- For a uniform rod, the moment of inertia about an axis through the centre is
Therefore, the moment of inertia about an axis through one end can be obtained.
From
Moment of inertia of a sphere of radius R and mass M about an axis through the point on its circumference can be obtained.
| IG | = | 2MR2/5 | ||
| From | parallel | axes | theorem | |
| I | = | IG + 2Mh2 | = | 2MR2 |
- Moment of inertia of a disc of radius r and mass m about an axis through a point on the circumference can be obtained.
IG = 2MR2/5
From the parallel axes theorem:
I = IG + Mh2
But h = R, so
I = MR2 + 2MR2 = 3MR2
PERPENDICULAR AXES THEOREM
The perpendicular axes theorem states that “the moment of inertia of a 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.”
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 the wheel‘s centre of mass, is moving horizontally with speed VG.
Figure 10 (a)
At the same time, the wheel is rotating about its centre of mass with angular speed ω.
Figure 10 (b)
Thus, rolling is the superposition of two motions:
- Translation of the object’s centre of mass at velocity VG along a straight line.
- Rotation about the center of mass at angular velocity ω.
Therefore, the wheel possesses both translational and rotational kinetic energy.
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 kinetic energy of an object undergoing both translational and rotational motion about its centre of mass axis is equal to the sum of the translational kinetic energy of the centre of mass and the rotational kinetic energy about the centre of mass.
ROLLING OBJECTS
Consider a rigid body rolling without slipping down a 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.
Figure 11
By the conservation of energy we have:
Total energy at A = total energy at B.
K.EA + P.EA = …
1. Rolling a Hollow Sphere
I = MR2
From
But v = Rω
Then
MgH = MRω2 + MRω2/2
gH = MR2/2
v = √(gH)
This is the velocity of a hollow sphere.
2. Rolling a Solid Sphere
From
v = Rω
For acceleration:
From the equation of motion:
v² = u² + 2aL
Assuming the body started from rest, u = 0
Then v² = 2aL
Also v² = gH
2aL = gH
This is the acceleration of the hollow sphere.
Then from
3. Rolling of a Solid Cylinder
From
I = 1/2 MR²
v = Rω
For acceleration:
From the equation of motion:
v² = u² + 2aL
u = 0
Then v² = 2aL
This is the acceleration of the 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.
For rotational motion about the centre of mass, the only force that produces the torque is the frictional force.
If I is the moment of inertia and α is the angular acceleration about the axis of rotation, then torque on the cylinder is FSR.
Equating torque and moment of inertia times angular acceleration:
Since the cylinder rolls without slipping:
Then
Putting the value of FS into equation (i):
For a solid cylinder, the moment of inertia about its symmetry axis is:
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.
Minimum Required for Rolling
From
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.
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.
Figure 12 (b)
Linear Acceleration of Point Mass (m)
We first examine the forces on the point mass m as shown below:
Looking at the cylinder end, we see that tension in the string exerts a torque on the cylinder.
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:
By equating equations (ii) and (iii):
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:
Putting the value of T into equation (i):
NEWTON’S LAWS OF ROTATIONAL MOTION
- 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”. - 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 in angular momentum takes place in the direction of the applied torque”. - Third law
To every external torque applied, there is an equal and opposite restoring torque.
SIMPLE PENDULUM
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.
Suppose at any time t, the bob is at point P making an angle θ with the vertical. Let ω 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.
The negative sign is used since this is a restoring torque, i.e., a torque which forces the bob toward θ = 0°.
The angular momentum of the bob about the axis of rotation at the considered instant (at point P) is:
L = mvr, where v = ωr
Therefore, L = mωr²
But r = l sin θ, where θ = 90º
Therefore, L = mωl²
For simple harmonic motion (S.H.M):
When θ = 0º, the bob has zero angular acceleration at the equilibrium position.
The negative sign shows that the tangential acceleration is directed toward the mean position.
If θ is small, sin θ ≈ θ
Time period
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