1206 - Concepts in Physics

Wednesday, November 18th 2009

Notes• Assignment #6 due today !!! (before NOON)

• Assignment #7 due next Wednesday

• Please pick up Assignments #1 - #5 ....

• Still have midterms (first round) to be picked up

• Second round of midterms won’t be available this week.

Example: circular motion with constant angular speed

A particle rotates counterclockwise in a circle of radius 3.00 m with a constant angular speed of 8.00 rad/s. At t = 0, the particle has an x coordinate of 2.00 m and is moving to

the right.

We have a particle in uniform circular motion, as the shadow we discussed last lecture for simple harmonic motion and the reference circle.

So, we can use x = A cos(ωt + Φ) to determine the x coordinate of the rotating particle, with ω = 8.00 rad/s and A = 3.00 m:

x = A cos(ωt + Φ) = (3.00m) cos(8.00t +Φ)

We can find Φ by using the initial condition x = 2.00 m and t = 0: 2.00m = (3.00m) cos(0 + Φ)

Therefore Φ = cos-1(2.00m/3.00m) = cos-1(0.667) = -48.2° = -0.841 rad the minus sign is correct, because it was given that the particle moves to the right.

Therefore the x coordinate in dependence of time is: x = (3.00 m) cos(8.00t -0.841)In a similar way the velocity and acceleration can be expressed in dependence of time.

ϕ is called the phase of the movement and describes where in the cycle we start.

PendulumA simple pendulum consists of a particle of mass m, attached to a frictionless pivot P by a cable of length L an negligible mass. When the particle is pulled away from its equilibrium position by an angle Θ and released, it swings back and forth. By attaching a pen to the bottom of the swinging particle and moving a strip of paper beneath it at a steady rate, we can record the position of the particle as time passes. The graphical record reveals a pattern that is similar (but not identical) to the sinusoidal pattern for simple harmonic motion. The force of gravity is responsible for the back-and-

forth rotation about the axis at P. The rotation speeds up as the particle approaches the lowest point on the arc and slows down on the upward part of the swing. Eventually the angular speed is reduced to zero, and the particle swings back. From earlier discussion, we know that a net torque is required to change the angular speed. The gravitational force mg produces this torque. The tension T in the cable creates no torque, because it points directly at the pivot P and, therefore, has a zero lever arm l, so that τ = -(mg)l. The minus sign is included since the torque is a restoring torque: that is, it acts to reduce the angle Θ. The angle Θ is positive (counterclockwise) , while the torque is negative (clockwise). The lever arm l is the perpendicular distance between the line of action of mg and the pivot P.

P

Θ

T

L

l

slever arm l is nearly equal to the

arc length s (for small angles)

P

Θ

T

L

l

s

Let’s look at the similarity between s and l for small angles Θ (< 10°). If Θ is expressed in radians, the arc length and the radius L of the circular path are related, according to s = LΘ. Under these conditions, it follows that l ≈ s = LΘ, and the torque created by gravity is

τ ≈ -mgL Θwe can summarize mgL as one constant k’. Therefore for small angles the torque that restores the pendulum to its vertical equilibrium position is proportional to the angular displacement Θ.

The expression τ = -k’Θ has the same form as Hooke’s law restoring force for an ideal spring, F = -kx. Therefore, we expect the frequency of the back-and-forth movement of the pendulum to be given by an equation analogous to ω = 2πf = sqrt(k/m). In place of the spring constant k, the

constant k’ = mgL will appear, and as usual in rotational motion, in place of the mass m, the moment of inertia I will appear:

ω = 2πf = sqrt(mgL/I) (small angles only)The moment of inertia of a particle of mass m, rotation at a radius r = L about an axis, is given by

I = mL2. Substituting this expression, we obtain ω = 2πf = sqrt(g/L) (small angles only) . Note! The mass has been eliminated algebraically, so only the length L and the acceleration due to gravity g determine the frequency of a simple pendulum. For large angles, the pendulum does

not exhibit a simple harmonic motion.

Using a pendulum to keep timeDetermine the length of a simple pendulum that will swing back and forth in simple

harmonic motion with a period of 1.00 s.

When a simple pendulum is swinging back and forth in simple harmonic motions its frequency f is given by f = 1/(2π) sqrt(g/L), where g is the acceleration due to gravity and L is the length of the pendulum. We also know that the frequency is the given by 1/T. We can replace f by 1/T and then solve for L.

The length of the pendulum is

L = T2g/(4π2) = (1.00 s)2(9.80 m/s2)/(4π2) = 0.248 m

If the object performing the movement is extended, the pendulum is called a physical pendulum. For a rigid object, the proper moment of inertia must be used. The length L for a physical pendulum is the

distance between the axis at P and the center of gravity of the object. The next example deals with an important type of physical

pendulum.

Pendulum Motion and WalkingWhen we walk, our legs alternately swing forward about the hip joint as a pivot. In this

motion the leg is acting approximately as a physical pendulum. Treating the leg as a uniform rod of length D = 0.80 m, find the time it takes for the leg to swing forward.

The time it takes for the leg to swing forward is one-half of the period T, which is related to the frequency f by f = 1/T. For a physical pendulum the frequency is given by f = 1/(2π) sqrt(mgL/I), where the moment of inertia for a thin rod of length D rotation about an axis perpendicular to one end is given a I = 1/3mD2. Since we are treating the

leg as a thin rod, the center of gravity is a the center L = 0.40 m.

We find f = 1/T = 1/(2π) sqrt(mgL/I) or T = 2π sqrt(I/mgL)

T = 2π sqrt{(1/3mD2)/(mgL)} = 2πD/sqrt(3gL) = 2π (0.80 m)/sqrt(3x9.8m/s2x0.40m) = 1.5 s

The desired time is one-half of the period or 0.75 s.

Example: A swinging rod

FG = mg

cm

PivotA uniform rod of mass m and length L is pivoted about one end and oscillates in a vertical plane. Find the period of

oscillation if the amplitude of the motion is small.

Since the rod (swinging forth and back) is not a point particle, we use what we have learned about a physical

pendulum. The moment of inertia for a uniform rod about an axis through one end is 1/3mL2. The distance d from

the pivot to the center of mass of the rod is L/2. Period T = 2π sqrt(I/mgL)

Using the expressions describes above:

T = 2π sqrt{(1/3mL2)/(mgL/2)} = 2π sqrt{2L/3g}

To think about at home: In one of the Moon landings, an astronaut walking on the Moon’s surface had a belt hanging from his space suit, and the belt oscillated as a

physical pendulum. A scientist on the Earth observed this motion of television and used it to estimated the acceleration due to gravity on the Moon. Which quantities

would he have to estimate (measure) and how did he make this calculation?

Example for completeness: Torsional PendulumA rigid object suspended by a wire attached at the top to a

fixed support forms such a device. When the object is twisted through some angle Θ, the twisted wire exerts on the object

a restoring torque that is proportional to the angular position. So, the torsion τ = -κΘ (κ is torsion constant).

(Real life example is swings twisted - they will rotate back)

Applying Newton’s second law fro rotational motion, we can obtain the equation for simple harmonic oscillation with

the period T = 2π sqrt(I/κ). This looks very similar to the “linear” pendulum and is called a torsional pendulum. There is also a small angle restriction here, the situation has to be

within the elastic limit of the wire.

Damped harmonic motionIn simple harmonic motion, an object oscillates with a constant amplitude, because there is no mechanism for dissipating energy. In reality, however, friction or some other energy-dissipating mechanism is always present. In the presence of energy

dissipation, the amplitude of oscillation decreases as time passes, and the motion is not longer simple harmonic motion. Instead, it is referred to as damped harmonic motion, the decrease in amplitude being called “damping”.

There are different degrees of damping that can exist. The graph on the left shows the Amplitude over time. The red curve shows a so-called underdamped version, it

is just slightly damped, but still clearly as oscillation visible.

The damping can be so strong, that the oscillation does not appear. The smallest degree of damping that completely eliminates the oscillations is termed “critical damping”, and

the motion is said to be critically damped (see green curve). When the damping exceeds the critical value, the motion is said to be overdamped, and it will take a long

time to return to the equilibrium position (see blue curve).

Driven harmonic motion and resonance

When energy is continually added to an oscillating system, the amplitude can be increased significantly - we will look at this effect. Let’s remember, that we need an agent to apply a force that stretches or compresses a spring initially to start the dimple harmonic motion. Suppose this force is applied at all times, not just for a brief initial moment. The force could be provided, for example, by a person who simply pushes and pulls the object back and forth. The resulting motion is known as driven harmonic motion, because the additional force drives or controls the behavior of the object to a large extent. The additional force is identified as the driving force. There is one particular important case of driven harmonic motion. When the driving force has the same frequency as the spring system, then it always points in the direction of the object’s velocity. The frequency of the spring system is f = (1/2π)sqrt(k/m) and is called a natural frequency, because it is the frequency at which the spring system naturally oscillates. Since the driving force and the velocity always have the same direction, positive work is done on the object at all times, and the total mechanical energy of the system increases. As a result, the amplitude of the vibration becomes larger and will increase without limit, if there is no damping force to dissipate the energy being added by the driving force. The situation is known as resonance.

ResonanceRESONANCE

Resonance is the condition in which a time-dependent force can transmit large amounts of energy to an oscillation object, leading to a large amplitude motion. In the

absence of damping, resonance occurs when the frequency of the force matches a natural frequency at which the object will oscillate.

The role played by the frequency of a driving force is a critical one. The matching of this frequency with a natural frequency of vibration allows even a relatively weak force

to produce a large amplitude vibration, because the effect of each push-pull cycle is cumulative. Resonance can occur with any object that can oscillate, and springs need

not be involved. The greatest tides occur in the Bay of Fundy, which lies between New Brunswick and Nova Scotia. The difference between water level at high an low tides averages in some locations 15 m. This phenomenon is partly due to resonance. The

time, or period, that it takes for the tide to flow into and ebb out of a bay depends on the size of the bay, the topology of the bottom, and the configuration of the shoreline. The ebb and flow of the water in the bay of Fundy has a period of 12.5 hours, which is

very close to the lunar tidal period of 12.42 hours. The tie then “drives” water into and out of the Bay of Fundy at a frequency that nearly matches the natural frequency.

The result is the extraordinary high tide. (You can create a similar effect in bath tub by moving back and forth in synchronism with the waves you’re causing).

Bay of Fundy Any object has a natural frequency. In the

construction of bridges, you need to know this.

You want to avoid choosing a natural

frequency that is close to typical interactions.

Tacoma Narrow BridgeIn 1940, turbulent winds set up torsional vibrations in the Tacoma Narrows Bridge,

causing it to oscillate at a frequency near one of the natural frequencies of the structure. This resonance condition led to the bridge’s collapse

Several versions of this movie are available on you tube ...

Waves ...

• A wave is a traveling disturbance

• A wave carries energy from one place to another

Water waves have two features in common:

Basic types of wavesThere are two basic types of waves, transverse and longitudinal. We can use Slinky to

demonstrate them.

transversal longitudinal

TRANSVERSE and LONGITUDINAL waves:

A transverse wave is one in which the disturbance occurs perpendicular to the direction of travel of the wave.

A longitudianl wave is on in which the disturbance occurs parallel to the line of travel of the wave.

Transverse waves - examples:

radio waveslight wavesmicrowaves

instrument strings (guitar, banjo, ...)

Longitudinal waves - examples:

sound wave

A wave can be a mixture of both types, water waves certainly are