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Work
Introduction
The concept of work-energy is universal and is applicable almost in all fields of physics, engineering chemistry, biology etc.
However here we analyze the application of work concept in Mechanics. ‘Work’ has much more in it then just a language tool. For
example, if a person is holding an object, he gets tired but still does no work. Here we will analyze such myths and also explore the
term power. We will also gain an insight into energy approach for solving mechanics problems which were tedious to solve using
Newton’s
law.
WORK:
When ever force acting on a body is able to actually move it through some distance in the direction of force, the work is said to be
done
by
the
force.
The
force
performing
work
may
be
constant
or
a
variable.
Work
done
by
CONSTANT
FORCE:
An object undergoes displacement‘s’ along a straight line while acted on by a force F, the angle between
done
isq. Then work
is
W=
Fig (1)
a)
Work
done
as
such
has
b)
Work
is
scalar
c)
Unit
of
d) Dimension of work is [ML2T-2]
no
relevance
quantity
work
is
until
(as
the
commenced
force
is
indicated
by
dot
Joules
1J
=
maintained.
product).
1Kgm 2/s2
Other units of Work:
Fig. (2)
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fluid mechanics
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umang
77
24/02/2007 10:53
rao_prabhat
elasticity
9
kash001
124
24/02/2007 10:44
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stoke's law
3
umang
43
24/02/2007 10:40
devnexus
basic summary of kinematics
0
devnexus
6
24/02/2007 10:36
devnexus
straight line motion
1
nrki99
25
24/02/2007 01:38
truly
FOR EXPERTS
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deep01
31
24/02/2007 01:24
truly
kinematics
7
mandar5227
73
24/02/2007 01:14
truly
constrain relation
0
joydas2514795
18
23/02/2007 21:36
joydas2514795
know the answer but not the reason
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tibu
68
23/02/2007 21:16
aditya_arora04
plzzz help-rotation prob
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anubhav07
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23/02/2007 21:00
vish0001
axis of rotation query- please answer quickly !!
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vish0001
28
23/02/2007 20:39
vish0001
rod
9
neeraj_agarwal_1990
91
23/02/2007 19:08
ankur.kkhurana
mechanics
3
sucess_seeker
93
23/02/2007 18:24
bhupesh.mohanty
rotational motion
2
swapna.sugunendran
57
23/02/2007 18:19
bhupesh.mohanty
QUESTIONS FOR THE EXPERTS
1
deep01
47
23/02/2007 18:07
anuragasp
Illustration:
1) A block of mass M is pulled along a horizontal surface by applying a force at an angle
with horizontal. Co-efficient of friction
between block and surface is . If the block travels with uniform velocity, find the work done by this applied force during a
displacement
d
of
the
block.
Solution: The force acting on the block is shown in figure.
Fig (3)
As
the
FCosq
Solving
block
=
FSinq+N=
moves
with
uniform
velocity
the
force
N
Mg
(1)
and
add
up
-------------------------------
to
zero.
(1)
(2)
(2)
FCosq
F
Work
=
done
by
W
WORK
If
the
=
a
displacement
variable
(Where
(i.e.
remains
VARYING
then
changes)
constant
for
S
done
interpretation
=
Area
need
to
d
is
=
BY
a
(If
Graphical
Work
Area
during
F.dCosq
DONE
is
force
=
force
this
=
(Mg-FSinq)
work
done
displacement
Ds)
0
of
under
)
above
vs.
with
F
add
FORCE:
is
result:
graph.
sign.
S
Fig (4)
SPRING:
The
spring
F
=
force
-KX,
varies
as
where
Work
done
(Initial
compression/elongation
a
K
lines
is
function
of
stiffness
in
is
X0
compression
constant
or
of
elongation.
spring.
on
and
final
spring:
compression/elongation
is
X1)
Proof:
(Assuming
force
is
parallel
to
displacement)
=
Illustration:
A block of mass m released from rest onto an ideal non-deformed spring of spring constant ‘K’ from a negligible height. Neglecting
the
air
resistance,
find
the
compression
‘d’
of
the
spring.
Solution:
Fig (6)
Here
there
are
two
forces
(Work
Net work
K=
W mg
POWER:
Power is
Power
done
involved,
done
must
be
equal
to
change
0-0
in
as the
can
and
kinetic
energy
(theorem
to
=
rate
at
be
force.
which
work
is being
average
weight)
be
W spring
time
spring
by
+
defined
weight
discussed
in
next
=
done
or
energy
or
is
being
section).
0
0
transferred.
instantaneous
Proof:
Remember power is never negative. dW stands for small work done and not difference in work because nothing called ‘difference
of
work’
exists.
Practical
unit
of
power
is
horse
power
(hp)
1hp = 746W.
Illustration:
An elevator has a mass of 1000kg and carries a maximum load of 600kg. A constant frictional force of 4000N retards its upward
motion, as shown in figure.


What must be the minimum power delivered by the motor to lift the elevator at a constant speed of 3 m/sec.
What power must the motor deliver at any instant of time, if it is designed to provide an upward acceleration of 100 m/s2
Fig
(7)
Solution:
Let
From
a)
force
applied
by
T-W-f
motor
f.b.d
0
=
T
be
T
of
[because
W+f
=
=
P
b)
V
=
Ma+W+f
=
=
TV
of
elevator
=
at
3
Watts
104
=
=
Where
v
is
instantaneous
speed
(Without ‘v’, value of P cannot be evaluated).
f
lift
is
constant]
(1000+600)´9.8+4000
19680N
x
T-W-f
P
be
T.VCos
19680
x
T
(8)
force
=
=
5.9040
=
Fig
frictional
and
which
power
Ma
=1600´1+1600´9.8+4000
21280
21280v
Watts.
needs
to
be
determined
ENERGY:
A body is said to possess energy if it has the capacity to do work. If some work is done by the body then it looses energy.
Energy
and
work
are
mutually
interconvertible.
Various forms of Energy are:
1)Heat 2) Electrical 3) Chemical 4) Mechanical 5) Nuclear.
MECHANICAL
ENERGY:
Kinetic Energy: Kinetic energy of a body is the energy possessed by a body by virtue of its motion. It is the energy associated with
moving
body.
,
Where,
m-mass
of
body,
v-velocity
of
Kinetic Energy is measured by the work done by moving against external impressed force, before coming to rest.
Relation
between
Kinetic
Energy
and
body.
Momentum:
How?
A ball falls under gravity from a height 10m, with an initial velocity V0. It hits the ground, looses 50% of its energy after collision and
it
rises
to
the
same
height.
What
is
the
value
of
V0?
Solution:
Let
Let
V1
Ratio
[
V
be
of
the
K
be
velocity
before
50%
the
velocity
after
impact
and
when
and
after
it
reaches
impact
of
it
hits
the
the
same
height
ground
10m
=
K
is
lost]
Potential
Energy:
Potential Energy of a body is the energy possessed by the body by virtue of its position or configuration. Potential energy is due to
the interaction between bodies.




Physically potential energy is applicable only to the class of forces where work done against the force gets ‘stored up’ as
energy. Such forces are called CONSERVATIVE FORCES.
Conservative forces perform work whose value is path independent. Thus work done by conservative force in a closed
path is zero.
Forces which do not satisfy path independent nature are called NON-CONSERVATIVE FORCES.
Mathematically conservative forces F(x) can be written in terms of their Potential energy forces V(x) as
For
3-Dimensional
Consequently
potential
energy
for
case
single
dimension
is
given
by
Potential
F
=
V-0
-Kx
here
Energy
natural
length
of
configuration
is
considered
Stretched
as
zero
of
potential
spring:
energy.
=
Illustration:
A spring of normal length ‘ ’ and spring constant K is fixed on the ground and the other is filled with a smooth ring of mass m
which slides on a horizontal rod fixed at a height also equal to l (see fig). Initially the spring makes an angle of 53 0 with horizontal
when
the
system
is
released
from
rest
then
what
is
the
speed
of
the
ring?
Solution:
In
the
initial
position
of
the
ring
as
shown
in
Fig
figure
the
length
of
the
(13)
spring
Extension
Energy
stored
in
spring
=
This stored energy when released becomes Kinetic energy of the ring, if V is the velocity of the ring, Kinetic energy when it is
vertical
Equating
=
two
energy
WORK
ENERGY
THEOREM:
Work Energy theorem states that, the net change in Kinetic Energy of a particle (system) is equivalent to the sum of work done by
all
types
of
external
forces
be
it
conservative
or
non-conservative.
Dumb Question:

Velocity time graph of particle of mass 2kg moving in a straight line is as shown in figure. Find the work done by all the
forces on the particle.
fig.14
Solution:
Initial velocity of particle V0 = 20 m/s,
Final
velocity
of
particle
vf,
From
Work
energy
theorem
=
0-400
Best thing about work energy theorem is that nature and value of force is not at all required for work equation.
Proof
of
Work
(Multiplying
Energy
y
and
(Segregating
Integrating
=
both
dividing
the
400.
Theorem:
by
‘ds’)
term)
side
Illustration:
Two bodies A and B connected by a light rigid bar of 10m long and moving in two frictionless guides as shown in the figure. If B
starts from rest when it is vertically below A, find the velocity of B when x=6m. Assume m A = mB =200kg and mC = 100kg.
Fig
Solution:
At
any
instant
Where
Including
C
move
Work done by
Final
Kinetic
when
the
velocity
of
all
down
gravity on
6m
system =
energy
of
Velocity
from
gravity
only
of
B
at
is
A
three
since
{Work done
the
(Speed
Now
(W ext)net Here
bar
system
of
(16)
as
shown
and
bodies
B
in
=
B
Kf
=
and
the
required
moment
the
----------
(2)
+
6m
Work
(KA)
C
force
(16)
(1)
velocity
work
external
constitute
figure
---------
moves
down A
moving
the
in
is
CONSERVATION
OF
Kinetic
and
Potential
energy
Total Mechanical energy = K+V changes only if
in
+
of
our
along
done in
(KB)
must
energy
which
=
MECHANICAL
mechanical
+
be
is
=
of
(KC)
same)
doing
6.9m/s
energy
B
system
x-axis
moving C}
theorem
work.
Ans.
ENERGY:
system.


Energy is converted into other forms like heat, sound etc.
Work is being done by non-conservative forces like, frictional force
Proof: Let a body undergo displacement Dx under the action of conservative force F. Then from work energy theorem
Kf
–
Ki
=
F(x) x
--------------(1)
Since
force
is
conservative
hence
its
potential
energy
function
can
also
be
written
as
Vf-Vi=
-F(x) x
----------------(2)
Adding
(1)
and
(2)
(Vf+Kf)(Ki+Vi)
=
F(x)
Dx
F(x)
Dx
=
0
Vf+Kf
=
Ki+Vi
Initial
Mechanical
energy
Important Points regarding Conservation Principle:



=
Final
Mechanical
energy.
If friction and drag are present then mechanical energy will not be conserved
Conservation of mechanical energy applies only to isolated systems.
Initial and final states of system must be clearly identified before applying conservation principle.
Illustration:
A block of mass m is pushed against a spring of spring constant K fixed at one end to a wall. The block can slide on a frictionless
table as shown in figure. The natural length of the spring is L 0 and it is compressed to half its natural length when the block is
released.
Find
the
velocity
of
the
block
as
a
function
of
its
distance
x
from
the
wall.
Solution:
Fig (18)
When the Block is released the spring pushes it towards right. The velocity of the block increases till the spring acquires its natural
length. There after the block loses contact with the spring and moves with constant velocity. Initially the compression in the spring
=
When
Using
the
distance
the
of
block
from
principle
the
wall
becomes
of
its
where
x<
conservation
the
compression
of
is
Solving
When
(L 0-x)
energy
this
the
spring
acquires
We
have
There after the block continues with this velocity.
its
natural
then
length
x
=
L0
CIRCULAR
MOTION:
Uniform Circular motion: Angular velocity w is constant throughout the motion. The magnitude of velocity also remains constant.
Non-Uniform
Circular
Relation
Angular
motion:
Between
Taking
a
moves an angle
velocity
v, ,
small
time
‘ ’
changes
r,
with
time.
:
interval
t
in
which
body
(very small)
Fig (19)
Distance
Considering
moved
this
time
by
interval
it
to
on
be
so
v
circumference
small
that
is
the
=
distance
(
is
)
almost
=
Actual
=
straight
s
line.
r
vector
equation:
CENTRIPETAL
AND
TANGENTIAL
FORCES:
The forces acting on a body could be resolved into two components, one in radial direction and one in tangential direction.
The force in radial direction is called CENTRIPETAL FORCE and the other is called TANGENTIAL FORCE.
Fig
In
First
part
is
tangential
equation
acceleration
(20)
and
second
is
centripetal
acceleration.
Illustration:
A 0.1 Kg block is undergoing circular motion. What is the range of ‘ ’ for which the particle can perform the circular motion?
Solution:
f.b.d
For
of
body
for
wmin,
vertical
the
particle
will
Equilibrium
Similar
value
of
‘ ’
friction
the
move
and
R
body
will
=
will
have
be
analysis
(22)
downwards.
body
2R
(1)
Now
For
maximum
Direction
to
of
For body to rotate in circle, net force towards center of circle must be equal to m
From
Fig
tendency
have
(2)
h
tendency
in
to
move
downward
tan450=h
upwards
direction.
yields.
Derivation:
Motion
in
a
vertical
circle:
A particle of mass m is attached to inextensible light string of length l and it is imparted a velocity u in horizontal direction lowest
point.
Let
v
be
its
velocity
at
point
B
of
circle.
Fig
Here
h
By
centripetal
Now
So
R
(1-Cosq)
conservation
Necessary
The
=
of
force
three
string
substituting
is
mechanical
provided
by
the
condition
does
T
not
=
0
energy
resultant
arise
stuck
at
and
(23)
(1)
------------
tension
T
and
depending
highest
=
of
point
for
if
completing
mgCos
on
T
0
the
circle
u,
at
= ,
in
(B)
(At
And
So
highest
h
from
point)
2R
(2)
=
equation
Or
Or
the
particle
Substituting q = 00 and v = u =
position is 6mg.
will
complete
the
circle.
in equation (3) we get T = 6mg. So in the critical condition tension in the string at lowest
Fig (24)
Putting
T
=
0
Substituting
in
Velocity
0
Now the particle
T
=
h1<h2
before
reaching
highest
equation
of
(1)
particle
becomes
point
for
we
zero
=
will
0
leave
the
but
circle
if
v
tension
0
in
the
string
this
becomes
is
zero but velocity is not
possible
only
get
when
u2-2gh
zero or,
when
Therefore
The
V
h2
particle
=
if
will
Further if h1 = h2,
oscillate
0
the
if velocity
but
of
T
the
particle
0
particle
becomes
this
<
zero
leaves
but
is
tension in the
possible
the
circle.
string is
only
not
zero.
when
h1
and tension and velocity both becomes zero simultaneously.
Illustration:
A ball of mass m slides without friction down a path from height h and then moves in loop of radius
exerted
on
the
ball
by
the
track
at
B
and
Find the force
at
C
Fig
(27)
Fig
(28)
Solution:
For
On
maintaining
applying
circular
energy
motion
net
conservation
On
centripetal
between
force
must
point
A
be
greater
and
C
than
we
get
solving
From
On
Similarly
f.b.d
of
body
FB=3mg
point
solving
for
mg
[h-(R+RCosa)]
=
(1+Cosb)
B
(energy
equilibrium)
FB = 3mg (1-Cosa)
Banking
of
Roads:
When vehicle go through turnings, they travel along a nearly circular arc. There must be some force to produce the required
centripetal acceleration. Centripetal force is provided to the vehicle by following three ways.



By friction only
By banking of roads only
By friction and banking of roads both.
1)
Let
a
And
car
of
f L=
mass
m
is
limiting
For
moving
value
a
at
a
By
speed
of
f
safe
v
is
=
at
horizontal
N
turn
So if m and r is fixed the speed of the vehicle should not exceed
circular
=
arc
of
friction:
radius r.
mg
(N=mg)
without
sliding,
and if v and r is fixed then coefficient of friction should be
greater than
2)
To
By
avoid
friction
outer
part
of
banking
road
is
some
of
what
lifted
roads
compared
to
the
inner
only:
part.
Fig
From
figure
From
At
this
speeds
two
(29)
NSin
we
car
does
(29)
=
get,
not
and
Tan
down
slide
NCos
=
mg
=
even
if
track
is
smooth.
3)
By
friction
and
banking
of
road
both:
If a vehicle is moving on a circular road which is rough and banked also then magnitude of N and direction plus magnitude of
friction mainly depends on the speed of the vehicle V-

f is outward if v = 0

f is inward if v>

f is outward if v <

f is zero if v =
a)
When
b)
Now
the
when
the
car
car
is
is
given
at
a
small
rest
speed
v,
f
Fig
-------
is
equation
upward-
(1)
(30)
becomes
Illustration:
A turn of radius 20m is banked for the vehicle of mass 200kg going at a speed of 10m/s. Find the direction and magnitude of
frictional force acting on a vehicle if it moves with a speed a) 5 m/s b) 15 m/s assume the friction is sufficient to prevent slipping ( g
=
10m/s2).
Solution:
a)
As
The
turn
the
is
banked
speed
is
for
speed
v
decreased
=
10
force
m/s
of
therefore
friction
f
acts
upwards
Fig
Substituting
b)
Substituting
,
Here
v=5
force
v=15
m/s,
of
m/s,
(31)
friction
,
m=200kg,
f
m=200kg
will
and
r=20m
act
r=20m
downwards
we
get
PROBLEMS
(EASY)
1) Force acting on a particle varies with x as shown in figure (33) .Calculate the work done by the force as the particle moves from
x = 0 to x = 6m.
Fig (33)
Solution: The work done by the force is equal to the area under the curve from x=0 to x=6.0m. This area is equal to the area of the
rectangular section from x = 0m to x = 4.0m plus the area of the triangular section from x = 4.0m to x = 6m.
2) The mass system is kept on sphere. Ball 1 is slightly disturbed. What is the velocity of these balls when it is making angle ‘q’
with
horizontal
(friction
is
absent
everywhere).
Fig (34)
Solution:
Since they are connected by inextensible string, therefore at any stage the velocity of both particle will be same. Let it be ‘V’.
From
work
energy
theorem:
(Only
3) What
is
the
minimum
value
of
‘u’
for
completing
external
circular
motion
force
of
particle
Fig
Solution:
as
is
shown
gravity)
in
figure
(35)?
(35)
Choosing
If
V1
zero
is
of
the
velocity
potential
at
energy
‘A’
then
at
from
level
energy
of
A.
conservation
4) A block is projected horizontally on rough horizontal floor. The coefficient of friction between the block and the floor is m. The
block strikes a light spring of stiffness K with velocity v0. Find the maximum compression of the spring.
Fig (36)
�������
Solution:
Since the block slides and the spring is compressed through a distance ‘x’ the net retarding force acting on it.
=
Þ
F
Work
=
done
by
(kx
net
+mN)
force
for
Fig
=
the
(37)
(mmg+kx)
displacement
x,
5) In figure 38-A and 38-B AC, DE and EF are fixed inclined planes BC = EF and AB = DE = y. A small block of mass m is released
from rest from the point A. It slides down AC and reaches C with a speed VC. The same block is released from rest from the point
D, it slides down DEF and reaches the point F with speed VF. The coefficient of kinetic friction between the block and the surface
AC
and
DEF
is
a,
calculate
VC
and
VF.
Fig
Solution:
a)
ME
This
Loss
at
A
in
=
mgY+0
ME
is
and
equal
(As
b)
In
if
to
(38)
VC
work
is
done
Cosa
this
the
velocity
against
=
at
friction
C
i.e.
x/s)
situation,
6)
A
the
Solution:
locomotive
of
mass
m
starts
moving
so
that
its
velocity
v
is
according
to
the
law
where a is constant and s is distance covered. Find the total work done by all the forces acting the locomotive during
first
t
seconds
after
the
beginning
of
motion.
Given,
w.r.t
differentiating
Accelerating
now
force
Here
‘t’
we
acting
on
locomotive
u
Now
get
=0
using
we
have
7) The Kinetic energy of a particle moving along a circle of radius R depends on the distance covered S and T = as2, where a is
constant.
Find
the
force
acting
on
the
particle
as
a
function
of
S.
Solution:
Differentiating
both
sides
w.r.t
‘t’
we
have
a
Now
force
=
acting
on
the
particle
is
given
by
8) A ball suspended by a string of length 20 cm is fixed to the free end of the pivoted rod of length 40 cm as shown in the figure
(39). The rod is made to rotate in a horizontal plane with constant angular speed. The string makes an angle q = 300 with the
vertical axis. Find the angular speed of the rotation?
Fig (39)

= angular speed
T = Tension in the string.
Ball
moves
in
horizontal
circle
TCosq
=
mg
Along
radial
direction,
TSinq
Using
(1)
with
radius
=
mrw2
r
=
L1+L2Sinq
and
along
vertical
direction,
----------(1)
----------(2)
(2)
w
=
3.398
rad/s
9) A smooth, light rod AB can rotate about a vertical axis passing through its end A. The rod is fitted with small sleeve of mass m
attached to the end A by a weightless spring of length l0, stiffness k. What work must be performed to slowly get this system going
and
the
angular
velocity
w?
Fig
Solution:
The
mass
Centripetal
\W
=
m
rotates
force
change
in
in
a
on
circle
of
radius
=
m
KE
of
(40)
l,
m+
which
k
energy
is
extended
(l-
stored
length
l0)
of
in
=
the
spring.
mw2l
the
spring
10) A spring gun having spring constant 100 N/m, a small ball of mass 0.1Kg is placed in its barrel by compressing the spring
through
0.05m
as
shown
in
figure
(41)
a)
Find
the
velocity
of
the
ball
when
spring
is
released
b) Where should a box is placed on ground so that ball falls in it, if the ball leaves the gun horizontally at a height of 2m above the
ground.
Fig
(41)
Solution:

When the spring is released its elastic potential energy is converted into kinetic energy.

As vertical component of velocity of ball is zero.
Time
taken
by
the
ball
to
reach
the
ground,
So,
the
horizontal
distance
traveled
by
the
ball
in
this
time
PROBLEMS
(MODERATE)
1) A hemispherical bowl of radius R = 0.1m is rotating about its own axis (which is vertical), with an angular velocity w. A particle of
mass 10-2Kg on the frictionless inner surface of the bowl is also rotating with the same w. The particle is at a height h from the
bottom
of
the
bowl.
a) Obtain the relation between h and w. What is the minimum value of w needed, in order to have a non-zero value of h?
b) It is designed to measure g (acceleration due to gravity) using this step-up, by measuring h accurately. Assuming that R and w
are known precisely and that the least count in the measurement of h is 10-4m, what is its minimum possible error Dg in the
measured value of g?
Solution:
Fig
a)
Þ
N
Along
Using
mRw2Cosq
b)
\Dg
|Dg|min
Along
x-direction,
NSinq
=
y-direction,
NSinq
=
=
mRw2
NCosq
(1)
m
m
=
(AC)
(RSinq)
----------------------------
mg
and
=
From
=
(3),
=
|Dh|
g
w2min
=
-Dh
=
(R-h)
(42)
w2
w2
(1)
(2)
(2)
mg
2
w2
2) A small body is placed on the top of a smooth hemisphere of radius R. When the sphere is given a uniform horizontal
acceleration a0 the body starts sliding down. a) Find the velocity of body relative to sphere at the instant of loosing contact b) Find
the angle Æ between radius vector drawn to the body from center of sphere at the time of loosing contact if a0 = g
Fig
(43)
a) Drawing f.b.d in frame of reference of particle.
From
For
just
Eliminating
b)
2g
work
loosing
a0
Equating
(1-Sinq)
energy
contact
N
from
+
=
0
then
(1)
(1)
2a0Cosq
theorem
using
(1)
(2)
and
and
=
(2)
gSinq
-
and
(2)
gives
gives
a0Cosq
3) A small bar A resting on a smooth horizontal plane is attached by threads to a point P as shown in figure. And by P means of a
weightless pulley to weight B possessing the same mass as the bar itself. Beside, the bar is also attached to a point O by means of
a light non deformed spring of length l0 = 50cm and with spring constant K, then find the velocity?
Fig
(44)
Solution:
Fig (45)
Drawing
N
Þ
From
the
f.b.d
=
of
KxCosq
KxCosq
(1)
A
for
=
just
breaking
=
off
=0
mg
and
(2)
mg
Mass
Since
Now
they
B
are
would
have
connected with inextensible
from
moved
string hence
same
velocity of
energy
A
distance
and B would
downwards
be same.
conservation,
4) A horizontal plane support a stationary vertical cylinder of radius R and disc A attached to the cylinder by a horizontal thread AB
of length l0. An initial velocity v0 is imported to the disc as shown. How long will it move along plane until it strikes against the
cylinder?
The
friction
is
assumed
to
be
absent.
Fig
(46)
Solution:
Fig (47)
ds
If
the
=
is
the
angle
path
turned
by
(l0-R
string
traversed
till
the
disc
comes
by
in
)
contact
disc
d
with
cylinder
the
length
of
Dumb Question:

How is l0 = R
?
Ans: Suppose the final picture is fig.(48). The whole thread rolls over the cylinder, then as obvious the length l 0 = R
.
Fig (48)
How the motion does become a projectile motion?
Ans: Note that at point B, the string becomes slack, so there is no tension acting over the particle. Now the particle has
some velocity and gravity is acting over it, so it starts projectile motion at B and ends at A.
PROBLEMS
(HARD)
1) A particle is suspended by a string of length ‘l’. This is projected with such a velocity v along the horizontal such that
after
the
string
becomes
slack
it
flies
through
its
initial
position.
Find
V?
Solution:
Fig (49)
Let the velocity be v1 at B where the string becomes slack and the string makes angle q with horizontal, by the law of
conservation
of
energy.
Here
From
glSinq
At
B
Here
ax
\lCosq
Þ
\
centripetal
=
the
force
is
provided
by
equation
(2)
and
v2-2gl
(1+Sinq)
particle
becomes
a
projectile
of
Ux
=
v1Sinq
and
=
0
and
=
2Sin3q+3Sin2q-1
=
½
Sinq
component
of
(3)
velocity
v
we
---------90-q
=
=
at
Uy
ay
v1Sinq
=
is
weight
get
(4)
horizontal
v1Cosq
-g
t
0
solution
acceptable
2) System consists of two identical slabs each of mass m linked by compressed weightless spring of stiffness K as shown
in figure. The slabs are also connected by a thread which is burnt at a certain moment. a) Find at what value of Dl the
initial compression of spring, the lower slab will bounce up after the thread is burned through b) If the upper block is
compressed
by
than
what
is
the
max
displacement
of
center
of
mass
in
upward
direction?
Fig
Solution:
When
Kx
the
lower
block
is
just
(50)
about
to
jump,
then
mg
=
From
energy
conservation
b) When the lower block is just about to move up, center of mass will be displaced to maximum for this
Work
energy
theorem
gives,
On
yields
Only
solving
force
acting
on
it
this
is
2mg
(both
masses
taken
single
system)
KEYWORDS:

















Work
Spring force
Power
Energy
Mechanical Energy
Kinetic Energy
Potential Energy
Conservative forces
Work Energy Theorem
Conservation of Mechanical Energy
Uniform and Non-Uniform Circular motion
Angular Velocity
Angular Acceleration
Centripetal and Tangential forces
Radius of Curvature
Vertical Circular motion
Banking