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- 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.

Figure | 9 | |||||||

IG 5 | = | 2MR2 | ||||||

From | parallel | axes | theorem | |||||

I = | IG | + | 2 Mh | h | = | R |

- 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.

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”

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”

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.

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 objects 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 K.E.

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.

Figure. 11

By the conservation of energy we have

Total energy at A = total energy at B E .

EA = EB

K.EA +

P.EA =

1. Rolling a Hollow Sphere

I = MR2

From

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

Is the velocity of a hollow sphere

From the figure

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

Is the acceleration of hollow sphere

Then from

2.

ROLLING A SOLID SPHERE From

2 2 2

v = R ω

2

7V = 10gH

2

V = 10gH

7

Then

Is the velocity of solid sphere

For acceleration

From,

2 v = u2 + 2aL

But u = 0

2

v = 2aL

gH = 2aL

a = 5 . gH

7 L

3. ROLLING OF A SOLID CYLINDER

From

2

I = 1 MR

2

2 2 2

v = R ω

Is the velocity of solid cylinder.

For acceleration

From,

2 2

v = u + 2aL

u = 0

2 v = 2aL

But,

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.

For rotational motion about the centre of mass, the only force that produce the torque is. Torque on cylinder

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

Equating equation (ii) and (iii)

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

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.

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 of 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.

……………………..(iii)

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)

- The tension T in the string is always less than mg.
- As moment of inertia increases T approaches mg.

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 angular momentum takes place in the direction of the applied torque”.

where k = 1.

- Third law

To every external torque applied, there is an equal and opposite restoring torque.

SIMPLE PENDULUM

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.

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.

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

From equation (i) and (ii) we have

dω = α

dt

Now , the angular acceleration of the bob at time t

This governs the oscillation of simple pendulum,in the vertical plane.

- When θ = 0º

Therefore, 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

Also

From

For S.H.M

From

Parallel axis theorem

Then

Then