Rotational Motion Notes

Chapter 5 Circular Motion; Gravitation

Today Chapter 5: Circular Motion / G

Monday No lab – no class

Tuesday -Chapter 5 – continue Circular Motion -Chapter 4 in class test

Friday -Finish Chapter 5

Class will start shortly

Notes on schedule • Chapter 4 homework and

chapter test (online) are assigned today (Monday, Nov. 6) They will be due on Tuesday, Nov. 14 at 5.30pm

• In class on Tuesday, Nov. 14 will be Chapter 4 test (in class)

• About the final exam (on Thursday Dec. 14 at 1-3pm): it will just be our last in class chapter test

Frothy Fissix Donald Glaser — don’t google him yet

Donald Glaser, Physics Nobel 1960

The real story…

Newton’s first law revisited

• A girl attaches a rock to a string, which she then swings counter-clockwise in a horizontal circle. The string breaks at point P in the figure, which shows a bird's-eye view (as seen from above). Which path (A- E) will the rock follow?

Contents of

Chapter 5

1. Kinematics of Uniform Circular Motion 2. Dynamics of Uniform Circular Motion 3. Highway Curves, Banked and

Unbanked 4. Nonuniform Circular Motion 5. Newton’s Law of Universal Gravitation 6. Gravity Near the Earth’s Surface 7. Satellites and “Weightlessness” 8. Planets, Kepler’s Laws, the Moon, and

Newton’s Synthesis 9. Something about Moon rising 10. Types of Forces in Nature

Exercises

• You don’t need to hand these in—but you may want to try a few (there are written solutions on D2L—so check your answers!)

• From the chapter: 5 23 35 44 45 50 58 61 68 84 87 88

Vocabulary of circular motion

On completing this section, you should be able to: • Describe rotary motion and convert between radians,

degrees, and rotations for angular measurement • Define angular displacement, velocity, and

acceleration • Define period and frequency of periodic motion

This section is not in this chapter in Giancoli—but I believe it will help you

understand the rest of the chapter

Angular measurements

• Once around a circle is measured as: – 1 revolution (aka rev) – 360º – 2π radians

• The actual distance covered along the edge is 2πr

1 rev = 360° 1° = 60' 1' = 60"

Measuring the angle swept out

In addition, any other arc length s (s represents arc length and not displacement in this chapter) must subtend an angle:

The angle θ through which an object rotates is called the angular displacement.

Note that the radian unit is a ratio of two lengths; therefore, it really has no physical dimensions. The radian unit merely indicates that an angle is being considered.

θ = s r

rad

θ is the Greek letter “th”, theta

Example 8-2

A car wheel with a diameter of 72.0 cm makes 235 complete revolutions. How far does the car travel if the wheels do not slip?

Solution

θ = 235 rev = 235 x 2π rad, r = 36.0 cm = 0.360 m, s =?

The total distance traveled by a point on the rim (in a circular path) must equal the distance travelled by the car: (0.360 )(235 2 ) 532s r m rad mθ π= = ⋅ =

A note about rads when multiplied by any unitsAlways use radians

Converting between units

units Case 1 Case 2 Case 3

Revolutions 1

Degrees 540°

Radians 15

Converting between units

units Case 1 Case 2 Case 3

Revolutions 1

Degrees 360° 540°

Radians 2π = 6.28 15

540 1.5 360

° =

°

2540 9.43 360 π

°⋅ = °

36015 859 2π

° ⋅ = °

115 2.39 2π ⋅ =

Angular velocity

• The angular velocity ω of a rotating object is defined as its time rate of change of angular displacement.

• If an object rotates through some angular displacement Δq in an elapsed time Δt, its average angular velocity:

ω is the Greek letter “oo” called omega-- it’s actually two o’s stuck together!

t θω ∆

= ∆

rev 2 rad1 min 60 s

π = Units of angular

velocity: rad/s

Technically: ω is a vector. What do

you think it’s direction is?

Example 8-3

A drive shaft rotates at 2750 rpm. Determine (a) the angular velocity in radians per second, and (b) the number of revolutions completed by the drive shaft in 12.0 s.

solution

a)

b) t = 12.0 s = 0.200 min and

A drive shaft rotates at 2750 rpm. Determine (a) the angular velocity in radians per second, and (b) the number of revolutions completed by the drive shaft in 12.0 s.

rev 2 rad2750 2750 x 288 rad/s min 60 s

π = =

ω θ =

t 2rev2750 (0.200 min) = 5.50 x 10 rev

min tθ ω  = =  

 

Angular Acceleration

• The angular acceleration α of a rotating object is defined as its time rate of change of angular velocity.

• If an object rotates through some angular velocity ω in an elapsed time t, its average angular acceleration:

t ωα ∆

= ∆

Units of angular acceleration: rad/s2

Technically: α is another vector.

What do you think its direction is?

α is the Greek letter “alpha”

Angular Motion

Even though the motions are quite different, similar equations and definitions are used to describe both linear and rotary motions.

There will be similarity between the equations used—use your familiarity with linear motion to help you solve rotary motion problems!

Kinematics of angular motion

2 2 1 0 2 x− = ∆v v a

21 20x t t∆ = +v a

a v v =

−1 0

t

21 20t tθ ω α= +

1 0

t ω ωα −

=

2 2 1 0 2ω ω αθ− =

Linear Motion For constant a

Angular Motion For constant α

These look familiar?!

A word about θ – two different uses

5-1 Kinematics of Uniform Circular Motion

Uniform circular motion: motion in a circle of constant radius at constant speed

Instantaneous velocity is always tangent to circle.

Uniform rotary motion

• Uniform rotary motion when angular velocity constant

• Consider a rigid object that rotates through some angular displacement θ in t. A particle located a distance r from the axis of rotation, moves in a circular arc through a distance s, its average speed:

s rv r t t

θ ω= = =

Circular Motion

• For constant speed and motion around a fixed point is called uniform circular motion

• Going around once is called a “cycle” • The time to complete one cycle is called the period,

T • The frequency (how many cycles per second) is

called

• For N cycles, T = t/N and f = 1/T = N/t

e.g., if an object goes around in 1/2 second, the frequency 2 Hz

1f T

= units of Hertz, Hz, or s-1.

Let’s do an experiment

• Object on a string. Time how long it takes to make a revolution. What is the frequency? What is the angular velocity (rpm, rad/s, degrees/second)? What is the speed?

s rv r t t

θ ω= = =

? ? ? ?

f T

r ω

= = = =

Let’s try out our new knowledge

How fast are the passengers moving? Ignore the rotation of the “basket” for now

? ? ? ?

f T

r ω

= = = =s rv r

t t θ ω= = =

The beast is 43m

Do we need a force to go in a circle at constant speed?

No, of course not… err… yes, for sure… well, I’m not sure…

5-2 Dynamics of Uniform Circular Motion

We can see that the force must be inward by thinking about a ball on a string:

Let’s look at the change in velocity vector

Looking at the change in velocity in the limit that the time interval becomes infinitesimally small, we see that the change in the velocity vector points into the circle.

Centripetal Force and Acceleration

• But since we know the acceleration is:

• We know the acceleration must point into the circle too, but even further:

• So the force is into the circle to—directed towards the center

The “c” stands for centripetal

So even though the magnitude of the velocity is constant…

• We now know we need a force to go around a circle

• What’s the magnitude of the force? There is a complicated pseudo-proof in the textbook but here’s the result:

2

c va r

= Where v is the speed and r is the radius of motion

Important equation?! You bet!

2

c va r

=

5-2 Dynamics of Uniform Circular Motion

For an object to be in uniform circular motion, there must be a net force acting on it. We already know the acceleration, so can immediately write the force:

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5-2 Dynamics of Uniform Circular Motion

There is no centrifugal force pointing outward; what happens is that the natural tendency of the object to move in a straight line must be overcome.

If the centripetal force vanishes, the object flies off tangent to the circle.

Example 5-1 Acceleration of a revolving ball

Example 5-2 Moon’s centripetal acceleration

Circular motion criteria

• If you want to go around in a circle at constant speed and at a fixed radius you need to supply a force:

2

c vF ma m r

= = What happens if I let

go? The ball no longer has a force and it

continues straight—now we understand!

Scalar equation!

What if the force isn’t exactly right?

Example 5-3 Force on revolving ball (horizontal)

Centrifugal force

• Not a real force, just the inertia of a body wanting to continue in a straight line.

• But centripetal force is real and calculable

Imagine you are riding on the back of a flat bed truck—if you watched the load slide when you rounded a corner, you would think that there was a “force” on the load (but there wasn’t!)

Earth around the sun

Going over a hill

What happens when we go over a hill?

2 2

y N N v vF F mg m F m g r r

  = − = − ⇒ = − 

  ∑

Your weight goes down!

Example 5-4 Revolving ball (vertical circle)

0topT ≈

Roller coaster problem

• How fast do you need to go to “loop the loop”?

2

2

2 : /

0 / y N

N

N L F F mg mv R

F mg mv R v gR

= − = −

≈ ⇒ − = − ⇒ ≥

∑

Example 5-5 Tetherball

Clay on a potter’s wheel

It must be held by static friction

2

2 2

2 : 0

/

/

y N y N

x c N c

s

N L F F mg ma F mg

F F F mg ma mv R

vmg mv R gR

µ µ

µ µ

= − = = ⇒ =

= = = = =

= ⇒ =

∑ ∑

Cornering about friction and centripetal force

Looking at a wheel rotating

• Let’s take a look at the interaction between the surface and the wheel.

Rolling without slipping

5-3 Highway Curves, Banked and Unbanked

When a car goes around a curve, there must be a net force towards the center of the circle of which the curve is an arc. If the road is flat, that force is supplied by friction.

Curves in roads

As long as the tires do not slip, the friction is static. If the tires do start to slip, the friction is kinetic, which is bad in two ways: 1. The kinetic frictional force is smaller than the static. 2. The static frictional force can point towards the center of

the circle, but the kinetic frictional force opposes the direction of motion, making it very difficult to regain control of the car and continue around the curve.

Q: Why do they say to turn into the skid? Increasing r in turn decreases the ac

and hence the force necessary to go around a corner.

2vF m r

=

5-3 Highway Curves, Banked and Unbanked

If the frictional force is insufficient, the car will tend to move more nearly in a straight line, as the skid marks show.

Example

Determine the centripetal force on a 1450 kg car that makes a turn of radius 315 m at a constant speed of 60.0 km/h.

Solution

3km 60.0 x 10 m60.0 16.7 m/s h 3600 s

v = = =

F mv rc

2 kg)(16.7 m / s) 315 m

N= = = 2 1450 1280(

m = 1450 kg, r = 315 m

Example

Determine (a) the minimum friction force and (b) the minimum coefficient of friction

between the tires and a flat horizontal road that would enable a 1450 kg car to make a turn of radius 135 m at a constant speed of 55.0 km/h.

x

y Solution

We draw a FBD

a) minimum friction force:

b) Therefore,

( )22 15.3 / 1450 2510

135 m svF m kg N

r m = = ⋅ =

0y N NF F mg F mg= − = ⇒ =∑ 2 2

0.177x N v vF F mg m r rg

µ µ µ= = = ⇒ = =∑

Example 5-6 Skidding on a curve

5-3 Highway Curves, Banked and Unbanked

Banking the curve can help keep cars from skidding. In fact, for every banked curve, there is one speed where the entire centripetal force is supplied by the horizontal component of the normal force, and no friction is required. This occurs when:

5-4 Nonuniform Circular Motion

If an object is moving in a circular path but at varying speeds, it must have a tangential component to its acceleration as well as the radial one.

Example 5-8 Two components of acceleration

5-5 Newton’s Law of Universal Gravitation

If the force of gravity is being exerted on objects on Earth, what is the origin of that force?

Newton’s realization was that the force must come from the Earth.

He further realized that this force must be what keeps the Moon in its orbit.

5-5 Newton’s Law of Universal Gravitation

The gravitational force on you is one-half of a Third Law pair: the Earth exerts a downward force on you, and you exert an upward force on the Earth.

When there is such a disparity in masses, the reaction force is undetectable, but for bodies more equal in mass it can be significant.

5-5 Newton’s Law of Universal Gravitation

Therefore, the gravitational force must be proportional to both masses.

By observing planetary orbits, Newton also concluded that the gravitational force must decrease as the inverse of the square of the distance between the masses.

In its final form, the Law of Universal Gravitation reads:

where G = 6.67 × 10−11 N·m2/kg2

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5-5 Newton’s Law of Universal Gravitation

The magnitude of the gravitational constant G can be measured in the laboratory.

This is the Cavendish experiment.

Example 5-9 Can you attract another person gravitationally?

Example 5-10 Spacecraft at twice the earth radius

5-6 Gravity Near the Earth’s Surface

How does universal gravitation (G) compare to local gravity (g) at the Earth’s surface?

We know the radius and mass of the Earth

245.98 10earthM kg= × 2

11 26.67 10 N mG

kg − ⋅

= ×

2 earth

G e

GM mF R

=

gF mg=

2 earth

e

GM g R

⇒ 29.81 /m s=

66.371 10earthR m= ×

5-6 Gravity Near the Earth’s Surface

The acceleration due to gravity varies over the Earth’s surface due to altitude, local geology, and the shape of the Earth, which is not quite spherical.

Example 5-11 Gravity on Everest

5-7 Satellites and “Weightlessness”

Satellites are routinely put into orbit around the Earth. The tangential speed must be high enough so that the satellite does not return to Earth, but not so high that it escapes Earth’s gravity altogether.

Let’s figure this out ourselves right now

1. Near earth orbit (time for orbit)

2. Geosynchronous

mass of Earth = 5.97219 × 1024 kilograms

Earth radius=6.3781×106 m

G = 6.67 × 10−11 N·m2/kg2

5-7 Satellites and “Weightlessness”

The satellite is kept in orbit by its speed—it is continually falling, but the Earth curves from underneath it.

5-7 Satellites and “Weightlessness”

Objects in orbit are said to experience weightlessness. They do have a gravitational force acting on them, though!

The satellite and all its contents are in free fall, so there is no normal force. This is what leads to the experience of weightlessness.

5-7 Satellites and “Weightlessness”

More properly, this effect is called apparent weightlessness, because the gravitational force still exists. It can be experienced on Earth as well, but only briefly:

Example 5-12 Geosynchronous satellite

We did this already!?! Let’s see if you can get it yourself?

5-8 Planets, Kepler’s Laws, the Moon, and Newton’s Synthesis

Kepler’s laws describe planetary motion.

1. The orbit of each planet is an ellipse, with the Sun at one focus.

5-8 Planets, Kepler’s Laws, the Moon, and Newton’s Synthesis

2. An imaginary line drawn from each planet to the Sun sweeps out equal areas in equal times.

5-8 Planets, Kepler’s Laws, the Moon, and Newton’s Synthesis

The ratio of the square of a planet’s orbital period is proportional to the cube of its mean distance from the Sun.

5-8 Planets, Kepler’s Laws, the Moon, and Newton’s Synthesis

Kepler’s laws can be derived from Newton’s laws. Irregularities in planetary motion led to the discovery of Neptune, and irregularities in stellar motion have led to the discovery of many planets outside our Solar System.

Example 5-13 Where is Mars?

Example 5-14 The Sun’s mass determined

5-9 Types of Forces in Nature

Modern physics now recognizes four fundamental forces:

1. Gravity

2. Electromagnetism

3. Weak nuclear force (responsible for some types of radioactive decay)

4. Strong nuclear force (binds protons and neutrons together in the nucleus)

5-9 Types of Forces in Nature

So, what about friction, the normal force, tension, and so on?

Except for gravity, the forces we experience every day are due to electromagnetic forces acting at the atomic level.

The connection to general relativity

• Generally, we discuss Newton’s law of universal gravitation here in more advanced courses.

• Gravity is seen as warping space-time fabric • Rotating frames are non-inertial

Summary of Chapter 5

• An object moving in a circle at constant speed is in uniform circular motion.

• It has a centripetal acceleration

• There is a centripetal force, which is the mass multiplied by the centripetal acceleration.

• The centripetal force may be provided by friction, gravity, tension, the normal force, or others.

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Summary of Chapter 5

• Newton’s law of universal gravitation:

• Satellites are able to stay in Earth orbit because of their large tangential speed.

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WE HAVE NOW FINISHED THIS CHAPTER.

THIS CHAPTER’S HOMEWORK AND TEST (ONLINE) ARE ASSIGNED TODAY

DUE IN ONE WEEK (SEE D2L) AT 5.30PM

WE WILL HAVE THE CHAPTER TEST (IN CLASS) NEXT WEEK

GOOD LUCK!

Next Topic – Chapter 6 – Work, Energy and Power