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CHAPTER NINE
RIGID BODY STATICS
Moment of a Force or Torque
The torque or moment of a force about a reference point is the vector given by
= rF
where r is the position vector of the point of application of the force relative to the
reference point.
r
F
O
Fig 9.1
Note that the angle between the two vectors is the angle between their directions. Thus,
we could redraw Fig. 9.1 as

F
r
O
Fig. 9.2
The magnitude of the moment is |  | =  | r  F || r | | F | sin  , where  is the angle
between the two vectors.
Notice that the moment of a force tends to make a body the force is acting upon to rotate
about the reference point O. It is important, therefore, to refer to the reference point. You
would recall that you were taught at the secondary school level, about clockwise and
counter-clockwise (or anticlockwise) moment. This is because, at least at this level, the
moment makes the body rotate either clockwise (as in the case above) or
counterclockwise about the reference point.
Let us take a look at opening a door. Do you push the door close to the hinges? Do you
push the narrow edge of the door parallel to the door through the line through the hinges?
No. It is easier to push the door at the farthest end (from the hinges) and perpendicular to
the length of the door. Thus, if the door is of length r and the force you apply is F, in
case, the moment of the force is rF sin 90  rF . Of course, if   0 , the moment of the
force is zero.
The moment of a force is a vector, the direction of which is given by that of the righthanded cockscrew. It follows that if your hand should move from r to F contained on the
plane of the paper, sweeping through the angle  (clockwise in this case), the vector
moment of the force points away from you, into the paper. The vector for anticlockwise
moment would point towards you.
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Equilibrium of a Rigid Body
Earlier, we saw what it takes for a point mass to be in equilibrium – the net force on it
must be zero. For a rigid body (an extended body), we require, additionally, that the net
moment on the body be zero. This is summed up in two laws:
(i)
The net force acting on the body in any direction is zero. You have been
taught the equivalent of this force: sum of upward forces is equal to the sum
of downward forces. This is just only one part of it. Indeed, there cannot be a
net force in any direction, otherwise, the body would move in that particular
direction.
(ii)
The net moment about any point on the body must be zero. This you saw as:
sum of clockwise moment = sum of anticlockwise moment. This applies to
any point at all, as if there is any point at which the net moment is not zero,
that point would rotate about the reference point.
Example 1:
As an example, consider a uniform beam of negligible weight resting on a fulcrum as
shown in Fig. 9.3. It is of length 1 m, and is resting on a knife edge placed at the 40 cm
mark. There is a 50 g mass hanging from the 80 cm mark. Clearly, the beam cannot be
balanced. There must be a force on the other side of the 50 g mass to ensure balance. Let
us decide to hand a 200 g mass on that side. Where exactly should we put it?
R
0.2m
0.4m
0.4m
1m
Let us place the 40 g mass at a point x m from the left end of the beam.
xm
R
0.4m
0.4m
200g
1m
0.2m
50g
Fig. 9.4
Condition 1: The sum of upward forces equals the sum of downward forces. The reaction
at the support equals the weight of the beam. In this case, indeed, we have not bothered
with the weight of the beam as it is negligible (otherwise, we would have had its weight
pointing down from the center of gravity, same as center of mass in this case).
Condition 2: Net Moment about the fulcrum is zero, that is, sum of clockwise moments
minus sum of anticlockwise moments is zero:
0.05kg  g  0.4m  [0.2kg  g  (0.4  x)m] = 0
9.1
Hence,
0.05
0.4  x 
 0.25
0.2
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x  0.4 – 0.25 = 0.15 m
Example 2:
A metre rule of mass 30 g is supported at the 40 cm and the 85 cm marks. Two masses 45
g and 52 g are hung from the 42 cm and the 90 cm marks respectively. Find the reaction
at the supports.
R1
R2
0.85 m
0.4 m
.42 m
45 g
m = 30 g
52 g
0.9 m
cm
Fig. 9.5
(iii)
(iv)
The sum of upward forces = sum of downward forces:
R1  R2  (.045  0.03  0.052) 10  1.27 N
Taking moments about R1 :
R2  0.45  0.045  g  0.02  0.03  g  0.1  0.52  g  0.4
9.2
9.3
From equation 9.3, we can obtain R2 . Putting this in equation 9.2 gives R1 .
Example 3:
A man of mass m is climbing a uniform ladder of weight W and length L m resting on a
rough floor and a smooth wall, as shown in Fig. 9.6. Find the reaction at the wall. How
far up the ladder can he go without the ladder slipping if the coefficient of static friction
is  s ?
F1
L
H
F2
W
F2 x
Fig. 9.6
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F2 y
Solution
The forces acting on the ladder are already shown. F2 is the force the ladder exerts on the
floor, and is a resultant of what the reaction would have been without friction F1 y , and
the frictional force F2 x . There is no friction at the wall. Hence, only the normal reaction
F1 needs be reckoned with.
Condition 1: Sum of upward forces = sum of downward forces
F2 y  W  mg
9.4
9.5
F1  F2 x
Let the man be x m up along the ladder. Then, if the angle the ladder makes with the floor
is  , then taking moments about the foot of the ladder,
L
9.6
F1 H  W cos   mgx cos 
2
If the center of gravity of the beam had been one-third of the way up the ladder, we
would have replaced L / 2 by L / 3 . Indeed, we replace L / 2 by L / p , if the center of
gravity is one-pth of the distance up the ladder.
From equation 9.4, we can calculate F2 y , since W and mg are known. From equation 9.6,
we can get find F1 and hence, F2 x (they are equal) in terms of x. But we recall that this
force can be written as F2 x   s F2 y . We can then get x since each term on the right is
now known. As an example, consider the following:
A man of mass 55 kg is climbing a uniform ladder of weight 70 kg and length 10 m
resting on a rough floor and a smooth wall. If the ladder touches the wall at a height of 7
m, how far up the ladder can the man go without the ladder slipping, given that the
coefficient of static friction between the ladder and the rough floor is 0.3?
F2 y  W  mg = 70 10  55 10  1250N
9.7
9.8
F1  F2 x
Let the man be x m up along the ladder. Then, if the angle the ladder makes with the floor
is  , then taking moments about the foot of the ladder,
L
F1 H  W cos   mgx cos 
2
10
7
7


F1  7  70  10  cos sin 1   55  10 x cos sin 1 
2
10 
10 


7F1  3500  .7141  550x  .7141
 2499.35  392.755x
F1  357.05  56.108x = F2 x =  s F2 y , where  s is the coefficient of static friction
between the ladder and the rough floor.
F1  357.05  56.108x  0.3 1250  375
Hence, x  0.3199 m.
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CHAPTER TEN
RIGID BODY DYNAMICS
A rigid body is indeed such that the relative position of the constituent masses remain
constant. Thus, if we consider Fig. 10.1., then
mn
r1
rn
m1
r2
m2
r3
m3
Fig. 10.1
Let the body rotate about O with angular frequency  . Then, the total kinetic energy of
the rigid body is
n
1
1
1
1
1
2
2
2
2
2
KE  m1v1  m1v1  m1v1  ...  mn v n   mi vi
2
2
2
2
i 1 2
n
n
1
1
  mi ( i ri ) 2   mi ( ri ) 2
i 1 2
i 1 2
(since the body is rigid, each constituent rotates at the angular frequency  )
Hence,
n
1
1
2
10.1
KE   2  mi ri  I 2
2
2
i 1
n
where I   mi ri is called the moment of inertia of the rigid body.
2
i 1
You would observe that the expression for the kinetic energy looks like that for kinetic
1
energy mv 2 if we make the identification v   and m  I . Thus, moment of inertia
2
plays the role of the equivalent of mass for the rotational motion of a rigid body.
Moment of Inertia of a Continuous Body
We have seen that the moment of inertia of a rigid body with discrete components is
n
I   mi ri . We shall now see the situation where the body is of continuous distribution.
2
i 1
For one, the summation becomes an integral, I   r 2 dm , and the integral evaluated as
the problem demands.
Examples
1.
A thin rod of length rotating about an axis perpendicular to one of its ends
50
dm
x
L
Fig. 10.2
Clearly, the elemental mass dm contributes an elemental moment of inertia x 2 dm .
The total moment of inertia is then,
L
L
I one end   x 2 dm   x 2  Adx
0
0
since dm   A dx , the product of the density and the volume of the element. Note that
the volume is A dx , where A is the cross-sectional area and dx is the (elemental) length of
the elemental mass.  and A are constant. Hence, we can write,
I one end   A
L
0
x3
x dx   A
3
L
2
0
L3
A
3
10.2
But  AL  M , the mass of the cylinder, or   M / AL . Putting this in the expression
for I one end ,
I one end 
M L3 1
A  ML2
AL 3 3
10.3
Let us now see what happens when the bar rotates around its center about an axis
perpendicular to the bar.
dm
x
L/2
L/2
Fig. 10.3
I middle  2  A
L/2
0
x3
x dx  2  A
3
L/2
A
2
0
L3
12
But  AL  M , the mass of the cylinder, or   M / AL . Putting this in the expression
for I middle ,
I middle 
M L3
1
A  ML2
AL 12 12
10.4
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Parallel Axis Theorem
You would notice that we could have got equation 10.3 from equation 10.2 as follows:
1
1
3 1
1
L
I one end  I middle  M    ML2  ML2 
ML2  ML2
12
4
12
3
2
This is indeed true in the general case of an axis through a point at a distance h from the
axis through the center of mass,
I  I cm  Mh 2
2
Assignment
Find the moment of inertia of uniform rod through an axis parallel to its length, passing
through a point one-third to one of its ends.
A body could have both rotational and translational motion. Such a body would then have
both translational kinetic energy and rotational kinetic energy, so we could write the total
kinetic energy as,
1
1
KE  Mv 2  I 2
10.5
2
2
Thus if a uniform solid cylinder of radius R and mass M should roll down an inclined
plane from height h from rest, the gravitational potential energy Mgh is fully converted
1
to the translational kinetic energy Mv 2 and rotational kinetic energy
2
1 2 11

I   MR 2   2 at the bottom of the incline, where v and  have their values at
2
22

the bottom of the incline. The moment of inertia of a solid cylinder about an axis through
1
its center, parallel to its length is I  MR 2 .
2
The total kinetic energy is then
1
1
1
1
2
1
3
KE  Mv 2  MR 2  2  Mv 2  Mv 2  Mv 2  Mv 2  Mv 2
2
4
2
4
4
4
4
3
Mgh  Mv 2 ,
4
or
4
v
gh
3
In the absence of rotation, all the potential energy is converted to translational kinetic
energy,
1
Mgh  Mv 2
2
or
v  2 gh
2 gh 
4
gh
3
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This is because some of the potential energy has been converted to rotational motion.
Assignment
A solid cylinder of mass 3 kg and radius 0.2 m is propelled up an inclined plane with an
initial velocity of 25m/s. How far up the incline can the cylinder travel?
ANGULAR MOMENTUM AND TORQUE
The angular momentum of a body is the moment of its momentum. Thus, if the
momentum of the body is p , then, its angular momentum about a reference point is,
10.6
L  rp
It is therefore, a vector the direction of which is given by the right-handed cockscrew
rule.
Just as the moment of a force gives the rotational effect of the force, so does the angular
momentum give the rotational effect of a momentum.
We can write the magnitude of the angular momentum as
L  rp sin   mvr sin 
where  is the angle between the two vectors.
10.7
If the vectors are perpendicular, L = r p = mvr, and zero if the vectors are parallel. Makes
sense, right? Just as it was in the case of the moment of a force, the rotating effect is zero
if the two vectors are in the same direction.
Let us now take the time-derivative of the angular momentum.
d
dp dr
L  r

p
dt
dt dt
= r  F  v  p  r  F  v  mv  r  F  mv  v = r  F = 
Remember we said the moment of a force about a reference point is the torque.
10.8
Thus, we conclude that the torque a body in angular motion is the time rate of change of
its angular momentum, that is,
dL
=
dt
10.9
Note that differentiating a cross product means that we keep the order of appearance of
the vectors involved, since cross product is not commutative, that is, r  p  p  r .
dL
dL
 0 , and the angular
, what happens if the torque is zero? Then,
dt
dt
momentum is constant. Does this equation look like the one you saw earlier? That is,
dp
 0 . Newton’s first law! In just the same way, in angular motion, a body remains in
dt
angular motion with constant angular momentum until a net external torque acts on it.
dp
Can you also see that equation 10.9 is like F 
, Newton’s second law? You can then
dt
Since  =
53
see again that torque plays in angular motion the equivalent of force in translational
motion.
In uniform circular motion,   0 , hence, there is no torque on a body in uniform circular
motion. This is why the angular momentum remains L  rp  mvr  mr 2 , because  is
constant in uniform circular motion.
In circular motion, r and p are perpendicular (remember?), because the velocity is along
the circumference. Hence, we can write | L | L  rp  mvr  mr 2 . Then, the torque is,
d
d
  (mr 2 )  mr 2
 I (r is constant, but  is not)
dt
dt
Example
A flywheel of radius 1.2 m and moment of inertia 12.5 kgm2 starts rotating from rest and
uniformly attains an angular speed of 10 rad/s2 in 5 seconds. Find the force applied along
the rim of the flywheel.
Solution
The torque on the flywheel is   I  rF , where r is the radius of the wheel.
The angular acceleration is given by,
  0 10

 = 5 rad/s2
t
2
Then,
I 12.5  2
F

= 20.8333 N
r
1.2
Notice that the moment of inertia of a thin rod rotating about an axis through the middle
is less than that of the same rod through an axis at the end of the rod for instance. This is
why a rod rotating about an end would be more effective or destructive than the one
rotating through the middle if they are both rotating at the same angular speed -   I .
The more the moment of inertia, the more the torque causing the motion, or the amount
of destruction the rod can cause.
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