Transcript Rotational Motion and Gravity

Rotation
• Imagine watching a spinning bicycle wheel:
• How would you describe the position of a point (a
reflector, for example) on this wheel with time?
• You could keep track of the (x,y) coordinates of this
point:
y
(x,y)
x
Angular Position
• Instead, it’s more convenient to describe the angle q
that the reflector point makes with the x axis on a
fixed coordinate system: y
q
(angular position)
x
• Defining a positive rotation direction defines the sign
(+ or –) of coordinate q:
+
q negative
+
y
y
q positive
q
q
x
x
• We use radians (rad) as the unit to
measure angle q
q = 1 rad when it subtends arc length s = r
Angular Velocity
• Thus q (in radians) = s / r or s = rq
• To convert between radians and degrees, use a
conversion factor like: 180° = p radians

– To determine the number of radians in 32°:
 (x rad)(180°) = (p rad)(32°)  x = 0.5585 rad
180 p rad


32
x rad
• Rotational motion can be described in analogy to
linear motion:
At time t = t1:
At time t = t2:
q1
q2
x
x
• We can thus define an average angular velocity wave
q 2  q1 q
to be:
(depends on angular
wave 

t 2  t1
t
displacement q2  q1)
Angular Velocity
• We can also define an instantaneous angular velocity
w similar to the linear case:
q
w  lim
 lim wave
t 0 t
t 0
• The units of angular velocity are radians/sec
– 1 rev/s = 2p rad/s so 1 rev/min = 1 rpm = 2p/60 rad/s
• The direction of w is given by the Right–Hand Rule:
– Curl fingers of right hand in direction of the rotation
– Thumb points in direction of w
– Thus w points parallel to axis of rotation (not in direction of
rotation)
• Note that all points on a rotating object move in a
circle centered about the axis of rotation
Angular Acceleration
• The average time rate of change of w is called the
average angular acceleration aave:
w2  w1
w
a ave 

t2  t1
t
• The angular acceleration at a particular instant in
time is the instantaneous angular acceleration a:
w
a  lim
 lim a ave
t 0 t
t 0
• When the rotation axis is fixed, a lies along that
axis (like w), but could point in the same or opposite
direction as w (i.e. parallel or antiparallel to w)
Qualitative Relationships Between w and a
• The qualitative behavior of w and a can be
summarized by the following table:
Same sign
Opposite
sign
w
a
Rotational motion
+
+
Speeding up
(faster rotation in + dir.)
–
–
Speeding up
(faster rotation in – dir.)
+
–
Slowing down
(slower rotation in – dir.)
–
+
Slowing down
(slower rotation in + dir.)
• These were the same relationships between linear
velocity and acceleration in Chapter 2!
Constant Angular Acceleration
• For special case of constant angular acceleration,
then w vs. t graph becomes a straight line:
w
 Constant angular acceleration
means constant rate of increase for w
(a ave  a )
t
• In analogy to motion with constant linear acceleration:
w  w0  at
1 2
q  w0t  at
2
w 2  w02  2aq
 w0  w 
q  
t
 2 
• Notice the similarity between linear (x, v, a) and
angular (q, w, a) quantities and relationships
Relating Linear and Angular Kinematics
• Consider the reflector on the rotating bike wheel
v
again:
r q
x
Axis of rotation
• The angle q (in radians) and arc length s subtended
by q are related by s = rq
– Thus linear displacement s and angular displacement q
are related by: s  rq (think of string unwinding from a
wheel)
– Called the “non-slip” condition
• A similar relationship holds between the linear (or
tangential) speed v and the angular speed w:
– The farther a point is from the rotation axis, the
greater its linear speed for a given value of w
vt  rw
Relating Linear and Angular Kinematics
• This relationship holds true for every point in a rigid
body
– Rigid body = object that has perfectly definite and
unchanging shape and size
– All points thus have the same angular velocity w
• Taking this a step further, we see another similar
relationship between tangential acceleration and
angular acceleration:
at  ra
– In general, a tangential acceleration changes the
linear speed of a point on a rotating object (i.e. the
magnitude of the velocity)
Example Problem #7.12
A 45.0-cm diameter disk rotates with a constant
angular acceleration of 2.50 rad/s2. It starts from rest
at t = 0, and a line drawn from the center of the disk to
a point P on the rim of the disk makes an angle of
57.3° with the positive x–axis at this time. At t = 2.30 s,
find (a) the angular speed of the wheel, (b) the linear
velocity and the tangential acceleration of P, and (c)
the position of P (in degrees, with respect to the
positive x–axis).
Solution (details given in class):
(a) 5.75 rad/s
(b) 1.29 m/s, 0.562 m/s2
(c) 76.2° CCW from the +x–axis
CQ1: Interactive Conceptual Example:
Which animation shows the correct path of
the ball once the string breaks?
A)
B)
C)
D)
E)
Animation A
Animation B
Animation C
Animation D
Animation E
PHYSLET Exercise 7.1.2, Prentice Hall (2001)
Uniform Circular Motion
• Circular motion of a puck on a frictionless horizontal
table with constant speed:
Top view:

v

v
 
a F

a

F 
a
R
F

a

F

v

v tangent to the circle at each pt.

 Changing direction of v

means an acceleration a directed
toward center of circle

v
Uniform Circular Motion

v

v
 
a F

a

F 
a
R
F

a

F

v

v

 Magnitude of a = arad = v2 / R (“centripetal”
acceleration)
 Or, since v = Rw: arad = R2w2 / R = Rw2

 Source of arad: centripetal force F
 From Newton’s 2nd law: F = Fnet = marad = mv2 / R
Circular Motion in a Vertical Plane
• Consider the circular motion of an object swung in a
vertical circular path:
mg
T
T
mg
• At the top: Fc = Fnet = –mg – T
• At the bottom: Fc = Fnet = T – mg
Analysis of Swinging Pail of Water
Top:
vt
+y
Bottom:
vb
Free–body diagrams of water:
Fp
(top)
mg
Fp
(bottom)
mg
Force exerted on water by pail at top:
S Fy = may = m(–vt2 / r)  – Fp – mg = m(–vt2 / r) 
Fp = m (vt2 / r) – mg
Analysis of Swinging Pail of Water
Minimum value of vt for water to remain in pail:
 Minimum force pail can exert on water is zero, so
set Fp = 0 (arad = –g) and solve for minimum speed vt,min:
0  mg = m(–vt,min2 / r)  vt,min2 = rg  vt,min = (rg)1/2
Force exerted on water by pail at bottom:
S Fy = may = m(vb2 / r)  Fp – mg = m(vb2 / r) 
Fp = m(vb2 / r) + mg
*Remember that centripetal force is not a special
external force acting on a body – it is just the name of
the net external force directed toward the center of the
circular motion
Design of a Loop–the–Loop Roller Coaster
Suppose we wish to design the following Loop–the–
1
Loop roller coaster:
0
H – 2R
2
y
y1 = H
R
y2 = 2R
What is the minimum height H necessary for the roller
coaster car to remain on the track at the top of the
loop? (Car falls under the influence of gravity only.)
Conservation of mechanical energy:
½ mv12 + mgy1 = ½ mv22 + mgy2
Assume that roller coaster starts from rest at top of
hill. Then we have: mgH = ½ mv22 + mg(2R)
v22 = 2mg(H – 2R) / m = 2g(H – 2R)
Design of a Loop–the–Loop Roller Coaster
For car to make it safely over the loop:
acar = arad  g (remember water in bucket)
arad = v22 / R  g
2g(H – 2R) / R  g
H  R / 2 + 2R = 5R / 2
H  5R / 2
 So, for example, if R = 22 cm then H needs to be at
least 55 cm.
Example Problem #7.75
In a popular amusement park ride, a rotating
cylinder of radius 3.00 m is set in rotation
at an angular speed of 5.00 rad/s (see
figure at right). The floor then drops
away, leaving the riders suspended
against the wall in a vertical position.
What minimum coefficient of friction
between a rider’s clothing and the wall is
needed to keep the rider from slipping?
Solution (details given in class):
ms = 0.131
Example Problem #7.31
A 40.0-kg child takes a ride on a Ferris wheel that
rotates four times each minute and has a diameter
of 18.0 m. (a) What is the centripetal acceleration
of the child? (b) What force (magnitude and
direction) does the seat exert on the child at the
lowest point of the ride? (c) What force does the
seat exert on the child at the highest point of the
ride? (d) What force does the seat exert on the
child when the child is halfway between the top
and bottom?
(Original figure from University
Solution (details given in class):
Physics, Young and Freedman,
11th Ed., Pearson Addison
Wesley)
(a) 1.58 m/s2
(b) 455 N upward
(c) 329 N upward
(d) 397 N directed inward at 80.8° above the horizontal
CQ2: Interactive Example Problem:
Racecar Safety
What is the maximum linear speed of the car
before it starts skidding off the track?
A)
B)
C)
D)
E)
20 m/s
30 m/s
50 m/s
100 m/s
400 m/s
(ActivPhysics Online Problem #4.5, Pearson/Addison Wesley)
Gravitation
• We are already familiar with concept of gravitational
attraction from concept of weight (gravitational
attractive force exerted by Earth):

g
W = mg
• Newton generalized this force to that acting between
any 2 bodies in his Law of Gravitation (1687):
– In Newton’s words: I thereby compared the force requisite to
keep the Moon in her orb with the force of gravity at the surface
of the Earth, and found them answer pretty nearly.
• Magnitude of gravitational force:
Gm1m2
Fg 
r2
– G = constant = 6.67  10–11 Nm2 / kg2
– Very small value  gravity is a weak (the weakest!) force
– Gravity very important for large masses, however
m1
Gravitation
• Direction of the gravitational force:
Demo
Fg
r
– Along line joining 2 particles (attractive force) m Fg
2
– m1 exerts force of equal magnitude, opposite direction as
m2 exerts on m1 (Newton’s 3rd Law)
• Newton also showed (using integral calculus) that:
m1
Fg
Fg
m2
=
m1
Fg
r
(spherically symmetric masses)
Fg
m2
r
(point masses concentrated at centers)
– True of points outside the matter distribution
• Inside a sphere, it’s a different story – for a solid
sphere of uniform density, Fg is proportional to r
Gravitation
• Graph of |Fg| vs. r for a solid sphere of uniform density:
|Fg|
R
~r
~ 1/r2
r
• When 1 of the masses is the Earth (having mass mE):
Fg = GmEm / r2 where r = dist. from m to Earth’s center
• At Earth’s surface: Fg = GmEm / RE2 (RE = radius of
Earth)
• This is just the weight of mass m at the surface:
W = Fg = GmEm / RE2 (again, measured at Earth’s
surface)
Weight
• But, we also defined weight as W = mg, so evidently:
g = GmE / RE2
= (6.67  10–11 Nm2/kg2)(5.97  1024 kg) / (6.38  106 m)2
= 9.8 m/s2 (in agreement with the value we have
been using thus far)
• We can now see how both the weight and the
gravitational acceleration decrease with distance
from Earth:
• W = GmEm / r2
• a = Fg / m = GmE / r2
• So why does an astronaut in a spacecraft orbiting
Earth experience a feeling of weightlessness?
CQ3: If M is the mass of the earth, m is the mass of the
moon, and d is the distance between their centers,
which of the following gives the instantaneous velocity
of the moon as it orbits the earth? (The universal
gravitational constant is given by G.)
A)
GM
d
B)
GMm
d
C)
Gm
d
D)
GM
md
CQ4: If the radius of the orbit of a satellite
orbiting the earth is reduced by a factor of 2, the
gravitational force on the earth will:
A)
B)
C)
D)
decrease by a factor of 2.
remain the same.
increase by a factor of 2.
increase by a factor of 4.
Example Problem #7.41
A satellite moves in a circular orbit around Earth
at a speed of 5000 m/s. Determine
(a) the satellite’s altitude above the surface of
Earth and
(b) the period of the satellite’s orbit.
Solution (details given in class):
(a) 9.57  106 m
(b) 5.57 h