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# PHYLect3

advertisement ```Angle measurement
In physics, we will often be required to measure angles.
Angles can be measured either in degrees or in radians.
The measurement of angles in
degrees is purely historical and
there is nothing fundamental
about it. The circumference of a
circle is arbitrarily divided into 360
degrees (o). A right angle, for
example, corresponds to 90o.
Each degree is divided into 60 minutes (') and each
minute into 60 seconds (''). An arbitrary angle is then
expressed in degrees, minutes and seconds of arc (e.g.
25o56'42'').
An angle (θ) measured in radians
is defined as the ratio between the
length of the subtended arc (s) to
the radius of the circle (r):
&quot; #
!
arc length AB
s
=
radius of circle
r
(rad)
This is particularly useful when we are interested in the
length of an arc. Note that the above relation is not valid
if θ is measured in degrees. Also note that the radian
has no physical dimensions – it is the ratio of two
lengths. We use rad to indicate angular measurement.
15
Since the circumference of a circle is 2 &quot; r , an angle of
2&quot; r
360o corresponds to
= 2 π rad = 6.28 rad.
r
!
• angle of 180o = π rad
!
&quot;
• angle of 90o =
rad
2
Also, we can relate radians to degrees by using
!
180o
1 rad &times;
= 57.30o = 57o18' .
&quot; rad
Trigonometric functions
!
We will also often make use of the basic trigonometric
functions.
For any right angled triangle, these are defined as
follows:
sin&quot; =
cos &quot; =
a
h
tan&quot; =
o
sin&quot;
#
a
cos &quot;
!
!
16
!
o
h
Some common values for trigonometric functions
Angle θ (o)
sin θ
cos θ
tan θ
0
30
45
90
0
0.5
0.707
1
1
0.866
0.707
0
0
0.577
1
∞
Relationship between radians and trigonometric
functions
&quot; #
arc length AB
s
=
radius of circle
r
AC
r
!
sin&quot; =
!
In general, θ (measured in radians) and sin θ are quite
different. However, for small angles, the arc AB
becomes nearly equal to AC.
As θ → 0
Also:
sin θ → θ
&quot;2
cos θ → 1 –
2
tan θ → θ
!
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For example, if we take θ = 30o
sin θ = 0.500
while
θ = 0.524 rad.
The two values are quite different – percent difference ≈
5%.
If we reduce the angle, say to θ = 10o, then
sin θ = 0.174
while
θ = 0.175 rad
The difference between the two is much smaller –
percent difference ≈ 0.6%.
For an even smaller angle, say 3o,
sin θ = 0.05234 while
θ = 0.05236 rad
the difference is so small that we can take sin θ ≈ θ.
Attempt assigned problems in Problem Sheet 1
Chapter
1
Assigned Problems
3,7,8,9,14,17,23,38,51
18
Scalar and vector quantities in physics
Physical quantities are classified in two categories:
scalars or vectors.
A scalar is a quantity that is completely specified by a
number (size), expressed in appropriate units.
e.g.
volume
time
temperature
energy
charge
3 m3
10 s
20 oC
5 J (Joules)
7 C (Coulombs)
Vectors, on the other hand, require that, in addition to
the size and units, we also specify a direction.
e.g.
displacement ≡ distance travelled in a particular
direction
Here, saying that the displacement is 10 m is not
enough:
moving 10 m North
is not the same as moving 10 m East
19
Force is another example of a vector quantity. To
describe the force on an object, we must specify both
the magnitude (size) of the force and the direction it is
applied:
A vector quantity is graphically represented by an arrow.
The arrow points in the direction of the vector, and its
length is proportional to the magnitude (size) of the
vector quantity.
Notation: We will indicate
r r rvector quantities by placing an
arrow on top (e.g., F, v , a ) and refer to their magnitude
by omitting the arrow (e.g., F, v, a).
!
Sum of scalar
and vector quantities
Scalar quantities are added using normal arithmetic (just
make sure that the units for all quantities are the same!).
Vectors, on the other hand, must be added in a special
way to take their direction into account.
To illustrate how it is done, we will use displacement as
a typical vector quantity (same applies for force, velocity,
acceleration, etc.).
20
If an object moves from
A to B (represented by
r
displacement vector d1), and then r from B to C
(represented by displacement vector d2 ), the result is
equivalent
to a single displacement from A to C (vector
r
d in the figure).!
!
!
r
r
r
We say that vector d is the vector sum of d1 and d2 , or
r
r r
d = d1 + d2 .
!
!
!
Note that this is a vector equation, different from
!
d = d1 + d2 ,
which refers to the magnitudes of the vectors, and does
not hold in this case (check using a ruler).
!
In fact, if the vectors are not along the same line, we
have that d &lt; d1 + d2 (magnitude of resulting vector is
smaller than the sum of the magnitudes of the separate
vectors).
!
21
The above example gives us a method to graphically
sum two (or more) vectors.
Imagine that we want to find the sum of the following two
forces
We take one of the vectors, and place the second one
with its tail at the tip of the first one, as shown below
The arrow going from the tail of the first one to the tip of
the second is the resulting sum vector
22
The process can be extended to more than two vectors.
In this case, the sum vector is found by connecting the
tail of the first one to the tip of the last one.
Note that the order in which the vectors are added is not
important
Let us look now in more detail to some special cases.
Case 1: The vectors are parallel to each other
In this case the sum vector is a vector in the same
direction as the vectors being added and we have
d = d1 + d2
18 m = 6 m + 12 m
23
!
Case 2: The two vectors are at right angles
Imagine that we want to sum the forces below, where
F1 = 10 N (newtons) [east direction]
F2 = 5 N [north direction]
We could do the sum graphically
using a ruler to
r
measure the magnitude of F and a protractor to
measure the angle.
Alternatively, we can make
use of the right triangle and
r!
find the magnitude of F using Pythagoras theorem:
2
2
2
2
→ F = F1 + F2 .
F 2 = F1 + F2
!
r
Since F is a vector, we also need to specify its direction.
We can calculate the angle θ (with respect to east)
!
!
using:
!
F
tan &quot; = 2
F1
#1 \$ F2 '
&quot; = tan &amp; ) .
% F1 (
→
24
!
!
Case 3: The two vectors are not at right angles
It can be shown that
F =
2
sin &quot; =
!
2
F1 + F2 + 2 F1 F2 cos &quot;
F2
sin # .
F
Subtraction of two vectors
!
r
r
Given a vector F , we define
– F vector as a vector with
r
the same magnitude as F , but opposite in direction.
!
!
!
r r
Note that with this definition, F − F = 0.
25
!
!
The subtraction of two vectors
r
r r
F = F1 &quot; F2
is then defined as
!
r
r
r
F = F1 + (&quot;F2 ) .
Graphically,
!
Components of a vector on the x and y axes
A much more powerful way to work with vectors is to
resolve them into suitable components.
These components are normally chosen to be along two
perpendicular directions or axes.
26
r
To see how this works, imagine a certain vector F (e.g.,
a force)
!
We introduce two axes (x and y), chosen as shown in
the figure below and project the vector on each of the
axes by drawing lines perpendicular to the axes:
r
The vector F is equal to the vector sum of the two
components, i.e.
r
r
r
F = Fx + Fy .
!
From the graph:
cos &quot;! =
sin&quot; =
!
Fx
F
Fy
F
Fx = F cos &quot;
→
Fy = F sin&quot;
→
!
27
!
!
Vector addition by the resolution of each vector into
its x and y components (recommended method)
r r r
Consider the addition of three vectors ( F1, F2 , F3 )
!
!
F1x = F1 cos &quot;1
F1y = F1 sin&quot;1
F2x = F2 cos &quot;2
F2y = F2 sin&quot;2
F3x = F3 cos &quot;3
F3y = F3 sin&quot;3
!
The
! magnitude of the vector!sum given by
!
Fx = F1x + F2x + F3x &quot; !
%
#
Fy = F1y + F2y + F3y \$
and its direction by tan &quot; =
!
Fy
Fx
28
!
F =
#
2
Fx + Fy
2
% Fy (
&quot; = tan '' **
&amp; Fx )
\$1
Note: If you measure your angles with respect to the
horizontal (+x axis), the sign of the components will
already take into account the orientation of the vector
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