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Physics Beyond 2000 Chapter 6 Angular Momentum

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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.

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

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

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

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

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

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

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The Kinematics of Rotation Average angular acceleration -

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The Kinematics of Rotation Instantaneous angular acceleration -

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

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

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

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The Kinematics of Rotation Constant angular acceleration α

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

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

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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

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

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Tangential acceleration a t = r. α It changes the angular velocity. atat r O

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Radial acceleration atat r arar O

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Linear velocity and angular velocity rArA O A B rBrB Points A and B have the same angular velocity but different linear velocities. vAvA vBvB ω

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Linear acceleration and angular acceleration rArA O A B rBrB Points A and B have the same angular acceleration but different linear tangential accelerations. vAvA vBvB ω aAaA aBaB

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Example 3 Find the tangential acceleration. equator A r

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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. axis of rotation ω

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Kinetic energy of a rotating object The i th particle has mass m i and speed v i The distance of the i th particle from the axis of rotation is r i ω mimi riri vivi Note that

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Kinetic energy of a rotating object ω mimi riri vivi The kinetic energy of the i th particle is

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Kinetic energy of a rotating object ω mimi riri vivi The kinetic energy of all N particles is

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Kinetic energy of a rotating object ω mimi riri vivi The rotational kinetic energy K r of the rigid body is

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Kinetic energy of a rotating object ω mimi riri vivi Define I is called the moment of inertia of the body about this axis of rotation.

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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 mimi riri axis of rotation

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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

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Radius of gyration For a rotating body, its I M. So we cab write I = M.k 2. The k is known as the radius of gyration of the body about the given axis.

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Example Find the radius of gyration k.

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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.

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Hoop about cylindrical axis I = MR 2 Hoop about any diameter I = MR 2 Table for Moment of Inertia M = mass of the hoop R = radius of the hoop

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Solid Cylinder about cylindrical axis I = MR 2 Solid Cylinder about central diameter I = MR 2 + ML 2 Table for Moment of Inertia M = mass of the cylinder R = radius of the cylinder L = length of the cylinder

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Thin Rod about axis through centre perpendicular to its length I = ML 2 Thin Rod about axis through one end and perpendicular to its length I = ML 2 Table for Moment of Inertia M = mass of the rod L = length of the rod

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Solid sphere about any diameter I = MR 2 Hollow sphere about any diameter I = MR 2 Table for Moment of Inertia M = mass of the sphere R = radius of the sphere

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

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

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Parallel Axes Theorem G G New axis of rotation P h IGIG IPIP Example 7

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

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Perpendicular Axes Theorem I Z = I X + I Y Example 8

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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 axis of rotation O

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

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Moment of force Γ Γ = F d Top view O F 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 r P θ

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Moment of force Γ Γ = F d 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

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

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

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

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

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

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

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

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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.

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

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Example 10 Torque and angular acceleration

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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

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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 m 2 s -1 or Nms. L is a vector. Its direction is determined by the direction of the angular velocity.

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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 = mr 2, L = I and v = r v r m axis of rotation

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Example 11 Find the angular momentum of a solid sphere.

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

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

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Example 12 Find the change in angular momentum.

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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

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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

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Torsional pendulum wire disk and

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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.

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

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Typical examples of second law Smooth pulley with moment of inertia I and radius r. r m1m1 m2m2 α a a Find the linear acceleration a of the two masses in terms of m 1, m 2, I and r.

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Typical examples of second law Smooth pulley with moment of inertia I and radius r. α m2m2 a T2T2 m2gm2g r T2T2 T1T1 m1m1 a T1T1 m1gm1g T 2.r-T 1.r = I.α T 1 -m 1 g = m 1 a m 2 g-T 2 = m 2 a a = rα

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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,

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The law of conservation of angular momentum For a system with initial moment of inertia I 1 and initial angular velocity ω 1, its initial angular momentum is I 1 ω 1. If the system changes its moment of inertia to I 2 and angular velocity ω 2, its final angular momentum is I 2 ω 2. If there is not any external net torque, then I 1 ω 1 = I 2 ω 2

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