7. Rotational Motion and the Law of Gravity 7.1. Angular Speed and Angular Acceleration

7.1.1. Speed is the magnitude of velocity A car that travels 100m in 20s has what speed?

!

v =100m20s

= 5 ms

7.1.1.1.This is linear motion 7.1.1.1.1. Motion along 1 line 7.1.1.1.2. Works great when treating things as a particle. 7.1.1.1.3. Not for looking at the larger picture.

7.1.2. Physics is about knowing motion and there are two types of motion 7.1.2.1.Linear and Rotational 7.1.2.2.Consider the wheel

x Or

| |----------x--------| If the wheel rolled 1 time (1 revolution) the position the wheel would have traveled would be the circumference.

7.1.3. The arc length “s” is the amount of distance swept out.

!

sfull = 2"rshalf = "r

squarter = 12 "r

When an object rotates, it moves about an angle

!

" .

!

" full = 2#"half = #

"quarter = 12 #

We can then write

!

" in terms of s.

!

" = sr

s = "r

The unit is “rad” for radians

7.1.3.1.Be certain to check your calculator modes when working with angular quantities.

7.1.4. Consider a rate of change per time

!

"#"t

=angular _ displacement

time= angular_velocity

Where

!

" “omega” is the variable to angular velocity

!

" ave =#$#t

" %t& 0lim#$

#t

A wheel rotates 3.5 revolutions in 4s. What is the angular speed?

Knowns Unknowns Equations

!

" = 3.5rev" = 3.5(2# )t = 4s

!

" = ?

!

" = #t

" = 7$4 s = 5.49 rad

s

Notice that the radius of the wheel was not considered. It doesn’t matter how large the wheel is that affects the angular velocity. The radius of the wheel effects the distance the wheel rolls in that time, not how fast. If the wheel above had a radius of 0.71m, how did the wheel roll?

!

s = "rs = 7# (0.71m)s =15.61m

7.1.5. Angular acceleration

7.1.5.1.Follow the same logic as acceleration, only use angular values now.

If

!

a ="vt

then angular acceleration,

!

" , can be written as…

!

" =#$#t

Where

!

" , “alpha”, is the variable for angular acceleration

!

"ave =#$#t

" %t& 0lim#$

#t

7.1.6. This gives us a linear and rotational comparison. Linear Rotational

!

x = x0 + v0t + 12 at

2

v = v0 + at

v 2 = v02 + 2a"x

!

" = "0 +# 0t + 12$t

2

# =# 0 + at# 2 =# 0

2 + 2$%"

7.2. Rotational Motion Under Constant Angular Acceleration A disc rotates, from rest, to an angular speed of

!

33.8 rads in 0.873s.

a) What is the angular speed of the disc? b) What angle does the disc rotate through? c) If the radius of the disc is 4.28cm, what is the tangential velocity of the disc at t=0.873s? d) What is the tangential acceleration of the disc at t=0.973s?

Knowns Unknowns Equations

!

" 0 = 0 rads

" = 33.8 rads

t = 0.873sr = 4.28cmr = 0.0428m

!

" = ?#$ = ?vt = ?at = ?

!

" = #$#t

% = %0 +$ 0t + 12"t

2

vt = r$at = r"

!

" =33.8 rad

s

0.873s= 38.72 rad

s2

#$ = 12 (38.72 rad

s2)(0.873s)2

#$ =14.75rad = 2.34rev = 845.11o

vt = (0.0428m)(33.8 rads ) =1.44 m

s

at = (0.0428m)(38.72 rads2) =1.66 rad

s2

7.3. Relations Between Angular and Linear Quantities

7.3.1. When something is rotating it also has a tangential velocity. 7.3.1.1.Tangential velocity

v

!

" =vtr

vt = r"

!

"

7.3.1.2.Tangential acceleration

!

" =atr

at = r"

7.4. Centripetal Acceleration 7.4.1. When ever some object turns, it has some angular velocity which in turn

can be used to find a tangential velocity. The direction of the tangential velocity is always changing requiring some acceleration.

!

ac =vt2

r

As a car drives around a circle, what direction is the angular acceleration, tangential acceleration, and acceleration?

!

" a

!

at So… Because

!

v = r"

!

ac =r2" 2

r= r" 2

The centripetal acceleration is always pointing toward the center of curvature for all rotating object. In this way, the centripetal acceleration and tangential acceleration are perpendicular to each other. A car moves from rest to a speed of

!

30.7 ms in 7.6s. The car is on a circular track

with a radius of 27.8m. Find

!

" ,

!

" , and

!

ac. Knowns Unknowns Equations

!

v0 = 0 ms

v = 30.7 ms

r = 27.8m

!

" = ?# = ?ac = ?

!

" = vr

" =30.7ms27.8m =1.2 rad

s

# = "t

# =1.2 rads7.6s = 0.15 rad

s2

ac = v 2r

ac =(30.7 ms )

2

27.8m = 33.9 ms2

Any object experiencing some form of rotation has both tangential acceleration,

!

at , and centripetal acceleration,

!

ac, at the same time. We use Pythagoreans theorem to solve for the total acceleration on the object.

!

a = at2 + ac

2

7.4.2. The Right Hand Rule 7.4.2.1.Right hand rule number 1 explains the cross product for a two vectors.

When two vectors are multiplied by each other, the product is a vector that is perpendicular to the two initial vectors. Consider the “x” and “y” axis. When these two vectors are multiplied, we get the “z” axis. Using your right, allow your pointer and thumb to represent the x and y axis,

now your palm points in the direction of the vector product. (We’ll see this again soon.)

7.4.3. Centripetal Force

Newton’s Laws are still working even though we have transitioned out of a linear pattern. If F=ma, then

!

Fc = mac = m v 2r = mr"

If a spaceship wants to simulate earth’s gravity by rotating a large cylinder of radius 29.8m, what speed would the cylinder have to be rotated at? We know that the centripetal acceleration needs to be the same as gravity.

Knowns Unknowns Equations

!

r = 29.8mac = 9.8 m

s2

!

v = ?" = ?

!

ac = v 2r = r" 2

v = (9.8 ms2 )(29.8m)

v =17.09 ms

" =9.8 m

s2

29.8m = 0.57 rads

7.5. Newtonian Gravitation Recall…

!

F =G m1m2r 2

Where G is the universal gravitational constant,

!

6.67x10"11N m 2

kg 2 , and F is the gravitational force between the two objects. For us and Earth, that force is best known as our weight, but for our planet and the sun, it is the force that keeps the planet from flying off into the universe. If we treat the first mass as the smaller mass, for any orbital relationship, we get…

!

F =G m1m2r 2 = m1ac

!

ac =G m2r 2

7.5.1. Recall Gravitational Potential Energy…

!

PE = mgy = "magy = "G Memr 2 r = "G Mem

r for

!

r > RE If we want to escape Earth, we have to overcome the force of gravity, we have to overcome the potential energy.

7.5.2. Escape Velocity

!

KE0 + PE0 = 12mv0

2 "G MemRE

When

!

v0 is large enough you can escape.

!

12mvo

2 "G MEmRE

= 0J

!

vescape =2GME

RE

For any planet

!

vescape =2GMP

RP

7.6. Kepler’s Laws 7.6.1. Geocentric Model Ptolemy

7.6.1.1.2nd century AD to 16th Century 7.6.2. Heliocentric Model Copernicus (1473-1543)

7.6.2.1.1543 to the Present 7.6.2.2.Tycho Brahe (1546-1601)

7.6.2.2.1. Gave data, over 20 years, showing evidence for accepted model of Solar System

7.6.2.3.Johannes Kepler 7.6.2.3.1. Brahe’s assistance

7.6.2.3.1.1.Used Brahe’s data to write a mathematical equation for planetary motion.

7.6.2.3.1.2. Took 16 years 7.6.2.3.1.3. 3 Laws

7.6.3. Kepler’s 3 laws 7.6.3.1. All planets move in elliptical orbits with the sun at one of the focal

points. 7.6.3.2. A line drawn from the sun to any planet sweeps out equal areas in

equal time intervals. 7.6.3.3. The square of the orbital period of any planet is proportional to the

cube of the average distance from the planet to the Sun.

!

T 2 =4" 2

GMs

#

$ %

&

' ( r3 = ksr

3

Where k is Kepler’s constant

!

k =4" 2

GMs

#

$ %

&

' ( = 2.97x10)19 s2

m 3

What is the average radius of Earth’s orbital around the sun?

We have to first figure out how many seconds are in one year. A year is 365.25 days and each day is 24 hours. An hour is 60 minutes which in turn is 60 seconds.

!

365.25days 24hours1day

"

# $

%

& ' 60min1hour"

# $

%

& ' 60s1min"

# $

%

& ' = 31557600s ( 3.156x107 s

Knowns Unknowns Equations

!

T = 3.156x107s

!

r = ?

!

T 2 = kr3

r =T 2

k3

r =(3.156x107 s)2

2.97x10"19 s2

m 3

3

r =1.51x1011m =1.51x108km