Rotational Motion and the Law of...
Transcript of Rotational Motion and the Law of...
Chapter 7
Rotational Motion and the Law of Gravity
• What is a Rigid Body?
• Rotational Kinematics
• Angular Velocity ω and Acceleration α
• Uniform Rotational Motion: Kinematics
• Uniform Circular Motion: Kinematics and Dynamics
• Applications
• Newton’s Law of Universal Gravitational Attraction
• Planetary motion
Rigid Body and its Motions
• Until now we considered only the motion of point-like objects. Objects with
extended size can be considered as a collection of many point-like particles.
• When these particles do not move with respect to each other, the system is called a
rigid body: it cannot be deformed.
• The general motion of a rigid body can be split into two types:
Translational (Linear) Motion
Rotational (Angular) Motion – we need an “angular” formalism
Object and terminology:
, , , , , ...r r v a F p
, , , , , ...L
Linear velocity
Translation velocity
Center or axis of rotation
Rigid body A particle of the body
• In purely rotational motion, all points on the object move
in circles around the axis of rotation (“O”), with each point
described by a vector position r
Convention:
• angles measured counterclockwise are positive
• angle measured clockwise are negative
x
y
O
θ′
θ r′ r
• In our approach to rotations, angles will be measured in
radians: 1 radian is the angle at the center of a circle subtending
an arc equal to the radius of the circle
• When the angle at the center is expressed in radians, the
length of the arc subtended is given by:
θ
θ = 1 rad
r
r
l = r
θ
r
r
l = rθ
l r
+
‒
Def: The angle θ made by the position vector r with respect
to an arbitrary axis (say x) is called angular position
Ex: see the two points on the adjacent bicycle wheel: notice that,
as long they are not on the same radius, the points on the rigid
body will have different angular positions
Angular Kinematics – Angular Position θ
Ex: The circumference 2πr of a complete
rotation subtends an angle of 2π
How can we use angular positions to describe rotations?
• Notice that even though the angular positions of
different points of a wheel are in general different, when
the wheel rotates, all points rotate through the same angle
2 1
0lim
t t
Def: The change in angular position of all the points on
a rotating rigid body is called angular displacement:
• Then, if we want to refer to how fast the angular position changes we have to define
first the average angular velocity as the angular displacement divided by time:
• Therefore, like in the linear case, the instantaneous angular velocity is given by:
x
θ2
θ1
Δθ
Arbitrary
radius
Ex: If the bicycle wheel makes two complete rotations
every second, we say that it has a constant angular
speed of 2×(2π/1 s) = 4π rad/s
Angular Kinematics – Angular Displacement and Velocity
2 1
2 1t t t
SI
rad s
• Hence we can define the average angular acceleration as the rate at which the
angular velocity changes with time:
2 1
2 1t t t
0lim
t t
2
SIrad s
Angular Kinematics – Angular Acceleration. Angular vector directions
The direction of
angular velocity
is given by a
right hand rule
The direction of
angular acceleration
is parallel or anti-
parallel with the
angular velocity
depending on
whether ω increases
or decreases 0 0
• Hence, the instantaneous angular acceleration:
• Although it is not as intuitive as in the translational case, the angular velocity and
acceleration are vectors, perpendicular on the circle of rotation:
Ex: If the bicycle wheel spins each
second through an angle larger and
larger by π, we say that each complete
rotation its average angular velocity is
π, so its instantaneous angular velocity
increases by 2π rad/s, so it has a
constant angular acceleration 2π rad/s2
Relating Linear and Angular Kinematics
• So, the angular displacement, velocity and acceleration characterize the entire rigid
body: all points have the same Δθ, ω and α
• However, the linear distance, velocity and acceleration of points at various radii r
from the axis of rotation are different: each point has a different Δl, v and a
lr
tv r
t
• So, as long as they are not at the same distance r from
the center of rotation, the different points on a rigid body
have different linear speeds, increasing from zero in the
center of rotation to a maximum value on the outer rim
of the rotating rigid body
x Δθ
Δl = rΔθ
ω
r r
• The linear kinematics of each point on a rigid body can
be related to the overall angular characteristics based on
the relationship l = rθ
• For instance, consider a wheel rotating with constant
angular speed ω. A point at distance r from the center of
rotation will rotate with constant linear speed v traveling
an arc Δl in a time Δt: therefore, we obtain
v v
ω
r1 v1 = ωr1
v2 = ωr2
v3 = ωr3
r2
r3
How about Acceleration?
• The acceleration is a bit more complex since in general the vector linear
acceleration of a particle in circular motion is not tangent to the trajectory.
• However, it can be considered as having two components – one tangent to the
trajectory (parallel or anti-parallel with the velocity) and one perpendicular on the
velocity:
at: tangent, describes how the magnitude of the linear velocity varies
ar: radial (or centripetal), describes how the direction of the velocity varies
at • The component at of a particle at distance r from the
axis of rotation can be easily related to the angular
acceleration α of the rigid body:
• The component ar (also called centripetal since it
always point toward the center of rotation) makes
necessary a separate discussion – a few slides further…
t
vr
ta r
t
x
ω
r
a
ar
Exercise : Components of acceleration
A particle moves as shown in the figure. Between points B and D, the path is a straight line.
Let’s figure out the net acceleration vectors in points A, C and E along the path represented
below, for each of the following cases:
a) the particle moves with steadily constant speed
b) the particle moves with steadily increasing speed
c) the particle moves with steadily decreasing speed
x
ω0
r
v0
x Δθ
Δl
r v
t0 = 0 At time t ω
• The linear-angular relationship provides an easy way to describe circular motion
with constant angular acceleration α, by simply noticing that each point on the
rotating rigid body accelerates uniformly along the respective arc of circle
• Assume that the motion starts at t0 = 0 when the rotation is characterized by θ0, ω0
210 0 2
0
2 2
0
10 02
2
t tt
t t
t
210 2
0
2 2
0
102
2
t
t
t
l v t a t
v v a t
v v a l
l v v t
The linear motion
of one particle of
the rigid body at
distance r from
the center of
rotation
if each linear
quantity is
divided by r,
we obtain…
The rotational
motion of the
entire rigid body
(valid for any of
its parts)
• Then, if the angular acceleration α is constant, at a later instant t,
Angular Kinematics – Uniformly accelerated rotation
Rolling Motion (Without Slipping)
Observation useful in problems:
• In figure (a), if the wheel of radius r is rolling
without slipping, the point P on the rim is at rest with
respect to the floor when it touches it, while the center
C moves with velocity v to the right
• In figure (b), the same wheel is seen from a reference
frame where C is at rest. Now point P is moving with
velocity –v. Since P is a point at distance r from the
center of rotation, we have:
• Therefore, when a wheel, or a sphere, or a cylinder
rolls, its translational speed (that is, the speed vcm of
its center of mass) is related to its angular speed ω by
• Caution: even though this has the same form as the
relation between linear speed of a point and the
angular speed, it is not the same thing.
cmv r
v
v
a) Wheel seen by someone on
the ground:
C
P
C
P
b) Wheel seen by someone on
the bike:
v r
ω
ω
r
srad 83srad s 60
22500rpm 2500
srad 150srad s 60
24500rpm 4500
1
0
Problems:
1. Uniformly accelerated rotation: An automobile engine slows down from 4500 rpm to
2500 rpm in 2.5 s. Calculate
a) its angular acceleration (assumed constant)
b) the total number of revolutions the engine makes in this time.
2. Rotational and linear motion: You are to design a rotating cylindrical axle to lift buckets
of cement from the ground to a rooftop. The buckets will be attached to a hook on the free end
of a cable that wraps around the rim of the axle; as the axle turns, the buckets will rise.
a) What should the diameter of the axle be in order to raise the buckets at a steady 2.00 cm/s
when it is turning at 7.5 rpm?
b) If instead the axle must give the buckets an upward acceleration of 0.400 m/s2, what should
be the angular acceleration of the axle be?
7.5 27.5 rpm rad s rad s
60 s 4
• This expression is valid for any particle
moving along a curved trajectory: if the
curvature of the path can be fitted locally by a
circle of radius r and the instantaneous speed
is v, the expression above gives the radial
component of the acceleration in the
respective point
• So, the radial (or centripetal) acceleration describes how the direction of the velocity
changes: that is, if the object moves in a straight line, ar = 0
• To see how ar is related to the speed v, we’ll consider a particle moving in a circle of
radius r with α = 0, that is, a constant speed (caution! the velocity is not constant)
2
r
va
r
Centripetal Acceleration
v
1v
Two similar triangles form, such that: 2
r
v v v v l vv l
l r r
va
rt t r
2v
~Δl
Δθ
1v
r
r
v
v • Assume that the particle travels an arc-distance Δl
subtending an angular displacement Δθ in a time Δt
• Then, looking at the corresponding change in
velocity of the revolving particle and using the
definition of acceleration, we find that the centripetal
acceleration is given by
path
path
1v
2v
ra ra
r1
r2
2
1
1
r
va
r
2
2
2
r
va
r
• Summarizing list of correspondences between linear and rotational quantities:
Linear Type Rotational Relation
x displacement θ x = rθ
v velocity ω v = rω
at acceleration α at = rα
ar acceleration - ar = rω2
1f T
2
rv r
T
• We can introduce a new set of parameters describing periodic motion:
1. Frequency f : the number of revolutions per time
2. Period T: time required to complete a revolution
• Frequency and period are the inverse of each other:
• Using the idea of period we can see once more how the relationship between the
linear velocity and the angular velocity makes sense and another relationship for ar:
22
r
va r
r
SI1 s Hertz, Hzf
SIsT
Some useful quantities and a summary…
Uniform Circular Motion – Kinematics
• The uniform circular motion is the motion of a particle in a circle of constant radius
at constant speed, such that α is zero
• Being always tangent to the circular path
the vector instantaneous velocity changes
direction, albeit its magnitude stays constant
• Hence, the tangential acceleration at is
zero, and the net acceleration is given only
by the centripetal acceleration ar pointing
everywhere perpendicular on the velocity
vr
path
Physical situation: a particle moving in a
circle:
ra
Comments:
• The magnitude of the centripetal acceleration is large if the speed is large
• The magnitude of the centripetal acceleration is large if the radius of rotation is small
Constant speed
Ex: if a car takes a turn at high speed, it will have a larger centripetal acceleration than
when taking it slowly
Ex: if a car takes a turn sharp turn (small radius), it will have a larger centripetal
acceleration than when taking a wide turn
2
r
va
r
• By Newton’s 2nd Law, since in the circular motion the acceleration is necessarily
not zero (since the velocity must change in direction), we see that 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:
• This “centripetal force” is not a new force: any net force
pointing radially inward the circular trajectory (perpendicular
on the velocity) can play the role of centripetal force, since it
has as a result a change in direction of velocity
2
rrr
maF mv
v
Comments:
• A common misconception is to assume that an object on a
curved trajectory is thrown out of it by an outward “centrifugal”
force. We now see that the force must be actually inward
• The objects taking turns are apparently pushed outward by their
inertia, while the centripetal force keeps it on the trajectory
• If the centripetal force vanishes, the object flies off tangent to
the circle, not outward as if a centrifugal force were present
Uniform Circular Motion – Dynamics
vr
path rF
Exercise 1: How Angelina Succumbed to Bad Physics…
In the movie Wanted, bullets are “curved” by skilled tattooed assassins. For instance, Angelina
kills herself by firing a bullet in a circle passing through the skulls of some bald dudes before
hitting her. Say that the bullet has a mass m = 4.2 g and a muzzle speed of 600 m/s. Also, say
that the dudes arranged themselves conveniently in a circle with radius r = 5.0 m.
a) How big should be the centripetal force keeping the bullet on the circular trajectory?
Meditate about the possible origin of such a force and how realistic is such a scenario…
b) How fast should Angelina toss the gun to the sweaty guy in the middle for the scene to make
sense?
Exercise 2: Ball revolving as held by a string
A ball of mass m is connected by a string of length r
and moved into a circular trajectory with constant
speed v. Let’s estimate the force a person must exert on
the string to make the ball revolve in a horizontal
circle.
a) What is the nature of the centripetal force exerted on the
ball?
T
v
r
b) Based on Newton’s 2nd Law, what is this force in terms of given quantities?
Problem
3. Tension as a centripetal force: Now let’s assume that the ball from the exercise above is
swung in a vertical circle, still with a constant speed.
a) Determine the tension in an arbitrary point of the circle where the radius makes an angle θ
with respect to the horizontal
b) Use the result from part (a) to determine the tension in the string when the string is
horizontal, on top of the circle, and at the bottom of the circle.
Problems:
4. Normal as a centripetal force: A small remote-control car
with mass m = 1.60 kg moves at a constant speed of v = 12.0 m/s
in a vertical circle inside a hollow metal cylinder that has a
radius of r = 5.00 m. What is the magnitude of the normal force
exerted on the car by the walls of the cylinder at
a) an arbitrary point on the loop
b) point A (bottom of the vertical circle)
c) point B (top of the vertical circle)
5. Conical pendulum: A bob of mass m is suspended from a
fixed point with a massless string of length L (i.e., it is a
pendulum). What tangential speed v must the bob have so that it
moves in a horizontal circle with the string always making an
angle θ from the vertical?
ta r
Nonuniform Circular Motion – Elementary observations
r
path
Physical situation: a particle moving in a
circle:
ra
Changing speed
• If the velocity of a revolving particle changes in magnitude as well, the objects is
said to be in a nonuniform circular motion
ta
a
2 2 4 2 2 2
r ta a a r r
v
• Accordingly, the net force acting on the object will contribute to the change in
velocity direction with a centripetal component, and to the change in speed with a
tangential component
• As any revolving object, the object has
a centripetal acceleration:
• However, in this case, the particle also
has a tangential acceleration:
22
r
va r
r
• Hence, at a certain moment when the instantaneous angular speed is ω, the
instantaneous net acceleration of the particle has a magnitude given by:
Newton’s Law of Universal Gravitation – The idea
• We learned that Earth exerts a force on any mass – the weight. We studied how the
weight contributes to the motion of the object, and we quantified its strength through
its effect when it acts alone: the gravitational acceleration g
• However, what is this force? For instance, is it acted only by Earth?
• The formulation of a coherent theory of gravity (consistent
with his mechanics) is another example of Newton’s many
fundamental contributions to Physics
• He realized that gravity acts between any masses: it is just
logical to assume that the downward force that pulls onto an
apple or a person must be have the same nature as the attraction
exerted by Earth on the Moon in order to keep it on its orbit
• Moreover, by studying the motion of the planets, he deduced
that the magnitude of the gravitational attraction must vary
inversely proportional to the square of the distance between the
interacting masses
Quiz: Based on what law can we infer that the gravitational force is not acted only by Earth?
weight
Every two particles in the universe attract each other with gravitational forces
directly proportional to the product of their masses m and M, and inversely
proportional to the square of the distance r between them.
2gF GmM
r
r m M
Fg
Comments:
• The force of attraction is along the line
connecting the centers of gravity of the objects
(which in the cases we are interested in coincides
with their centers of mass or symmetry)
• Hence, the distance r between objects is the
distance between their centers of gravity
• Note the presence of Newton’s 3rd law in this
law: mass m attracts mass M and vice-versa with
action-reaction forces
• The gravitational constant G is very small,
meaning that the gravitational attraction will be
hardly observable between small masses
Newton’s Universal Law of Gravitation – Quantitatively
Gravitational constant 11 2 26.673 10 N m kgG
Fg
Ex: Two particles:
Two spheres:
r
m M
Two square plates:
r
m M
• The first determination of the value of the gravitational constant G is sometimes
attributed (wrongly) to Sir Henry Cavendish
• Cavendish actually tried to find experimentally the density of Earth. His data was
only much later used to calculate G
• Cavendish’s
apparatus was based on
the observation of the
gravitational attraction
between relatively
small objects using a
torsion balance
• If the force of
attraction is known, G
can be calculated easily
Modern version of Cavendish’s experimental arrangement:
Newton’s Universal Law of Gravitation – Determining G
m
m m
a
3
1
2 a
Problems:
6. Principle of superposition: Three identical particles of mass m are positioned at three
corners of a isosceles right triangle of equal sides a, as in the figure. Calculate the total
gravitational attraction on mass 1.
a√2
Gravitational Acceleration – In a more generic context
rE
mE
h
r
m h
mg G
2 22 2
1
1
E E Eh
EE E
m m mg G G
r rr h h r
• Until now, we assumed the local gravitational acceleration
constant. However, wee see that this is just an approximation
sing the weight does depend on the distance r to the center of
attraction (Earth):
• On the surface of the Earth h → 0 so
• Therefore, the gravitational acceleration at altitude h above
the surface of the Earth can be written:
2
2 2
9.8 m s
1 1h
E E
gg
h r h r
2
0 2 2~ 9.8 m sE E
h
E E
mm mmg G g G
r r
• So, 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.
m
mg
mgh
• The motion of a satellite can be associated with a
projectile trajectory missing the surface of the Earth
due to the high speed.
• If the speed is too small, the satellite falls on the
surface (paths 1-2)
• In order to put a satellite on the orbit (paths 3-5), it
must have a minimum speed called satellite speed,
which can be estimated assuming a circular orbital
trajectory of radius r. Then
21initial 22
0 0E
Er r r
E
mmKE PE PE mv G
r 2 2 E
E
mv G
r
2
E 1
2r r
mm vmg ma G m
rr
Applications of the Law of Gravity – Satellites
• However, if the initial speed is too great, the satellite flies out into space and
becomes a spacecraft as it flies along open (paths 6-7)
• The minimum speed necessary for the object to escape the Earth’s gravity is called
escape speed and can be estimated by noticing that very far away from earth the
gravitational potential energy vanishes. Therefore
E1
mv G
r
mE
rE
Launching tower
For any two planets,
Applications of the Law of Gravity – Kepler’s Laws
1. The path of each planet about the Sun is an ellipse,
with the Sun at one focus.
Comment: This behavior is due to the specific 1/r2
dependence of the force on the distance.
3 2
1 1
2 2
T a
T a
2. The position of each planet about the Sun sweeps
out equal areas in equal times.
Comment: This behavior is explained by the
conservation of angular momentum (we’ll learn
about it in the next chapter)
3. The periods of the planets are proportional to 3/2
powers of the semi-major axis lengths of their orbits.
Comment: This is explained by Newton’s 2nd Law
applied to the orbiting body.