Download Lecture 7: Rotational Motion and the Law of Gravity

Chapter 7: Rotational Motion and the
Law of Gravity
Angular Speed & Acceleration
 A unit
of angular measure: radian
y
P
r
length of the arc from the x-axis s:
s = rq
where s,r in m, and q in rad(ian)
q
x
A complete circle: s = 2pr
360o = 2p rad
57.3o= 1 rad
Momentum and Impulse

Angular displacement and velocity
y
Angular displacement:
Dq = qf - qi in a time interval Dt = tf – ti
P at tf
Average angular velocity:
 av =
r
qf
P at ti
Dq qf - qi
=
Dt
tf - ti
rad/s
Instantaneous angular velocity:
qi
x
Dq dq
 = lim
=
Dt 0 Dt
dt
rad/s
 < (>) 0 (counter) clockwise rotation
Momentum and Impulse

Angular acceleration
y
Average angular acceleration:
av =
P at tf
D f - i
=
Dt
tf - ti
Instantaneous angular acceleration:
D d
=
Dt 0 Dt
dt
 = lim
r
qf
rad/s2
P at ti
qi
x
rad/s2
Momentum and Impulse

An example of a rigid body
• The distance of any two points in a
rigid object does not change when
the body is even in motion.
• When a rigid object rotates about
a fixed axis, every portion of the object
has the same angular speed and the
same angular acceleration.
Rotational Motion under
Constant Angular Acceleration

One-to-one correspondence between linear and angular
quantities
• Similarity between av and vav
Dq qf - qi
 av =
=
Dt
tf - ti
Dx xf - xi
vav =
=
Dt tf - ti
• Similar derivation used for linear quantities can be used for
angular quantities.
v = vi  at
 = i   t
1 2
Dx = vi t  at
2
v 2 = vi2  2aDx
1 2
Dq = i t  t
2
 2 = i2  2Dq
Rotational Motion under
Constant Angular Acceleration

An example
• Example 7.2 : A rotating wheel
A wheel rotates with a constant angular acceleration of 3.50 rad/s2.
If the angular speed of the wheel is 2.00 rad/s at ti=0,
(a) through what angle does the wheel rotate between t=0 and t=2.00 s?
1 2
Dq = i t  t = 11.0 rad
2
= (11.0 rad)(1.00 rev/2 p rad) = 1.75 rev
(b) What is the angular speed of the wheel at t=2.00 s?
 = i  t = 9.00 rad/s
Relations between Angular and
Linear Quantities

 and v
• Consider an object rotating about
the z-axis and a point P on it.
Ds
Dq =
r
tangent to circle
Dq 1 Ds
Dq 1
Ds
=
 lim
= lim
D
t

0
Dt r Dt
Dt r Dt 0 Dt
v
=
r
vt = r
tangential speed
v = vt
The tangential speed of a point on a rotating object equals
the distance of that point from the axis of rotation multiplied
by the angular speed.
Relations between Angular and
Linear Quantities

 and a
• Consider an object rotating about
the z-axis and a point P on it.
Dvt = rD
tangent to circle
Dvt
Dvt
D
D
=r
 lim
= r lim
D
t

0
Dt 0 Dt
Dt
Dt
Dt
at = r
tangential acceleration
The tangential acceleration of a point on a rotating object equals
the distance of that point from the axis of rotation multiplied
by the angular acceleration.
Centripetal Acceleration

Acceleration at a constant speed
• Consider a car moving in a circular
path with constant linear speed v.
 
v f - vi


Dv
aav =
=
t f - ti
Dt
Even though the magnitude of vi and
vf are the same, Dv can be non-zero if
their directions are different.
This leads to non-zero acceleration
called centripetal acceleration.
Centripetal Acceleration

Centripetal acceleration
• Consider a car moving in a circular
path with constant linear speed v.
 
v f - vi


Dv
aav =
=
t f - ti
Dt
p
p - Dq
2
2
=
• Triangle OAB and the triangle
in Fig. (b) are similar.
Dv Ds
v
=
 Dv = Ds
v
r
r
p - Dq
2
 Dq -
Dv v Ds
v 2 r 2 2
aav =
=
 ac =
=
= r 2
Dt r Dt Dt 0
r
r
Total acceleration
a = at2  ac2
p
2
=
Dq
2
-
p - Dq
Dq
2
2
Centripetal Acceleration

Vector nature of angular quantities
• Angular quantities are vector and their directions are defined as:
 points into the page  points out of the page

Centripetal Acceleration

Forces causing centripetal acceleration
• An object can have a centripetal acceleration only if some external
force acts on it.
• An example is a ball whirling in a circle at the end of a string. In this
case the tension in the string is the force that creates the centripetal
force.
Net centripetal force Fc is the
sum of the radial components
of all forces acting on a given
object.
v2
Fc = mac = m
r
• A net force causing a centripetal
acceleration acts toward the center
of the circular path. If it vanishes,
the object would immediately leave
its circular path and move along a
straight line tangent to the circle.
T=Fc
Centripetal Acceleration

Examples
• Example 7.7 : Buckle up for safety
Find the minimum coefficient of static
friction ms between tires and roadway
to keep the car from sliding.
v2
m = f s ,max = m s n
r
n - mg = 0  n = mg
v2
m = m s mg
r
v2
ms = = 0.366
rg
v=13.4 m/s
r=50.0 m
Centripetal Acceleration

Examples
• Example 7.8 : Daytona International Speedway
(a) Find the necessary centripetal acceleration on the banked curve
so that the car will not slip due to the inclination (neglect friction).
 


ma =  F = n  mg
y-component (vertical) :
mg
n cos q - mg = 0  n =
cos q
x-component (horizontal) :
mg sin q
Fc = n sin q =
= mg tan q
cos q
q=31.0o
r=316 m
mac = Fc  ac = Fc / m = mg tan q / m = g tan q = 5.89 m/s 2
(b) Find the speed of the car.
v 2 / r = ac  vc = rac = 43.1 m/s
Centripetal Acceleration

Examples
• Example 7.9 : Riding the tracks
(a) Find the speed at the top.
 

mac = n  mg ; n = 0 at the top.
m
2
vtop
R
= mg  vtop = gR
(b) Find the speed at the bottom.
1 2
1
Etop = mvtop  mgh = mgR  mg (2 R) = 2.5mgR
2
2
1 2
Ebot = mvbot
2
Etop = Ebot 
1 2
mvbot = 2.5mgR  vbot = 5 gR
2
R=10.0 m
Centripetal Acceleration

Examples
• Example 7.9 : Riding the tracks (cont’d)
(a) Find the normal force on a passenger
at the bottom if R=10.0 m
 

mac = n  mg
2
vbot
m
= n - mg
R
2
vbot
n = mg  m
R
5 gR
= mg  m
= 6mg
R
n does not depend on R!
Newtonian Gravitation

Law of universal gravitation
• Using accumulated data on the motions of the Moon and planets,
and his first law, Newton deduced the existence of the gravitational
force that is responsible for the movement of the Moon and planets
and this force acts between any two objects.
If two particles with mass m1 and m2 are separated by a distance r,
then a gravitational force acts along a line joining them with magnitude
m1m2
F =G 2
r
G = 6.673 10-11 kg -1m3s -2
constant of universal gravity
Newton’s 2nd law
Newtonian Gravitation

Law of universal gravitation (cont’d)
• The gravitational force exerted by a
uniform sphere on a particle outside the
sphere is the same as the force exerted
if the entire mass of the sphere were
concentrated at its center.
This is a result from Gauss’s law and
stems from the fact that the gravitational
force is inversely proportional to square
of the distance between two particles.
• The expression F=mg is valid only
near the surface of Earth and can be
derived from Newton’s law of universal
gravitation.
Newtonian Gravitation

Gravitational potential energy revisited
• Gravitational potential energy near Earth (approximation)
PE = mgh
• General form of gravitational potential energy due to Earth
mass of Earth radius of Earth
M Em
PE = -G
for r > RE
r
This is a special case where the zero level for potential energy
is at an infinite distance from the center of Earth.
The gravitational potential energy associated with an object is
nothing more than the negative of the work done by the force
of gravity in moving the object.
Newtonian Gravitation

Gravitational potential energy revisited (cont’d)
• Derivation of gravitational potential energy near Earth
M Em 
M Em 

PE2 - PE1 = -G
-  - G
( RE  h ) 
RE 
 1
1 
= -GM E m 

R

h
R
E 
 E
GM E mh
=
RE ( RE  h )
1
1
 2 for h << RE
RE ( RE  h) RE
PE2 - PE1 
GM E
GM E
mh
=
mgh
where
g
=
RE2
RE2
PE = mgh
Newtonian Gravitation

Escape speed
• If an object is projected upward from Earth’s surface with a large
enough speed, it can soar off into space and never return. This
speed is called Earth’s escape speed vesc.
The initial mechanical energy of the object-Earth system is:
KEi  PEi =
1 2 GM E m
mvi 2
RE
If we neglect air resistance and assume that the initial speed is
large enough to allow the object to reach infinity with a speed of
zero, this value of vi is the escape speed vesc.
1 2 GM E m
mv f = 0 = KEi  PEi ; v f = 0, r = 
2
r
4.3 km/s for Mercury
2GM E
11.2
for Earth
=
2.3
for Moon
RE
60.0
for Jupiter
KE f  PE f =
vesc
Newtonian Gravitation

Examples
• Example 7.10 : Billiards
(a) Find the net gravitational force
on the cue ball.
mm
Fy = F21 = G 2 2 1 = 3.75 10 -11 N
r21
Fx = F31 = G
m3m1
-11
=
6
.
67

10
N
2
r31
F = Fx2  Fy2 = 7.65 10-11 N
Fy
-1
q = tan
= 29.3
Fx
m1,2,3 =0.300 kg
Newtonian Gravitation

Examples
• Example 7.10 : Billiards (cont’d)
(b) Find the components of the force
of m2 on m3.
mm
F23 = G 2 2 3 = 2.40 10 -11 N
r23
cos  = 0.600 , sin  = 0.800
F23x = - F23 cos  = -1.44 10 -11 N
F23 y = F23 sin  = 1.92 10-11 N
m1,2,3 =0.300 kg
Newtonian Gravitation

Examples
• Example 7.12 : A near-Earth asteroid
m1,2,3 =0.300 kg
An asteroid with mass m=1.00x109 kg comes from infinity, and falls
toward Earth.
• Find the change in potential energy when it reaches a point 4.00x108
m from Earth. Find the work done by gravity. ri=0.
 1 1
GM E m  GM E m 
 = GM E m -  
DPE = PE f - PEi = -   r

rf
ri 
r
f
i



DPE = -9.97 1014 J = -Wgrav
(b) Find the speed of the asteroid when it reaches rf=4.00x108 m.
1
DKE  DPE = 0  ( mvb2 - 0) - 9.97 1014 J = 0  vb = 1.41103 m/s
2
(c) Find the work needed to reduce the speed by half.
1
W = DKE  DPE  W = [ m(vb / 2) 2 - 0] - 9.97 1014 J  W = -7.48 1014 J
2