今日課程內容 CH11 角動量角動量角動量守恆 CH14 振動簡諧振動單擺...

今日課程內容CH11 角動量

角動量角動量守恆

CH14 振動簡諧振動單擺阻尼振動強迫振動

11-1 Angular Momentum( 角動量 )—Objects Rotating About a Fixed Axis

The rotational analog of linear momentum is angular momentum, L:

Then the rotational analog of Newton’s second law is:

This form of Newton’s second law is valid even if I is not constant.

11-1 Angular Momentum—Objects Rotating About a Fixed Axis

In the absence of an external torque, angular momentum is conserved:

More formally,

the total angular momentum of a rotating object remains constant if the net external torque acting on it is zero.

0 and constant.dL

L Idt

11-1 Angular Momentum—Objects Rotating About a Fixed AxisThis means:

Therefore, if an object’s moment of inertia changes, its angular speed changes as well.

Example 11-1: Object rotating on a string of changing length.

A small mass m attached to the end of a string revolves in a circle on a frictionless tabletop. The other end of the string passes through a hole in the table. Initially, the mass revolves with a speed v1 = 2.4 m/s in a circle of radius R1 = 0.80 m. The string is then pulled slowly through the hole so that the radius is reduced to R2 = 0.48 m. What is the speed, v2, of the mass now?

Example 11-2: Clutch.

A simple clutch consists of two cylindrical plates that can be pressed together to connect two sections of an axle, as needed, in a piece of machinery. The two plates have masses MA = 6.0 kg and MB = 9.0 kg, with equal radii R0 = 0.60 m. They are initially separated. Plate MA is accelerated from rest to an angular velocity ω1 = 7.2 rad/s in time Δt = 2.0 s. Calculate (a) the angular momentum of MA, and (b) the torque required to have accelerated MA from rest to ω1. (c) Next, plate MB, initially at rest but free to rotate without friction, is placed in firm contact with freely rotating plate MA, and the two plates both rotate at a constant angular velocity ω2, which is considerably less than ω1. Why does this happen, and what is ω2?

Example 11-3: Neutron star.

Astronomers detect stars that are rotating extremely rapidly, known as neutron stars. A neutron star is believed to form from the inner core of a larger star that collapsed, under its own gravitation, to a star of very small radius and very high density. Before collapse, suppose the core of such a star is the size of our Sun (r ≈ 7 x 105 km) with mass 2.0 times as great as the Sun, and is rotating at a frequency of 1.0 revolution every 100 days. If it were to undergo gravitational collapse to a neutron star of radius 10 km, what would its rotation frequency be? Assume the star is a uniform sphere at all times, and loses no mass.

11-1 Angular Momentum—Objects Rotating About a Fixed Axis

Angular momentum is a vector; for a symmetrical object rotating about a symmetry axis it is in the same direction as the angular velocity vector.

11-1 Angular Momentum—Objects Rotating About a Fixed AxisExample 11-4: Running on a circular platform.

Suppose a 60-kg person stands at the edge of a 6.0-m-diameter circular platform, which is mounted on frictionless bearings and has a moment of inertia of 1800 kg·m2. The platform is at rest initially, but when the person begins running at a speed of 4.2 m/s (with respect to the Earth) around its edge, the platform begins to rotate in the opposite direction. Calculate the angular velocity of the platform.

11-2 Vector Cross Product; Torque as a Vector

The vector cross product is defined as:

The direction of the cross product is defined by a right-hand rule:

11-2 Vector Cross Product; Torque as a VectorThe cross product can also be written in determinant form:


Some properties of the cross product:


Torque can be defined as the vector product of the force and the vector from the point of action of the force to the axis of rotation:


For a particle, the torque can be defined around a point O:

Here, is the position vector from the particle relative to O.

r

Example 11-6: Torque vector.

Suppose the vector is in the xz plane, and is given by = (1.2 m) + 1.2 m) Calculate the torque vector if = (150 N) .


r

Fr

11-3 Angular Momentum of a ParticleThe angular momentum of a particle about a specified axis is given by:

11-3 Angular Momentum of a ParticleIf we take the derivative of , we find:

Since

we have:

L

11-4 Angular Momentum and Torque for a System of Particles; General Motion

The angular momentum of a system of particles can change only if there is an external torque—torques due to internal forces cancel.

This equation is valid in any inertial reference frame. It is also valid for the center of mass, even if it is accelerating:

11-5 Angular Momentum and Torque for a Rigid Object

For a rigid object, we can show that its angular momentum when rotating around a particular axis is given by:

11-5 Angular Momentum and Torque for a Rigid Object

Example 11-8: Atwood’s machine.

An Atwood machine consists of two masses, mA and mB, which are connected by an inelastic cord of negligible mass that passes over a pulley. If the pulley has radius R0 and moment of inertia I about its axle, determine the acceleration of the masses mA and mB, and compare to the situation where the moment of inertia of the pulley is ignored.

11-6 Conservation of Angular Momentum

If the net torque on a system is zero,

The total angular momentum of a system remains constant if the net external torque acting on the system is zero.


Example 11-11: Kepler’s second law derived.

Kepler’s second law states that each planet moves so that a line from the Sun to the planet sweeps out equal areas in equal times. Use conservation of angular momentum to show this.


Example 11-12: Bullet strikes cylinder edge.

A bullet of mass m moving with velocity v strikes and becomes embedded at the edge of a cylinder of mass M and radius R0. The cylinder, initially at rest, begins to rotate about its symmetry axis, which remains fixed in position. Assuming no frictional torque, what is the angular velocity of the cylinder after this collision? Is kinetic energy conserved?

Chapter 14Oscillations( 震盪、振動 )

If an object vibrates or oscillates back and forth over the same path, each cycle taking the same amount of time, the motion is called periodic. The mass and spring system is a useful model for a periodic system.

14-1 Oscillations of a Spring

We assume that the surface is frictionless. There is a point where the spring is neither stretched nor compressed; this is the equilibrium position. We measure displacement from that point (x = 0 on the previous figure).

The force exerted by the spring depends on the displacement:


• The minus sign on the force indicates that it is a restoring force( 回復力 )—it is directed to restore the mass to its equilibrium position.

• k is the spring constant.

• The force is not constant, so the acceleration is not constant either.


• Displacement is measured from the equilibrium point.

• Amplitude is the maximum displacement.

• A cycle is a full to-and-fro motion.

• Period is the time required to complete one cycle.

• Frequency is the number of cycles completed per second.


If the spring is hung vertically, the only change is in the equilibrium position, which is at the point where the spring force equals the gravitational force.


Example 14-1: Car springs.

When a family of four with a total mass of 200 kg step into their 1200-kg car, the car’s springs compress 3.0 cm. (a) What is the spring constant of the car’s springs, assuming they act as a single spring? (b) How far will the car lower if loaded with 300 kg rather than 200 kg?


Any vibrating system where the restoring force is proportional to the negative of the displacement is in simple harmonic motion (SHM), and is often called a simple harmonic oscillator (SHO).

Substituting F = -kx into Newton’s second law gives the equation of motion:

with solutions of the form:

14-2 Simple Harmonic Motion( 簡諧運動 )

Substituting, we verify that this solution does indeed satisfy the equation of motion, with:

The constants A and φ will be determined by initial conditions; A is the amplitude, and φ gives the phase of the motion at t = 0.

14-2 Simple Harmonic Motion

The velocity can be found by differentiating the displacement:

These figures illustrate the effect of φ:



Because then

Example 14-2: Car springs again.

Determine the period and frequency of a car whose mass is 1400 kg and whose shock absorbers have a spring constant of 6.5 x 104 N/m after hitting a bump. Assume the shock absorbers are poor, so the car really oscillates up and down.


The velocity and acceleration for simple harmonic motion can be found by differentiating the displacement:


Example 14-3: A vibrating floor.

A large motor in a factory causes the floor to vibrate at a frequency of 10 Hz. The amplitude of the floor’s motion near the motor is about 3.0 mm. Estimate the maximum acceleration of the floor near the motor.


Example 14-5: Spring calculations.

A spring stretches 0.150 m when a 0.300-kg mass is gently attached to it. The spring is then set up horizontally with the 0.300-kg mass resting on a frictionless table. The mass is pushed so that the spring is compressed 0.100 m from the equilibrium point, and released from rest. Determine: (a) the spring stiffness constant k and angular frequency ω; (b) the amplitude of the horizontal oscillation A; (c) the magnitude of the maximum velocity vmax; (d) the magnitude of the maximum acceleration amax of the mass; (e) the period T and frequency f; (f) the displacement x as a function of time; and (g) the velocity at t = 0.150 s.

Example 14-6: Spring is started with a push.

Suppose the spring of Example 14–5 (where ω = 8.08 s-1) is compressed 0.100 m from equilibrium (x0 = -0.100 m) but is given a shove to create a velocity in the +x direction of v0 = 0.400 m/s. Determine (a) the phase angle φ, (b) the amplitude A, and (c) the displacement x as a function of time, x(t).


We already know that the potential energy of a spring is given by:

The total mechanical energy is then:

The total mechanical energy will be conserved, as we are assuming the system is frictionless.

14-3 Energy in the Simple Harmonic Oscillator

If the mass is at the limits of its motion, the energy is all potential.

If the mass is at the equilibrium point, the energy is all kinetic.

We know what the potential energy is at the turning points:


The total energy is, therefore,

And we can write:

This can be solved for the velocity as a function of position:

where


This graph shows the potential energy function of a spring. The total energy is constant.


Example 14-7: Energy calculations.

For the simple harmonic oscillation of Example 14–5 (where k = 19.6 N/m, A = 0.100 m, x = -(0.100 m) cos 8.08t, and v = (0.808 m/s) sin 8.08t), determine (a) the total energy, (b) the kinetic and potential energies as a function of time, (c) the velocity when the mass is 0.050 m from equilibrium, (d) the kinetic and potential energies at half amplitude (x = ± A/2).


If we look at the projection onto the x axis of an object moving in a circle of radius A at a constant speed υM , we find that the x component of its velocity varies as:

This is identical to SHM.

14-4 Simple Harmonic Motion Related to Uniform Circular Motion

A simple pendulum consists of a mass at the end of a lightweight cord. We assume that the cord does not stretch, and that its mass is negligible.

14-5 The Simple Pendulum

In order to be in SHM, the restoring force must be proportional to the negative of the displacement. Here we have:

which is proportional to sin θ and not to θ itself.

However, if the angle is small, sin θ ≈ θ.


Therefore, for small angles, we have:

where

The period and frequency are:


A physical pendulum is any real extended object that oscillates back and forth.

The torque about point O is:

Substituting into Newton’s second law gives:

14-6 The Physical Pendulum and the Torsional Pendulum

14-6 The Physical Pendulum and the Torsional Pendulum

For small angles, this becomes:

which is the equation for SHM, with

Damped harmonic motion is harmonic motion with a frictional or drag force. If the damping is small, we can treat it as an “envelope” that modifies the undamped oscillation.

If

then

14-7 Damped( 阻尼 ) Harmonic Motion

This gives

If b is small, a solution of the form

will work, with

14-7 Damped Harmonic Motion

If b2 > 4mk, ω’ becomes imaginary(虛數 ), and the system is overdamped(過阻尼 ) (C).

For b2 = 4mk, the system is critically damped (B) —this is the case in which the system reaches equilibrium in the shortest time.


There are systems in which damping is unwanted, such as clocks and watches.

Then there are systems in which it is wanted, and often needs to be as close to critical damping as possible, such as automobile shock absorbers and earthquake protection for buildings.


Example 14-11: Simple pendulum with damping.

A simple pendulum has a length of 1.0 m. It is set swinging with small-amplitude oscillations. After 5.0 minutes, the amplitude is only 50% of what it was initially. (a) What is the value of γ for the motion? (b) By what factor does the frequency, f’, differ from f, the undamped frequency?


Forced vibrations occur when there is a periodic driving force. This force may or may not have the same period as the natural frequency of the system.

If the frequency is the same as the natural frequency, the amplitude can become quite large. This is called resonance.

14-8 Forced Oscillations; Resonance

The sharpness of the resonant peak depends on the damping. If the damping is small (A) it can be quite sharp; if the damping is larger (B) it is less sharp.

Like damping, resonance can be wanted or unwanted. Musical instruments and TV/radio receivers depend on it.



The equation of motion for a forced oscillator is:

The solution is:

where

and


The width of the resonant peak can be characterized by the Q factor:

10-8 Rotational Kinetic Energy

The torque does work as it moves the wheel through an angle θ:

10-9 Rotational Plus Translational Motion; Rolling

In (a), a wheel is rolling without slipping. The point P, touching the ground, is instantaneously at rest, and the center moves with velocity .

In (b) the same wheel is seen from a reference frame where C is at rest. Now point P is moving with velocity – .

The linear speed of the wheel is related to its angular speed:

v

v

10-10 Why Does a Rolling Sphere Slow Down?A rolling sphere will slow down and stop rather than roll forever. What force would cause this?

If we say “friction”, there are problems:• The frictional force has to act at the point of contact; this means the angular speed of the sphere would increase.

• Gravity and the normal force both act through the center of mass, and cannot create a torque.

10-10 Why Does a Rolling Sphere Slow Down?

The solution: No real sphere is perfectly rigid. The bottom will deform, and the normal force will create a torque that slows the sphere.

Sample problem: Newton’s 2nd Law in Rotational Motion

習題Ch10: 5, 19, 27, 30, 40, 42, 50, 58, 67, 73Ch11: 8, 15, 32, 36, 39, 41, 48, Ch14: 10, 13, 17, 24, 34, 37, 44, 60, 64

今日課程內容 CH11 角動量角動量角動量守恆 CH14 振動簡諧振動單擺...

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Transcript of 今日課程內容 CH11 角動量 角動量 角動量守恆 CH14 振動 簡諧振動 單擺...

今日課程內容 CH11 角動量角動量角動量守恆 CH14 振動簡諧振動單擺...

Transcript of 今日課程內容 CH11 角動量角動量角動量守恆 CH14 振動簡諧振動單擺...