# Chapter 6 Angular Momentum

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Chapter 6 Angular Momentum
Physics Beyond 2000 Chapter 6 Angular Momentum

Rotational Motion The body is rigid. (i.e. It does not suffer deformation by external forces.) The forces on the body may act at different points.

The Kinematics of Rotation
Axis of rotation – the body is rotating about a fixed axis. side view axis of rotation

The Kinematics of Rotation
Axis of rotation – the body is rotating about a fixed axis. ω axis of rotation top view

The Kinematics of Rotation
Angular displacement – The reference line moves an angle Δθ about the axis of rotation. ω Δθ axis of rotation top view

The Kinematics of Rotation
Average angular speed - ω Δθ axis of rotation top view

The Kinematics of Rotation
Instantaneous angular speed - ω Δθ axis of rotation top view

The Kinematics of Rotation
Example 1 – Find the angular speed.

The Kinematics of Rotation
Average angular acceleration -

The Kinematics of Rotation
Instantaneous angular acceleration -

The Kinematics of Rotation
Constant angular acceleration α ωo = initial angular velocity ω = final angular velocity θ = angular displacement t = time taken

The Kinematics of Rotation
Constant angular acceleration α ωo = initial angular velocity ω = final angular velocity θ = angular displacement t = time taken

The Kinematics of Rotation
Constant angular acceleration α ωo = initial angular velocity ω = final angular velocity θ = angular displacement t = time taken

The Kinematics of Rotation
Constant angular acceleration α

The Kinematics of Rotation
Note that the following quantities, except time t, are vectors.

The Kinematics of Rotation
We may use + and – signs to indicate the direction of the vectors.

The Kinematics of Rotation
Example 2 – to find the angular acceleration. The negative sign of α indicates that it is in opposite direction to the positive angular velocity. ω α O

Linear Acceleration When the object is rotating, it has two components of linear acceleration. Tangential acceleration at It is the linear acceleration along the tangent. Radial acceleration ar It is the centripetal acceleration pointing radially inwards.

Tangential acceleration
at = r. α It changes the angular velocity. at r O

Radial acceleration at r ar O

Linear velocity and angular velocity
vA vB ω rA O A B rB Points A and B have the same angular velocity but different linear velocities.

Linear acceleration and angular acceleration
vA vB ω rA O A B rB Points A and B have the same angular acceleration but different linear tangential accelerations. aA aB

Example 3 Find the tangential acceleration. A r equator

Kinetic energy of a rotating object
A rigid body of mass M is rotating about a fixed axis at angular speed ω. Treat the body as a composition of N particles. ω axis of rotation

Kinetic energy of a rotating object
The ith particle has mass mi and speed vi The distance of the ith particle from the axis of rotation is ri Note that ω vi mi ri

Kinetic energy of a rotating object
The kinetic energy of the ith particle is ω vi mi ri

Kinetic energy of a rotating object
The kinetic energy of all N particles is ω vi mi ri

Kinetic energy of a rotating object
The rotational kinetic energy Kr of the rigid body is ω vi mi ri

Kinetic energy of a rotating object
Define ω vi mi ri I is called the moment of inertia of the body about this axis of rotation.

Moment of inertia The value of I depends on the mass of the body
the way the mass is distributed the axis of rotation ω mi ri axis of rotation

Example 4 Find the moment of inertia and thus the rotational kinetic energy. Change the axis of rotation and find the moment of inertia. axis of rotation

Radius of gyration For a rotating body, its I  M.
So we cab write I = M.k2. The k is known as the radius of gyration of the body about the given axis.

Example Find the radius of gyration k.

Experiment to determine the moment of inertia of a flywheel
Supplement Ch.6 The gravitational potential energy of the weight is converted into the rotational kinetic energy of the flywheel and the kinetic energy of the weight. However there is loss of energy due to friction.

Table for Moment of Inertia
Hoop about cylindrical axis I = MR2 Hoop about any diameter I = MR2 M = mass of the hoop R = radius of the hoop

Table for Moment of Inertia
Solid Cylinder about cylindrical axis I = MR2 Solid Cylinder about central diameter I = MR2 + ML2 M = mass of the cylinder R = radius of the cylinder L = length of the cylinder

Table for Moment of Inertia
Thin Rod about axis through centre perpendicular to its length I = ML2 Thin Rod about axis through one end and perpendicular to its length I = ML2 M = mass of the rod L = length of the rod

Table for Moment of Inertia
Solid sphere about any diameter I = MR2 Hollow sphere about any diameter I = MR2 M = mass of the sphere R = radius of the sphere

Parallel Axes Theorem m is the mass of the object
G is the centre of gravity of the object G IG is the moment of inertia about the axis through the centre of gravity h IP is the moment of inertia about the axis through the point P. G P New axis of rotation

Parallel Axes Theorem G Note that the two axes are parallel. h G P
New axis of rotation

Parallel Axes Theorem Example 7 IG G IP h G P New axis of rotation

Perpendicular Axes Theorem
For a lamina lying in the x-y plane, the moment of inertia IX , IY and IZ, about three mutually perpendicular axes which meets at the same point are related by IZ = IX + IY

Perpendicular Axes Theorem
IZ = IX + IY Example 8

Moment of force Γ Moment of force (torque) It is the product of a force and its perpendicular distance from a point about which an object rotates. Unit: Nm F O axis of rotation

Moment of force Γ F axis of rotation O Top view O F Γ = F  d

Moment of force Γ The force F acts at point P of the object.
The distance vector from O, the point of rotation, to P is r. θis the angle between the force F and the distance vector.  Γ = F.r.sinθ Top view O F θ r P Γ = F  d

Moment of force Γ Moment of force Γis a vector.
In the following diagram, the moment of force is an anticlockwise moment. It produces an angular acceleration α in clockwise direction. Top view O F r P θ α Γ = F  d

Moment of force on a flywheel
A force F acts tangentially on the rim of a flywheel. F r Γ = F × r

Work done by a torque Suppose a force F acts at right angle to the
distance vector r. F r O

Work done by a torque What is the moment of force about O? Γ= F × r O

Work done by a torque The moment of force turns the object through
an angle θ with a displacement s. F F r θ O Γ= F × r s

Work done by a torque What is the work done by the force? W= F × s F O
θ O s W= F × s

Work done by a torque Express the work done by the force in terms
of Γ and θ. Use F = and s = r. θ F r θ O s W = F × s = Γ× θ

Example 9 Work done against the moment of friction is equal to the loss of rotational kinetic energy of the flywheel.

Torque and Angular acceleration
 = .   is the torque I is the moment of inertia  is the angular acceleration Compare to F = m.a in linear motion.

Torque and Angular acceleration
In an angular motion with uniform angular acceleration :

Example 10 Torque and angular acceleration

Conditions for equilibrium
A body will be in static equilibrium, if 1. net force is zero 2. net moment of force about any point is zero

Angular momentum L The angular momentum L of an object about an axis is the product of the angular velocity  and its moment of inertia. L = I. Unit of L: kg m2 s-1 or Nms. L is a vector. Its direction is determined by the direction of the angular velocity .

Angular momentum of a rotating point mass
A point mass m is rotating tangentially at speed v at a distance r from an axis. From I = mr2 , L = I and v = r  v r m axis of rotation

Example 11 Find the angular momentum of a solid sphere.

Newton’s 2nd law for rotation
The torque  acting on a rotating body is equal to the time rate of change of the angular momentum.

Newton’s 2nd law for rotation
If the net torque is zero, the angular momentum is a constant. The angular acceleration is zero.

Example 12 Find the change in angular momentum.

Torsional pendulum A disk is suspended by a wire.
The wire is twisted through an angle θ The restoring torque is Γ= c. θ where c is the torsional constant. wire disk

Torsional pendulum The restoring torque is
Γ= c. θ where c is the torsional constant. Prove that the torsional oscillation is a SHM with the equation wire disk

Torsional pendulum and wire disk

Typical examples of second law
Flywheel with moment of inertia I. r mass m axis α Find the angular acceleration α in terms of I, m and r.

Typical examples of second law
Flywheel with moment of inertia I. mass m T mg a r axis α T T.r = I. α a = r. α mg – T = ma

Typical examples of second law
Smooth pulley with moment of inertia I and radius r. α Find the linear acceleration a of the two masses in terms of m1, m2, I and r. r m1 m2 a a

Typical examples of second law
Smooth pulley with moment of inertia I and radius r. m2 a T2 m2g α m1 a T1 m1g r T2 T1 T2.r-T1.r = I.α T1-m1g = m1a m2g-T2 = m2a a = rα

The law of conservation of angular momentum
If external net torque = 0, the sum of angular momentum of the system is zero. If Γ=0,

The law of conservation of angular momentum
For a system with initial moment of inertia I1 and initial angular velocity ω1, its initial angular momentum is I1ω1. If the system changes its moment of inertia to I2 and angular velocity ω2, its final angular momentum is I2ω2. If there is not any external net torque, then I1ω1 = I2ω2