Transcript Rotational Motion

Warm-up: Centripetal Acceleration
Practice
 Assume a satellite in low Earth orbit has an acceleration,
caused by gravity, of 9.81 m/s2. If the radius of the orbit is
approximately the radius of the Earth, 6370 km, find the
velocity of the satellite and the time for one orbit.
Review: Centripetal acceleration
 A particle undergoing circular
motion is accelerating, even if its
speed is not changing.
 The acceleration of an object
moving in a circle is toward the
center of the circle.
 It is thus referred to as centripetal
acceleration.
 The magnitude of the centripetal
acceleration depends on the
velocity of the particle tangent to
the circle.
vt2
aC 
r
©2008 by W.H. Freeman and Company
Rotational Motion
AP Physics C
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Rotational Kinematics
 We will begin our study of
rotational motion by
considering the motion of a
rigid body about a fixed
axis, or an axis that is
moving parallel to itself
like a rolling ball.
Angular Position
 Because the object is rigid,
and the axis is fixed, all the
points in the object maintain
the same relative position.
 When the object rotates,
every point rotates through
the same angle.
 The angle θ, measured
relative to some starting
position, gives the angular
position of every point in the
object.
Angular Displacement
 When the object rotates on its
axis, each point undergoes the
same angular displacement
Δθ.
 If we measure the angular
displacement in radians, we
can easily find the distance
traveled by each point, since by
definition of radian measure
Length of an arc  r
where r is the radial distance
from the center.
Angular Displacement Example
 Two people ride on a carousel.
One rides on a horse located 5
meters from the center. The
other rides on a swan located 3
meters from the center.
 When the carousel goes
around ¼ of a revolution, how
far does each person travel?
Length of an arc  r
Angular velocity
 We define the angular velocity
as the rate of change of angular
position.
 Since radians are
dimensionless, the dimensions
of the angular velocity are T-1.
 The magnitude of the angular
velocity is the angular speed.
 Counterclockwise rotation
corresponds to a positive
velocity.
d

dt
Angular velocity practice
 What is the angular speed, in radians/second, of a motor that
spinning with 6000 rpm?
Angular acceleration
 The angular acceleration is the
rate of change of angular
velocity.
 Since radians are dimensionless,
the dimensions of angular
acceleration velocity are T-2.
 This acceleration refers to the
increase or decrease in
rotational speed of the particle.
d d 2

 2
dt
dt
Angular and Tangential Quantities
 Each angular quantity has a
corresponding tangential
quantity.
 The arc length s is the distance
travelled along the circle, in m .
 The tangential velocity is the
speed of the particle in the
direction tangent to the circle, in
m/s.
 The tangential acceleration is the
acceleration of the particle
tangent to the circle, in m/s2.
s  arc length  r
ds dr
d
vt 

r
 r
dt
dt
dt
dvt dr
d
at 

r
 r
dt
dt
dt
Angular equations of motion for
constant acceleration
 Begin with the definition of angular acceleration.
d

dt
 If the acceleration is constant, what is the angular velocity?
  t  C
  0  t
 Then what is the displacement?
   0  0 t  t
1
2
2
Rotational motion of a rigid body is
analogous to linear motion.
Straight line motion
Rotation about a fixed axis
 Linear position x.
 Angular position .
 Linear displacement x
 Angular displacement 
 Linear velocity v
 Angular velocity ω
 Linear acceleration a
 Angular acceleration α
Equations of motion for constant
acceleration
Straight line motion
x  x0  v0t  at
1
2
Rotation about a fixed axis
2
v  v0  at
v  v  2a( x  x0 )
2
2
0
   0  0t  t
   0  t
1
2
2
    2 (   0 )
2
2
0
Problem solving with rotational motion.
 Convert tangential velocities, speeds etc. to their angular
counterparts by dividing by the radius.
 Solve for the angular motion by using the angular equations
of motion just as we did with linear motion.
 Convert quantities to their tangential counterparts if needed.
Centripetal acceleration in angular
form.
 We can write the equation for centripetal acceleration in
terms of the angular velocity.
vt2 (r ) 2 r 2 2
aC 


r
r
r
aC  r
2
Angular kinematics examples
 Toast falling off a table usually starts to fall when it makes an
angle of 30 degrees with the horizontal, and falls with an
 0the
.956
g /l
initial angular speed of
wherel is
length
of one side. On what side of the bread will the toast land if it
falls from a table 0.5 m high? If it falls from a table 1.0 m
high? Assume l = 0.10 m, and ignore air resistance.
 When a turntable rotating at 33 rev/min is turned off, it
comes to rest in 26 s. Assuming constant angular
acceleration, find the angular acceleration and the angular
displacement. If the turntable is 0.20 m in radius, how far
would an ant riding on the outside edge have moved in that
time?
Rotational Dynamics: Equilibrium
Rotational Equilibrium
 What causes angular acceleration?
 F1 and F2 are forces equal in size but acting in opposite
directions.
 Will both situations result in angular acceleration?
Torque
 A force F acts on mass m
located with a rigid body.
 Only the tangential
component affects the
rotation of the disk.
 The farther away from the
axis of rotation the force is,
the greater the effect on
the rotation.
Torque
 The torque that acts on
mass m in the rigid body is
the tangential force FT
times the distance from the
axis of rotation r.
 Τ=FT r = (F sin φ) r
Torque
 Another way of looking at the
same thing is to say the torque
is caused by the entire force,
but the distance is only the
perpendicular distance of the
force to the line of action of the
force.
 This distance, which is
r sin
φ, is known as the lever arm l.
 So our formula becomes
 Τ=F l = F (r sin φ) = Fr sin φ
Conditions of Equilibrium
 We learned last year that there are two conditions for
equilibrium.
 The net external force acting on the body must be zero.
 The net external torque about any point must be zero.
Torque and Rotational Equilibrium
Examples
 A 700 N window washer is standing on a uniform scaffold
supported by a vertical rope at each end. The scaffold
weights 200.0 N and is 3.00 m long. What is the force in
each rope when the window washer stands 1.00 m from one
end?
 Does the Sun exert a torque on the Earth relative to the
Earth’s rotational axis? Discuss.
Rotational equilibrium example
 A 700 N window washer is standing on a uniform scaffold
supported by a vertical rope at each end. The scaffold
weights 200.0 N and is 3.00 m long. What is the force in
each rope when the window washer stands 1.00 m from one
end?
 Answers: 333 N and 567 N
Rotational Dynamics:
Newton’s 2nd Law for Rotation
Newton’s Second Law for Rotational
Motion
 Consider a mass m in a rigid
body that is free to rotate.
 The mass is acted on by a
force F.
 What does the radial
component do?
 What does the tangential
force do?
 The mass will undergo a
tangential acceleration at.
 This acceleration is in
addition to the constant
centripetal acceleration.
Newton’s Second Law for Rotational
Motion
 By Newton’s 2nd Law,
Ft  mat
 The tangential acceleration
is related to the angular
acceleration.
a

r

t
 Thus,
Ft  mr
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Newton’s Second Law for Rotational
Motion
 Multiplying both sides by
r:
Ft r  mr 
2
 The left-hand side of this
equation is the torque.
  (mr )
2
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Newton’s Second Law for Rotational
Motion
 For a single particle of
mass m
  (mr )
2
 The quantity in
parentheses is called the
rotational inertia.
 It tells how much torque
must be applied to produce a
given angular acceleration.
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Newton’s Second Law for Rotational
Motion
 We can add up the torques on every mass mi in the object to
derive an equation for the entire rigid object.
   mr 
2
i
i
 Since this is a rigid object, α is the same for every particle.
Thus we can remove it from the sum.
   m r 
2
i
i i
Newton’s Second Law for Rotational
Motion

   m r 
2
i
i i
 The internal torques add up to zero, so the left hand side of
the above is just the net external torque.
 The quantity in brackets above depends on the mass and
radius of each particle, both of which are fixed for a rigid
object and choice of axis. So, that sum is also fixed for a
given object and rotational axis. It is called the rotational
inertia.
Newton’s Second Law for Rotational
Motion

 NET , EXT  I
 Where I = rotational inertia
I   mi ri for a system of particles.
2
i
I   r dm for a continuous object.
2
 Rotational inertia is also called the moment of inertia.
Example: Rotational Inertia for a
System of Particles
 Calculate the rotational inertia of the following system of
particles about the given axis.
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Example: Rotational Inertia for a
System of Particles
 Calculate the rotational inertia of the following system of
particles about the given axis.
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Example: Rotational Inertia of a
uniform thin rod.
 Calculate the moment of inertia of this rod about an axis at
one end.
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©2008 by W.H. Freeman and Company
Parallel Axis Theorem
 The parallel-axis theorem states
I  I CM  Mh
2
 Where ICM is the rotational inertia about an axis containing the
Center of Mass
 I is the rotational inertia about an axis parallel to ICM
 M = total mass and h = distance between axes.
Parallel Axis Theorem Example
 Use the parallel-axis theorem to find the rotational inertia
about the center of mass of an uniform thin rod.
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Newton’s 2nd Law for Rotation
Example 1
 A bike is placed on a stand so it
can be used as a stationary bike.
The bike is pedaled so that a
force of 18 N is applied to the
rear sprocket at a distance of rs
= 7.0 cm from the axis. Treat
the wheel as a hoop (I = MR2)
of radius R=35 cm and mass M
= 2.4 kg. What is the angular
velocity of the wheel after 5.0 s?
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Newton’s Second Law for Rotation
Example 2
 A 2.5 kg, 11 cm radius cylinder, initially at rest, is free to
rotate about the axis of the cylinder. A rope of negligible
mass is wrapped around it and pulled with a force of 17 N.
Assume the rope does not slip, find the torque exerted on
the cylinder by the rope, the angular acceleration of the
cylinder, and the angular speed of the cylinder after 0.50 s.
 Answers: 1.9 Nm, 1.2 x 102 rad/s2, 6.2 x 102 rad/s
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Nonslip Conditions
 Many problems involve a
taut string wrapped around
a pulley. Assuming the
string does not slip, the
string will have the same
velocity as the tangential
velocity of the pulley over
which it passes.
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Newton’s 2nd Law for Rotation:
Example 3
 An object of mass m hangs on a
light string that is wound around
the rim of a pulley of rotation
inertia I and radius R. The wheel
bearing is frictionless and the
string does not slip. Find the
tension in the string and the
acceleration of the object.
©2008 by W.H. Freeman and Company
Rotational Dynamics:
Kinetic Energy
Rotational Kinetic Energy
 We can apply the formula for kinetic energy to rotation as
well. In a rotating rigid body, each particle moves with a
velocity vt=rω.

K ROT   12 mi vti2   12 mi (ri)2

K ROT  12  mi ri2 2

K ROT  12 I 2
Rotational Kinetic Energy Example
 Calculate the rotational kinetic energy of the Earth, and
compare that to the kinetic energy of the Earth’s motion
around the Sun. Assume the Earth is a sphere of radius 6.4 x
106 m and mass 6.0 x 1024 kg. The radius of the Earth’s orbit
is 1.5 x 1011 m.
Rolling Objects
Rolling Objects
 Consider an object with a
circular cross section that
rolls without slipping across a
flat surface.
 The point of contact between
the wheel and the ground
moves a distance s = Rφ.
 Since the center of mass
remains above the point of
contact, it also moves s=Rφ.
Rolling Objects
sCM  R
vCM
dsCM
d

R
 R
dt
dt
aCM
d
R
 R
dt
Rolling Objects
 For an object that rolls
without slipping, the motion
of the center of mass is
described by the same
equations as an object
rotating about a fixed axis.
 Newton’s Second Law for
rotation also holds, if the
torques are calculated about
an axis containing the center
of mass.
 τNET, CM = ICMα
Rolling Objects
 The total kinetic energy of a
moving object is the kinetic
energy of the center of mass
plus the kinetic energy of the
particles relative to the CM.
 Here, that means the rotation
kinetic energy plus the kinetic
energy of the CM motion.
K  I CM   mv
1
2
2
1
2
2
CM
A bowling ball of radius 11 cm rolls without slipping up a ramp. If the initial
speed of the ball is 2 m/s find the height h that the ball reaches when it
momentarily comes to rest.
A cue stick hits a cue ball horizontally a distance d above the center of the
ball. Find the value of d for which the cue ball will roll without slipping from
the beginning. Express your answer in terms of the radius R of the ball.