetal acceleration.
R
Hence Centripetal Force is F 
mv 2
 mR 2
r
ROTATING FRAMES OF REFERENCE
If a point is stationary in a frame of reference rotating at  about the origin,
then it is moving at the speed v = R in the stationary frame of reference. In
the rotating frame it moves as if fictitious forces act upon it.
mv 2
 mR 2
Fictitious Forces:
Radial, Centrifugal force F 
r
Tangential, Coriolis force F  2mv where v is radial
speed.
3. Rigid Bodies A general rigid body is a potato-shaped object, able to undergo
rotational and translational motion. It may be considered to be assembled out
of a large number of point masses.
Angular Momentum: The angular momentum L of a point mass about a point is its
linear momentum p = mv times the perpendicular distance to the point. In
  

vector notation, L  p  r where r is the position vector of the mass from the
point. The angular momentum of a rigid body is the sum of the angular
momenta of all its point masses. It is a conserved quantity..
Moment of Inertia: The moment of inertia I of a point mass about an axis is
I  mR 2
where R is the perpendicular distance to the axis. The moment of inertia of a
rigid body about an axis is the sum of the moments of inertia of all its point
masses.
Then L  I
The moment of inertia is always I  Mk 2 where k is some characteristic
length, or I  cMR 2 where R is the radius or length and c is some fraction.
Torque: The rate of change of angular momentum. The torque exerted by a force
  
about a point is   F  r . Its scalar value is force times perpendicular
distance, FR, and is often called the moment.
dp
Rdv
d
m
 mR 2
 I for force applied to a
dt
dt
dt
point mass at a radius R from the axis.
Then we have   RF  R
Rotational Kinetic Energy is E  ½ I2
Comparison of Dynamical Quantities in Translation and Rotation.
Translational
Position
Velocity
Acceleration
First Integral
Second Integral
Mass
Momentum
Kinetic Energy
Force
Power
Second Law
Rotational

s
Angle



 ds
v
dt

 d 2s
a 2
dt
Angular
acceleration

d

dt

 d 2
 2
dt
v  u  at
.....
  0  t
s  ut  ½at 2
.....
  0t  ½t 2
m


p  mv
p2
E  ½ mv 
2m
 dp
F
dt
 
P  F .v


F  ma
2
Angular
velocity
Moment of
Inertia
Angular
Momentum
.....
Torque
.....
.....
I


L  I
L2
E  ½ Iω 
2I


 dL
  F r 
dt

P  .


  I
2
4. Gravity and the Solar System
Kepler’s Laws. You should know that the second (equal areas) corresponds to
conservation of angular momentum, and that the third ( T 2  R 3 ) is easy to
derive, by equating centrifugal force to gravitational force.
Inverse-Square Law Gravity: F 
GMm
r2
Gravitational Potential Energy: E  
, an attractive, central, conservative force
GMm
r
Total Energy (kinetic plus potential) is negative for closed orbits (elliptical), zero for
parabolic and positive for hyberbolic open orbits.
You should understand that at two points on an orbit which are the same
distance from the sun, the speed of the planet is therefore the same (though the
direction is different).
Escape velocity: This is the velocity at which the total energy is zero, so that the
orbit will go to infinity. It is found from KE and PE:
GMm
2
½ mvesc

0
r
Orbital Speed: The speed of a satellite in a circular orbit is found by equating
centrifugal and gravitational forces:
mv 2 GMm
 2
r
r
v
END
GM
r