AP Physics B Rotational Motion and the Law of Gravity Lecture Notes 2008

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```					Rotational Motion and
the Law of Gravity

AP Physics B
Lecture Notes

Rotational Motion and the Law of Gravity
Rotational Motion and the Law of Gravity

Topics

7-04 Centripetal Acceleration
7-05 Newtonian Gravitation
7-06 Kepler’s Laws

Rotational Motion and the Law of Gravity
Centripetal Acceleration

Uniform circular motion: motion in a circle of constant
radius at constant speed
Instantaneous velocity is always tangent to circle.

v2

v1

Rotational Motion and the Law of Gravity
Centripetal Acceleration
Radial Acceleration:                    Similar Triangles
Δv v                vΔr
v2
         Δv 
Δr r                 r

Dt      -v1      Dq         v2
r2

Dq      Dr
v1                 Dv
r1
Divide by time
Δv     v  Δr 
      
Δt     r  Δt 
Centripetal
v2
Acceleration               ar 
r
Rotational Motion and the Law of Gravity
Centripetal Acceleration

In uniform circular motion the acceleration is called the
centripetal, or radial, acceleration. It is perpendicular to the
velocity and points towards the center of the circle.

v

v2
a                   ar 
r
r
r

Rotational Motion and the Law of Gravity
Rotational Motion and The Law of Gravity

Is it possible for an object moving with a constant
speed to accelerate? Explain.
(A) Yes, although the speed is constant, the direction
of the velocity can be changing.
(B) No, if the speed is constant then the acceleration
is equal to zero.
(C) No, an object can accelerate only if there is a net
force acting on it.
(D) Yes, if an object is moving it can experience
acceleration
Centripetal Acceleration Problem

A jet plane travelling 525 m/s pulls out of a
dive by moving in an arc of radius 6.00 km.
What is the plane’s acceleration?

Centripetal Acceleration:

v2        525 m/s 2     46 m/s 2
a       
r          6000 m

Rotational Motion and the Law of Gravity
Rotational Motion and The Law of Gravity

An object moves in a circular path at a constant speed.
Compare the direction of the object's velocity and acceleration
vectors.
(A) The vectors are perpendicular.
(B) Both vectors point in the same direction.
(C) The vectors point in opposite directions.
(D) The question is meaningless, since the acceleration is zero.
Rotational Motion and The Law of Gravity

What type of acceleration does an object moving with
constant speed in a circular path experience?
(A) free fall
(B) constant acceleration
(C) linear acceleration
(D) centripetal acceleration
Centripetal Acceleration

For an object to be in uniform circular motion, there
must be a net force acting on it.

 v2 
The radial force                         Fr  mar  m 
 r 
on the ball is                   v                  
provided by the string
 v2 
Fr  m 
 r 
a                     
r
Fr                There is no
This radial force                              centrifugal force
is called a                                 acting on the ball
centripetal force

Rotational Motion and the Law of Gravity
Centripetal Acceleration

The speed of an object in
Uniform Circular Motion      m
T             r

 F  ma          M

T  mar               Mg

mv 2
T  Mg          Mg 
r                    v2
ar 
r
Mgr
v
m
Rotational Motion and the Law of Gravity
Centripetal Acceleration Problem

A 0.45 kg ball, attached to the end of a horizontal cord, is rotated
in a circle of radius 1.3 m on a frictionless horizontal surface. If
the cord will break when the tension in it exceeds 75 N, what is
the maximum speed the ball can have?

F  ma
 v2 
FT  m 
 r 
 

v
FTr  75 N1.3 m 
m    0.45 kg
 15 m/s

Rotational Motion and the Law of Gravity
Centripetal Acceleration

Motion in a vertical circle
T   mg
The tension in the string
when the ball is at the top.            r

 F  ma                   v2
ar 
r
v2
mg  Ttop  m
r

mv 2
Ttop       - mg
r
Rotational Motion and the Law of Gravity
Centripetal Acceleration

Motion in a vertical circle

The tension in the string when
the ball is at the bottom.              r

v2
ar 
 F  ma              r                T

v2
Tbottom - mg  m
r
mg
mv 2
Tbottom        mg
r

Rotational Motion and the Law of Gravity
Centripetal Acceleration Problem

A bucket of mass 2.00 kg is whirled in a vertical circle of
radius 1.10 m. At the lowest point of its motion the tension
in the rope supporting the bucket is 25.0 N.
(a) Find the speed of the bucket.                         FT

 F  ma
 v2 
FT - mg  m 
 r 
 
Fg  mg
v
r FT - mg 

         
1.1 m 25 N - 2 kg 9.8 m/s 2   
m                    2 kg

v  1.7 m/s

Rotational Motion and the Law of Gravity
Centripetal Acceleration Problem (con’t)

A bucket of mass 2.00 kg is whirled in a vertical circle of
radius 1.10 m.
(b) How fast must the bucket move at the top of the
circle so that the rope does not go slack?

 F  ma
 v2 
FT  mg  m 
 r       FT  0
 
FT
v
r FT  mg 
m

 rg  1.1 m 9.8 m/s     2
                 Fg  mg

v  3.3 m/s

Rotational Motion and the Law of Gravity
Rotational Motion and The Law of Gravity

A pilot executes a vertical dive then follows a semi-
circular arc until it is going straight up. Just as the
plane is at its lowest point, the force on him is
(A) less than mg, and pointing up.
(B) less than mg, and pointing down.
(C) more than mg, and pointing up.
(D) more than mg, and pointing down.
Centripetal Acceleration

Maximum Speed in horizontal turn                 Fy  0
N - mg  0
Fx  ma x                      2
v max           N  mg
ax 
r
2
mv max                                        N
fmax 
r                        r

ax         fmax
2
mv max
mg 
r
fmax  N
vmax  gr
fmax  mg            mg

Rotational Motion and the Law of Gravity
Rotational Motion and The Law of Gravity

A car goes around a curve of radius r at a constant
speed v. What is the direction of the net force on the
car?
(A) toward the curve's center
(B) away from the curve's center
(C) toward the front of the car
(D) toward the back of the car
Rotational Motion and The Law of Gravity

A car goes around a curve of radius r at a constant speed v.
Then it goes around a curve of radius 2r at speed 2v. What is
the centripetal acceleration of the car as it goes around the
second curve, compared to the first?
(A) four times as big
(B) twice as big
(C) one-half as big
(D) one-fourth as big
Centripetal Acceleration Problem

How large must the coefficient of static friction be between
the tires and the road if a car is to round a level curve of
radius 85 m at a speed of 95 km/h?
FN
F  ma
Ff   s FN
 v2                            Ff
Ff  m                                          Ff   smg
 r 
 
 v2                     Fg  mg
 smg  m 
 r 
 

v2     95 km/h 2                 2
 1000 m   1 h 
2
 0.84
s 
rg


85 m 9.8 m/s 2      

       
1km   3600 s 

Rotational Motion and the Law of Gravity
Centripetal Acceleration
N
Friction on a banked road                                                                 v
q
 Fy  0  N cosq  - mg - f sinq                  a

mg  f sinq                                              q
N
cosq                                               f
q
 Fx  ma                2                  r
N sinq   f cosq  
mv                                                 mg
r
 mg  f sin q                          mv 2
                  sin q   f cosq  
   cosq                                 r
 v2                                          mv 2
m  cosq   mg sin q   f              f      cosq  - mg sinq 
 r 
                                              r
Rotational Motion and the Law of Gravity
Centripetal Acceleration

mv 2
Friction on a banked road                     f      cosq  - mg sinq 
r
N
v
q              f
a
When
mv 2
f                                 cosq   mg sinq 
r
When                         q

mv 2                                 mg       When
cosq   mg sinq 
mv 2
r                                                 cosq   mg sinq 
r
No Friction
Rotational Motion and the Law of Gravity
Centripetal Acceleration
N
v
Turning a banked curve                                   q
with no friction                           a

mv 2
f      cosq  - mg sinq 
r                                                q
mv 2                                                        mg
0      cosq  - mg sinq 
r
mv 2
cosq   mg sinq 
r
2                             v2 
tanq  
v                  q  tan -1  
 rg 
rg                             
Rotational Motion and the Law of Gravity
Rotational Motion and The Law of Gravity

The banking angle in a turn on the Olympic bobsled track
is not constant, but increases upward from the horizontal.
Coming around a turn, the bobsled team will intentionally
"climb the wall," then go lower coming out of the turn.
Why do they do this?
(A) to give the team better control, because they are able
to see ahead of the turn
(B) to prevent the bobsled from turning over
(C) to take the turn at a faster speed
(D) to reduce the g-force on them
Centripetal Acceleration

Weight

 Fy  0
Reading on scale
is the normal force                       N - mg  0
N
N  mg

Scale

mg

Rotational Motion and the Law of Gravity
Centripetal Acceleration

Apparent Weight at the Earth’s Surface
At the North Pole:
NN                           F  0
N N - mg  0
mg                                 N N  mg

v                             At the Equator:
NE
 F  mar
mg
mv 2
mg - N E 
R
mv 2
N E  mg -
R
Rotational Motion and the Law of Gravity
Centripetal Acceleration

A space station is in the shape of a hollow ring 450 m
in diameter. Gravity is simulated by rotating the ring.
Find the speed in revolutions per minute needed in
order to simulate the Earth’s gravity.

R
v

N
Rotational Motion and the Law of Gravity
Centripetal Acceleration

The speed in revolutions per minute

 F  mar              v2
ar 
r
mv 2
N  mg                                 v
R                                              N
v  gR                      2R
v      2Rf           R = 225 m
2Rf  gR                      T

g              9.8
f                              0.00332 rev/s
2
4 R           4 225 
2

rev  60 s              rev
0.00332                  1.99
s  min                min
Rotational Motion and the Law of Gravity
Newtonian Gravitation

Gravitational Force:
Gravitational Force is the mutual force of attraction between
any two objects in the Universe.

m
F

Mm
F           R                  FG
R2
M

Universal Gravitational Constant
Nm 2
G  6.67 x 10 -11
kg 2             Rotational Motion and the Law of Gravity
Newtonian Gravitation

Gravitational Potential Energy associated with an
object of mass m at a distance r from the center of
the Earth is
m

r

ME
m Em
PE  -G                    [7.21]
r

Rotational Motion and the Law of Gravity
Newtonian Gravitation

Escape velocity
v esc
An object projected upward from the Earth’s         RE ME
surface with a large enough speed will soar off
into space and never return.
This speed is called the Earth’s escape velocity.      11.2 km/h
25,000 mi/h
KEi  PEi  0

mv esc  GM Em 
2
-      0
2       RE 

2GM E
v esc 
RE
Rotational Motion and the Law of Gravity
Newtonian Gravitation Problem

Calculate the acceleration due to gravity on the Moon. The
Moon’s radius is 1.74 x 106 m and its mass is 7.35 x 1022 kg.

GM mm
F       2
 mg m
Rm

gm 
GM m

6.67 x 10-11 N  m2/kg 2 7.35 x 1022 kg
2
Rm                     1.74 x 10 m
6    2

g m  1.62 m/s 2

Rotational Motion and the Law of Gravity
Newtonian Gravitation Problem

A hypothetical planet has a mass 1.66 times that of Earth,
but the same radius. What is g near its surface?

Mp        1.66M E             GM E 
gp  G        G                 1.66 2 
2             2              R 
Rp            RE              E 


g p  1.66g E  1.66 9.8 m/s 2       16.3 m/s 2

Rotational Motion and the Law of Gravity
Newtonian Gravitation

Speed of a Satellite
v2
v
FG
Mm       Fm
R
m             R2

F
Mm    v2
G     m
2    R
R
F          R

M
GM
v
R

Rotational Motion and the Law of Gravity
Rotational Motion and The Law of Gravity

Two planets have the same surface gravity, but planet
B has twice the radius of planet A. If planet A has
mass m, what is the mass of planet B?
m
(A)                    (B ) m
2
(C ) 2 m               ( D) 4m
Newtonian Gravitation Problem

A certain neutron star has five times the mass of our Sun
packed into a sphere about 10 km in radius. Estimate the
surface gravity on this monster.

MN         5M S
gN  G         G
R2
N         R2
N

gN 
6.67 x 10-11 N  m2/kg 2 51.99 x 1030 kg 
1. x 10 m
4   2

g N  6.6 x 1012 m/s 2

Rotational Motion and the Law of Gravity
Newtonian Gravitation

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

Without
gravity

With
Earth                    gravity

Rotational Motion and the Law of Gravity
Rotational Motion and The Law of Gravity

Compared to its mass on the Earth, the mass of an
object on the Moon is
(A) the same.
(B) less.
(C) more.
(D) half as much.
Newtonian Gravitation Problem

Calculate the force of Earth’s gravity on a spacecraft 12,800 km
(2 Earth radii) above the Earth’s surface if its mass is 1350 kg.

Fg 
mg


1350 kg 9.8 m/s 2       1470 N
9           9

Rotational Motion and the Law of Gravity
Kepler’s Laws

Kepler’s Three Laws:

1. All planets move in elliptical orbits with the
Sun at one of the focal points.

Rotational Motion and the Law of Gravity
Kepler’s Laws

Kepler’s Three Laws:

2. A line drawn from the Sun to any planet sweeps
out equal areas in equal time intervals.

Area 1                            Area 2

Area 1 = Area 2

Rotational Motion and the Law of Gravity
Kepler’s Laws

Kepler’s Three Laws:

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
R

T2
3
 constant
R

Rotational Motion and the Law of Gravity
Kepler’s Laws

T
MS                    r       F

F  ma                                     4  2r 3
GM S 
T2
M Sm      v2
G          m
r2       r
r3          GM S
2                       
 2r                           2
4 2
                           T
M Sm
G 2 m  T 
r       r
Rotational Motion and the Law of Gravity

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