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  1. For a uniform rod the moment of inertia about an axis through the centre
is EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Therefore the moment of inertia about an axis through one end can be obtained
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
From
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Moment of inertia of a sphere of radius R and mass M about an axis through the point on its circumference can be obtained.
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Figure
9
IG
5
=
2MR2
From
parallel
axes
theorem
I =
IG
+
2
Mh
h
=
R
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
  1. 3 . Moment of inertia of a disc of radius r and mass m about an axis through a point on the circumference can be obtained.
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
2
IG = 2MR
5
From parallel axes theorem
2
I = IG + Mh
2 2
I = 1MR + MR But h = R
2
2 2
I = MR + 2MR
2
2
I = 3MR
2
PERPENDICULAR AXES THEOREM
The perpe
ndicular axis theorem states that “the moment of inertia of plane body about an axis perpendicular to its plane is equal to the sum of the moments of inertia of the plane body about any two mutually perpendicular axes in the plane of the body which intersect the first axis”
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
COMBINED ROTATIONAL AND TRANSLATIONAL MOTION
Consider a wheel rolling on a flat surface without slipping. Each particle of the wheel is undergoing two types of motion at the same time.
The centre of the wheel, which is wheel‘s centre of mass, is moving horizontally with speed VG.
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Figure 10 (a)
At the same time, the wheel is rotating about its centre of mass with angular speed ω.
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)

Figure 10 (b)
Thus rolling is the superposition of two motions
  1. Translation of objects centre of mass at velocity VG along a straight line
  2. Rotation about the center of mass at angular velocity ω.
Therefore the wheel possesses both translational and rotational K.E.
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
M = Mass of the object
VG = Velocity of object’s centre of mass
I = Moment of inertia of the object about an axis through the centre of mass.
The total K.E of an object undergoing both translational and rotational motion about its centre of mass axis is equal to the sum of translation K.E of the centre of mass and the rotational K.E about the centre of mass.
ROLLING OBJECTS
Consider a rigid body rolling without slipping down the plane inclined at an angle θ with the horizontal.
Required to determine the velocity, acceleration and minimum required coefficient of static friction (μs) for rolling.
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Figure. 11
By the conservation of energy we have
Total energy at A = total energy at B E .
EA = EB
K.EA +
P.EA =
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
1. Rolling a Hollow Sphere
I = MR2
From
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
2 2 2
But v = R ω
Then
2 2 2 2
MgH = 1 MR ω + 1 MR ω
2 2
2 2
MgH = MR ω
gH = MR2
2
v = gH
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Is the velocity of a hollow sphere
From the figure
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
For Acceleration
rd
From the 3 equation of motion
2 2
v = u + 2aL
Assuming the body started from rest
u = 0
Then
2 v = 2aL
Also
2
v = gH
2aL = gH
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Is the acceleration of hollow sphere
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Then from
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
2.
ROLLING A SOLID SPHERE From
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
2 2 2
v = R ω
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
2
7V = 10gH
2
V = 10gH
7
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Then
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Is the velocity of solid sphere
For acceleration
From,
2 v = u2 + 2aL
But u = 0
2
v = 2aL
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2) gH = 2aL
a = 5 . gH
7 L
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
3. ROLLING OF A SOLID CYLINDER
From
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
2
I = 1 MR
2
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
2 2 2
v = R ω
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Is the velocity of solid cylinder.
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
For acceleration
From,
2 2
v = u + 2aL
u = 0
2 v = 2aL
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
But,
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Is a acceleration of solid cylinder.
ALTERNATIVELY
Consider a solid cylinder of radius R and mass M rolling down a plane inclined at an angle θ to the horizontal.
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
For rotational motion about the centre of mass, the only force that produce the torque is. Torque on cylinder
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
If I is the moment of inertia and α is the angular acceleration about the axis of rotation, then torque on the cylinder is FS
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Equating equation (ii) and (iii)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Since the cylinder rolls without slipping
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Then
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Putting the value of FS into equation (i)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
For a solid cylinder, the moment of inertia about its symmetry axis is
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
All uniform rolling solid cylinders have the same linear acceleration down the incline irrespective of their masses and radii.
The linear acceleration is just 2/3 as large as it would be if the cylinder could slide without friction down the slope
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Minimum Required for Rolling
From
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
1.The frictional force is static, the cylinder rolls without slipping and there is no relative motion between the cylinder and the inclined plane at the point of contact. 2. If there were no friction between the cylinder and the incline, the cylinder would have slipped instead of rolled.
MOTION OF A MASS TIED TO THE STRING WOUND ON A CYLINDER
Consider a solid cylinder of radius R and mass M capable of rotating freely without friction about its symmetry axis.
Suppose a string of negligible mass is wrapped around the cylinder and a heavy point mass m is suspended from the free end.
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Figure 12 (a)
When the point mass m is released, the cylinder rotates about its axis and at the same time the mass falls down due to gravity and unwinds the string wound on the cylinder
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Figure 12 (b)
Linear Acceleration of Point Mass (m)
We first examine the forces of the point mass m as shown below
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Looking at the cylinder end; we see that tension in the string exerts a torque on the cylinder
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
If I is the moment of inertia of the cylinder about the axis of rotation and α is the angular acceleration produced in the cylinder.
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)……………………..(iii)
By equating equations (ii) and (iii)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)

EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Since the string unwinds without slipping the linear acceleration a of the point mass m and the angular acceleration α of the cylinder are related as
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Putting the value of T into equation (i)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
  1. The tension T in the string is always less than mg.
  2. As moment of inertia increases T approaches mg.
NEWTON’S LAWS OF ROTATIONAL MOTION
  1. First law
States that “everybody continues in its state of rest or of uniform rotational motion about an axis until it is compelled by some external torque to change that state”.
  1. Second law
States that “the rate of change of angular momentum about an axis is directly proportional to the impressed external torque and the change angular momentum takes place in the direction of the applied torque”.
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2) where k = 1.
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
  1. Third law
To every external torque applied, there is an equal and opposite restoring torque.
SIMPLE PENDULUM
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Figure. 13
It consists of a string of length l suspended from a fixed point O and carrying a bob of mass m at its free end.
When the bob is slightly displaced from the equilibrium position and then released, it executes rotational motion about an axis through the suspension point O.
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Suppose at anytime t, the bob is at point P making an angle θ with the vertical let be the angular velocity and α be the angular acceleration at that instant.
The only force which exerts a torque on the bob is that due to the weight of the bob.
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
The negative sign is used since this is a restoring torque i.e. a torque which force the bob toward θ = 0°.
The angular momentum of the bob about the axis of rotation at the considered instant (at point P)
L = mvr v = ωr
L = mωr.r
2
L = mωr
But r = lsinθ where θ = 90º

Therefore L = mωl2
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
From equation (i) and (ii) we have
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
= α
dt
Now , the angular acceleration of the bob at time t
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
This governs the oscillation of simple pendulum,in the vertical plane.
  1. When θ = 0º
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Therefore, the bob has zero angular acceleration at the equilibrium position.
  1. The negative sign shows that the tangential acceleration is directed toward the mean position.
  1. If θ is small, Sin θ = θ
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
  1. Time period
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Also
From
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
For S.H.M
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
From
Parallel axis theorem
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Then
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)
Then EcoleBooks | PHYSICS As LEVEL(FORM FIVE) NOTES - ROTATION OF RIGID BODIES(2)




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