PHYSICS - 1 (Lecture - 2)
Santabrata Das
Indian Institute of Technology Guwahati
sbdas@iitg.ernet.in
August 6, 2014
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Section 1
Newton’s Law of Motion
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First Law of Motion
Isolated Bodies Move With Uniform Velocity.
Common Notion: All moving bodies come to a halt.
Isolated Bodies
Inertial Frames
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Second Law of Motion
F =
d
dt
(mv) = ma
It is a law, not a definition!
Mass is constant and additive
Force is in the direction of change of velocity
Force: a spring force
Force = L − L0
L
Force = L’ − L0
L’
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Second Law of Motion
Mass is a measure of inertia
External agency that alters the state of motion
P
Force is a vector quantity F = Fi
i
F1
1
0
0
1
F1 + F2
F2
Arises out of interaction between systems.
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Second Law of Motion
Therefore, isolated bodies ⇒ No interaction ⇒ No force ⇒ motion with
uniform velocity.
Question: Is there laws for the forces themselves?
Answer: YES, in some cases.
Gravity
Electrostatic
Mass on a spring
But, for comlex systems, it is difficult to have a complete idea about
laws of the forces
three body motion
collision of two automobiles
molecular motion of a gas
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Third Law of Motion
Every action has equal and opposite reaction
Describes the nature of forces
Validates the concept that force is necessarily the result of
interaction between systems.
Is it always true?
There is some problem with electromagnetic forces.
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Fundamental Forces
Gravitational Forces
Electromagnetic Forces
Weak Nuclear Forces
Strong Nuclear Forces
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Gravitational Forces
Tycho Brahe and Kepler
Newton’s Law of Gravitation
F=G
mm0
b
r
r2
Force is along the line of centres and is attractive
Inertial or gravitational Mass?
Explains planetary Motion (Large distances)
Cavendish Experiment (On Earth)
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Electrical and Magnetic Forces
Coulomb’s Law (static charges)
F=
qq 0
b
r
r2
Magnetic Force (moving charges)
Fields E and B
Lorentz Force
F = q (E + v × B)
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Nuclear Forces
No known law for nuclear forces
Range of Nuclear Forces ∼ 10−13 cm
Scattering Experiments
Newtonian or Quantum Mechanics?
Quantum mechanical laws are valid with such small particles at a
tiny distance.
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Applications of Newton’s Laws
Mentally divide the system into smaller systems, each of which can
be treated as a point mass
Draw a force diagram for each mass
Fix the coordinate system
Use third law whenever necessary
Find out constraints and construct the relevent equation
Use second law
Identify the number of unknown quantities. There must be enough
number of equations ( Equations of motion + third law pairs +
constraint equations) to solve for all the unknown quantities.
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Everyday Forces
Tension in the ropes
Normal Forces
Frictional Forces
Viscous Forces
Springs
Atomic Forces
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Forces
Normal Force
Force exerted by a surface on a body in contact with it can be
resolved into two components
Component perpendicular to the surface is called Normal force
N → equal & opposite to the resultant of all other forces
perpendicular to the surface
Frictional force
For bodies not in relative motion, 0 ≤ f ≤ µN ; f opposes the
motion that would occur in it’s absence
For bodies in relative motion, f = µN ; f is directed opposite to
relative velocity.
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Constraints: Example I
A block moves on a wedge which in turn moves on a horizontal table.
The wedge angle is θ. How are the accelerations of the block and wedge
related?
Constraint - I: Vertical acceleration
is zero.
Constraint - II:
x
h
X
y
θ
x − X = (h − y) cot θ
ẍ − Ẍ = −ÿ cot θ
h−y
θ
x−X
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Coordinates are fixed on ground
Equations are independent of forces
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Example 2
The two blocks M1 and M2 shown in the sketch are connected by a
string of negligible mass. If the system is released from rest, find how
far block M1 slides in time t. Neglect friction.
x
Equations of motion:
M1
g
M1 ẍ1 = T
M2
and
M2 ẍ2 = W2 − T
Constraint equations:
L − x1 + x2 = const ⇒ ẍ1 = ẍ2
Unknowns: ẍ1 , ẍ2 , T and three equations.
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Example 2
Solve three equations simulteneously and obtain the solution as,
M2 g = (M1 + M2 )ẍ1
ẍ1 =
M2 g
M1 + M2
After integration,
x1 =
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M2 gt2
2(M1 + M2 )
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Example 3
A particle of mass m slides without friction on the inside of a cone. The
axis of the cone is vertical, and gravity is directed downward. The apex
half-angle of the cone is θ.
υ0
m
θ
The path of the particle happens to be a circle in a horizontal plane.
The speed of the particle is v0 . Draw a force diagram and find the
radius of the circular path in terms of v0 , g, and θ.
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Example 3
Force diagram is self-explanatory.
Equations of motion:
N sin θ = W
N cos θ = mrω 2
Dividing them yields
tan θ =
W
mg
g
=
=
mrω 2
mrω 2
rω 2
Speed of mass m is v0 = rω
tan θ =
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gr
v02 tan θ
⇒
r
=
g
v02
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Example 4
Masses M1 and M2 are connected to a system of strings and pulleys as
shown. The strings are massless and inextensible, and the pulleys are
massless and frictionless. Find the acceleration of M1 .
M1
M2
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Example 4
Equations of motion:
The fixed length of the string is a constraint.
M1 ẍ1 = T − M1 g
M2 ẍ2 = T 0 − M2 g
Unknowns: ẍ1 , ẍ2 , T, T 0 and we require four equations to solve the
problem.
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Example 4
Net force is 0 ⇒ 2T 0 = T
Then, from Eqns of motion, we have
T
M2 ẍ2
= M1 ẍ1 + M1 g
T
− M2 g
=
2
Constraint: Fixed length of the string is a constraint.
x1 + l1 + l10 +
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(x2 + l2 + l20 )
= constant
2
ẍ2 = −2ẍ1
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Example 4
Upon solving, we get
−4M2 ẍ1 = T − 2M2 g = M1 ẍ1 + M1 g − 2M2 g
(2M2 − M1 ) g
ẍ1
=
(4M2 + M1 )
Acceleration of M2 is twice the rate of M1 and the weight of M1 is
counterbalanced by twice the weight of M2 .
When
M1 M2 ⇒ ẍ1 ≈ −g and ẍ2 ≈ −2g
and if
M2 M1 ⇒ ẍ1 ≈
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g
2
and ẍ2 ≈ −g
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Example 5
A mass m is connected to a vertical revolving axle by two strings of
length l, each making an angle of 45o with the axle. Both axle and mass
are revolving with angular velocity ω. Gravity is directed downward.
Find the tension in the upper string, Tup , and lower string, Tlow .
ω
45°
l
m
45°
l
√
√
Radial distance of m from the axle is l/ 2 as cos 45o = sin 45o = 1/ 2
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Example 5
Equations of motion:
Tup
√
2
mlω 2
√
2
Tlow
= mg + √ → V ertical
2
Tup + Tlow
√
=
→ Radial
2
Unknows: Tup , Tlow and two equations.
Solving we get,
Tup =
Tlow =
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mlω 2 mg
+√
2
2
2
mlω
mg
−√
2
2
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Example 6
A disk rotates with constant angular velocity ω. Two masses, mA and
mB , slide without friction in a groove passing through the center of the
disk. They are connected by a light string of length l, and are initially
held in position by a catch, with mass mA at distance rA from the
center. Neglect gravity. At t = 0 the catch is removed and the masses
are free to slide. Find r̈A immediately after the catch is removed, in
terms of mA , mB , l, rA , and ω.
mB
ra
w
l
mA
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Example 6
Blocks are constrained by the groove
Tangential motion plays no dynamical role, neglected in force diag.
Force T on each mass is radially inward.
Equations of motion:
−T
= mA r̈A − rA ω 2
−T
= mB r̈B − rB ω 2
Constraints: rA + rB = l ⇒ r̈A = −r̈B
Solving,
mA
r̈A
r̈A − rA ω 2 = mB r̈B − rB ω 2 = mB −r̈A − (l − rA ) ω 2
=
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rA ω 2 −
mB lω 2
(mA + mB )
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Example 7: Whirling block
A horizontal frictionless table has a small hole in its center. Block A on
the table is connected to block B hanging beneath by a string of
negligible mass which passes through the hole. Initially, B is held
stationary and A rotates at constant radius r0 with steady angular
velocity ω0 . If B is released at t = 0, what is its acceleration
immediately afterward?
r̂
MA
T
θ
T
A
z
B
MB
z
WB
Two movable bodies and their free body diagram
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Example continued ...
Equations of motion:
−T
0 = MA (rθ̈ + 2ṙθ̇)
WB − T
Radial
= MA (r̈ − rθ̇2 )
= MB z̈
Tangential
Vertical
Constrain equations: r + z = ` ⇒ r̈ = −z̈.
Unknowns: ar , aθ , z̈, T . Four unknowns and four Eqs.
z̈ =
WB − MA rθ̇2
MA + MB
Immediately after B is released, r = r0 and θ̇ = ω0 . Hence,
2
WB − MA r0 θ˙0
z̈(0) =
MA + M B
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Example 8: Block on wedge with friction
A block of mass m rests on a fixed wedge of angle θ. The coefficient of
friction is µ. Find the value of θ at which the block starts to slide.
Constraint:
N
f
mÿ = 0
y
m
Eqns of motion:
x
θ
W
mẍ = W sinθ − f
mÿ = N − W cosθ
θ
= 0
Coordinate system is fixed on
the wedge.
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Example 8: Block on wedge with friction
When sliding is about to start: f = fmax = µN ; ẍ = 0.
Eqns of motion:
W sin θmax = µN
W cos θmax = N
tan θmax = µ
As the wedge angle is gradually increased from zero, the friction force
grows in magnitude from zero to its maximum value µN , since before
the block begins to slide we have
f = W sin θ
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θ ≤ θmax
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Example 8: Block on wedge with friction
Now, the wedge is given horizontal acceleration a. Assuming that
tan θ > µ, find the minimum acceleration for the block to remain on the
wedge without sliding.
Horizontal acceleration of the block is
g
mamin = N cos θ − f sin θ; f ≤ µN
(1)
a
θ
In the limit f = µN , we have
mamin = N (cos θ − µ sin θ)
Block has zero Vertical acceleration
N sin θ + f cos θ − mg = 0
N (sin θ + µ cos θ) = mg
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(2)
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Example 8: Block on wedge with friction
Eleminating normal force N from Eqns (1) and (2) we get
cos θ − µ sin θ
amin =
g
sin θ + µ cos θ
We can estimate the maximum acceleration also.
Horzontal acceleration of the block
mamax = N (cos θ + µ sin θ) ;
f = µN
Vertical acceleration of the block
N sin θ − f cos θ − mg = 0
As earlier,
amax =
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cos θ + µ sin θ
sin θ − µ cos θ
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g
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Examples 9
A 4 Kg block rests on top of a 6 Kg block, which rests on a frictionless
table. Coefficient of friction between blocks is 0.25. A force F = 10N is
applied to the lower block.
4 Kg
6 Kg
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F
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Examples 9: Force Diagrams
N
F1
4 Kg
Wb
N’
F1
F
6 Kg
Wa
N
N’
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Examples 9: Coordinate System and
Constraints
Fix the coordinate system to the table.
xB
yB
xA
yA
4 Kg
6 Kg
yA = const
yB = const
xA = xB + const
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Examples 9: Equations
Equations of Motion in Y direction.
mA ÿA = N 0 − WA − N
Constraints
mB ÿB = N − WB
ÿA = 0
ÿB = 0
Solution
N 0 = WA + WB
N
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= WB
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Examples 9: Equations
Equations of Motion in X direction.
mA ẍA = F − F1
mB ẍB = F1
Constraints
ẍA = ẍB
Solution
F
= 1 m/s2
mA + MB
= mB ẍB = 4N
ẍA = ẍB =
F1
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Examples 9: Example Continued ...
The force F1 < µN = 10 N, the maximum frictional force between
the blocks. Hence the solution is consistent with assumption.
What would be the motion if F = 40 N?
If the blocks move together then xB = 4 m/s and F1 = 16 N! More
than the maximum frictional force!
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Examples 9: Equations
Equations of Motion in X direction.
mA ẍA
= F − F1
mB ẍB
= F1
But,
F1 = µN
Solution
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ẍA
=
ẍB
=
F − µN
= 5 m/s2
mA
µN
= 2.5 m/s2
mB
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Example 10
A block of mass m slides on a frictionless table. It is constrained to
move move inside a ring of radius l fixed to the table. At t = 0 the
block is touching the ring and has a velocity v0 in tangential direction.
1111
0000
0000
1111
m
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
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1
0
0
1
0
1
0
1
0
1
0
1
0
1
l
0
1
0
1
0
1
0
1
0
1
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Example 10: Equations
Constraint Equation is r = l, that is ṙ = r̈ = 0.
Equations of Motion
m r̈ − rθ̇2 = −mlθ̇2 = −N
m rθ̈ − 2ṙθ̇ = mrθ̈ = −f
Eliminating N , we get
θ̈ = −µθ̇2
v(t) = lθ̇
=
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v0
1 + µv0 t/l
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Tension in a string
A string consists of long chains of atoms characteristic of the particular
material found in the string. When a string is pulled, we say it is under
tension. The long chains of molecules are stretched, and inter-atomic
forces between atoms in the molecules prevent the molecules from
breaking apart. A detailed microscopic description would be difficult
and unnecessary for our purpose. Instead, a macroscopic model would
be developed for the behavior of strings under tension.
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Block and String
A string of mass m attached to a block of mass M is pulled with force
F . What is the force F1 that the string “transmits” to the block?
Neglect gravity.
Equations of motion are:
F1 = M aM ;
F − F1 = mas
F
m
String is inextensible: as = aM
By Newton’s third law: F1 = F10
a = as = aM
F
=
M +m
M
aM
M
F1 F ′1
aS
F
and
F1 = F10 =
M
F
M +m
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Block and String
The force on the block is less
than F
The string does not transmit
the full applied force.
F
m
M
aM
M
F1 F ′1
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aS
F
If the mass of the string is
negligible compared with the
block, F1 = F to good
approximation.
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Continued...
A string is composed of short sections interacting by contact forces.
Each section pulls the sections to either side of it, and by Newton’s
third law, it is pulled by the adjacent sections. The magnitude of the
force acting between adjacent sections is called tension. There is no
direction associated with tension. In the sketch, the tension at A is F
and the tension at B is F 0 .
A
A
F
B
B
F
F′ F′
Although a string may be under considerable tension, for example a
string on a guitar, the net string force on each small section is zero if
the tension is uniform, and the section remains at rest unless external
forces act on it. If there are external forces on the section, or if the
string is accelerating, the tension varies along the string.
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Dangling Rope
A uniform rope of mass M and length L hangs from the limb of a tree.
Find the tension in the rope at distance x from the bottom.
The section is pulled up by a force
of magnitude T (x), where T (x) is
the tension at x. The downward
force on the section is its weight
W = M g(x/L). The total force on
the section is zero since it is at rest.
Hence
T (x) =
Mg
L x.
T(x)
L
x
x
W
At the bottom of the rope the tension is zero, while at the top where
x = L the tension equals the total weight of the rope M g.
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The Spinning Terror
The Spinning Terror is an amusement park ride—a large vertical drum
that spins so fast that everyone inside stays pinned against the wall
when the floor drops away. What is the minimum steady angular
velocity ω that allows the floor to be dropped safely?
ω
ω
R
Radial Equ. N = M Rω 2 .
f
By the law of static friction,
f ≤ µN = µM Rω 2
Since we require M to be in
vertical equilibrium, f = M g.
Therefore, M g ≤ µM Rω 2
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⇒ ω2 ≥
g
µR
PHYSICS - 1 (Lecture - 2)
M
N
W = Mg
⇒ ωmin =
q
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g
µR
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Example
Consider the force as a function of time, position, or velocity, and solve
the differential eq.
d2 x
= F (x, t, v)
dt2
to find the position, x(t), as function of time
m
Since we need to solve second order differential equations, we must have
two initial conditions, usually x0 ≡ x(t0 ) and v0 ≡ v(t0 ), to obtain the
final solutions.
Three special cases: F is function of t only, function of x only, and
function of v only.
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Example
F is a function of x only: F = F (x)
We will use
a=
dv
dx dv
dv
=
=v
dt
dt dt
dt
Following
F = ma
⇒ mv
dv
= F (x)
dx
F = −kx
or
F = −k/x3
As example,
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Example Continued ...
Now separate the variables and integrate both sides to obtain
ˆ
v(x)
m
ˆ
v 0 dv 0 =
v0
x
F (x0 )dx0
x0
Separating variables in dx/dt ≡ v(x), yields
ˆ
x(t)
x0
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dx0
=
v(x0 )
ˆ
t
dt0 .
t0
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Inverse-cube force field
A particle of mass m starts from rest at x0 (> 0) in an attractive
inverse-cube force field F = −k/x3 (k is a positive constant). Calculate
the time taken to reach the origin.
The equation of motion is,
dv
k
dv
= mv
=− 3
dt
dx
x
ˆ v(x)
ˆ x
dx
m
vdv = −k
3
0
x0 x
s
p
x20 − x2
dx
k
v(x) =
=−
dt
x
mx30
m
‘-’ sign for motion in ‘-’ x-direction.
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Inverse-cube force field
Integrating the velocity Eq. with respect to t,
ˆ
x(t)
x0
xdx
p
x20 − x2
s
= −
k
mx20
s
x(t) = x0
1−
ˆ
t
dt
0
k 2
t
mx40
‘+’ root as particle is to the right
qof x > 0.
For t = T0 , x = 0, then
T0 =
mx40
k
and
k
k 2 −1/2
v(t) = −
t 1−
t
mx30
mx40
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Free motion in viscous medium
A body of mass m released with velocity v0 in a viscous fluid is retarded
by a force Cv. Find the motion, supposing that no other forces act.
F = −Cv ⇒ −Cv = m
dv
v
ˆ v
dv
v0 v
v
log
v0
v
dv
dt
C
dt
m
ˆ t
C
= −
dt
0 m
C
= − t
m
= v0 exp(−C/m)t
= −
Let τ = m/C, then v = v0 e − t/τ . τ is a characteristic time for the
system; it is the time for the velocity to drop to e − 1 ≈ 0.37 of its
original velocity.
Santabrata Das (IITG)
PHYSICS - 1 (Lecture - 2)
August 6, 2014
54 / 56
Free motion in viscous medium
Therefore,
dx
= v0 e−t/τ
dt
Hence,
ˆ
t
v0 e−t/τ dt
0 −t/τ
x = v0 τ 1 − e
x =
As τ → ∞, C → 0, means no resistance. Then x = v0 t.
1 2 1 3
−x
e = 1 − x + x − x + ...
2
6
Santabrata Das (IITG)
PHYSICS - 1 (Lecture - 2)
August 6, 2014
55 / 56
Santabrata Das (IITG)
PHYSICS - 1 (Lecture - 2)
August 6, 2014
56 / 56