Transcript Chapter 10
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Chapter 10
Rotation of a Rigid Object about a Fixed Axis
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Rigid Object
A
rigid object
is one that is non-deformable.
The relative locations of all particles making up the object remain constant.
All real objects are deformable to some extent, but the rigid object model is very useful in many situations where the deformation is negligible.
Introduction
Angular Position
P
is located at (
r
, q ) where
r
is the distance from the origin to
P
measured counterclockwise from the reference line.
and q is the As the particle moves, the only coordinate that changes is q, and as the particle moves through q , it moves though an arc length s.
The arc length and r are related: s = q r
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Radian
This can also be expressed as: q =
s r
Comparing degrees and radians
1
rad
= 360 ° 2 p = 57.3
°
Converting from degrees to radians
q = p 180 ° q (
degrees
)
Section 10.1
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Angular Position, final
We can associate the angle q as with an individual particle.
with the entire rigid object as well Remember every particle on the object rotates through the same angle.
The
angular position
of the rigid object is the angle q the reference line on the object and the fixed reference line in space.
between The fixed reference line in space is often the x-axis.
The angle θ plays the same role in rotational motion that the position
x
does in translational motion.
Section 10.1
Angular Displacement
The
angular displacement
is defined as the angle the object rotates through during some time interval.
i
This is the angle that the reference line of length
r
sweeps out.
Section 10.1
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Average/Instantaneous Angular Speed
The
average
angular speed, ω avg , of a rotating rigid object is the ratio of the angular displacement to the time interval.
w
avg
=
t f
-
t i i
= D q D
t
The instantaneous angular speed is defined as the limit of the average speed as the time interval approaches zero.
w º lim
t
0 D q D
t
=
d
q
dt
Section 10.1
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Angular Acceleration
The average angular acceleration, a
avg
, of an object is defined as the ratio of the change in the angular speed to the time it takes for the object to undergo the change.
a
avg
=
t f
-
t i i
= D w D
t
The instantaneous angular acceleration is defined as the limit of the average angular acceleration as the time goes to 0.
a º lim
t
0 D w D
t
=
d
w
dt
Section 10.1
Directions, details
Strictly speaking, the speed and acceleration ( w , a ) are the magnitudes of the velocity and acceleration vectors.
The directions are actually given by the right-hand rule.
Section 10.1
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Rotational Kinematic Equations
The kinematic expression for the rigid object under constant angular acceleration are of the same mathematical form as those for a particle under constant acceleration.
Substitutions from translational to rotational are x → θ v → ω a → α
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Example
A wheel rotates with a constant angular acceleration of 3.50 rad/s 2 . If the angular speed of the wheel is 2.00 rad/s at t i = 0, through what angular displacement does the wheel rotate in 2.00 s?
Through how many revolutions has the wheel turned during this time period?
What is the angular speed of the wheel at t = 2.00s?
Relationship Between Angular & Translational Motion
The linear velocity is always tangent to the circular path.
Called the tangential velocity The magnitude is defined by the tangential speed.
v
=
ds dt
=
r d
q
dt
=
r
w Since r is not the same for all points on the object, the tangential speed of every point is not the same.
The tangential speed increases as one moves outward from the center of rotation.
Section 10.3
Relationship Between Angular & Translational Motion
The tangential acceleration is the derivative of the tangential velocity.
a t
=
dv dt
=
r d
w
dt
=
r
a An object traveling in a circle, even though it moves with a constant speed, will have an acceleration.
Therefore, each point on a rotating rigid object will experience a centripetal acceleration.
a C
=
v
2
r
=
r
w 2
Section 10.3
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Resultant Acceleration
The tangential component of the acceleration is due to changing speed.
The centripetal component of the acceleration is due to changing direction.
Total acceleration can be found from these components:
a
=
a t
2 +
a r
2 =
r
2 a 2 +
r
2 w 4 =
r
a 2 + w 4
Section 10.3
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Rotational Motion Example
Find the angular speed of the disc in revolutions per minute when information is being read from the innermost first track (r = 23 mm) and the outermost final track (r = 58 mm) The maximum playing time of a standard music disc is 74 min and 33 s. How many revolutions does the disc make during that time?
What is the angular acceleration of the compact disc over the 4473 s time interval?
Section 10.3
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Torque
Torque, t , is the tendency of a force to rotate an object about some axis.
Torque is a vector, but we will deal with its magnitude here: t =
r F
sin f =
F d
F is the force f is the angle the force makes with the horizontal
d
is the
moment arm
the force (or lever arm) of There is no unique value of the torque on an object.
Its value depends on the choice of a rotational axis.
Section 10.6
Net Torque
The force
F
1 will tend to cause a counterclockwise rotation about
O.
The force
F
2 will tend to cause a clockwise rotation about
O.
S t = t 1 + t 2 =
F
1
d
1 –
F
2
d
2
Section 10.6
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Example
What is the net torque acting on the cylinder about the rotation axis?
Suppose T 1 T 2 = 5.0 N R = 15 N, and R 2 starting from rest? 1 = 1.0 m, = 0.50 m. What is the net torque about the rotation axis, and which way does the cylinder rotate Figure 10.9
Torque and Angular Acceleration
The tangential force provides a tangential acceleration:
F t
=
ma t
The magnitude of the torque produced
å
F
t
through the center of the circle is
St
=
S
F t
r = (ma
t
) r
The tangential acceleration is related to the angular acceleration.
St
= (ma
t
) r = (mr
a
) r = (mr 2 )
a
Since mr 2 the particle,
St
= I
a **
is the moment of inertia of
The torque is directly proportional to the angular acceleration and the constant of proportionality is the moment of inertia.
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Moment of Inertia
The definition of moment of inertia is
I
=
i
å
i i
The dimensions of moment of inertia are ML 2 and its SI units are kg .
m 2.
We can calculate the moment of inertia of an object more easily by assuming it is divided into many small volume elements, each of mass D m i.
Mass is an inherent property of an object, but the moment of inertia depends on the choice of rotational axis.
Moment of inertia is a measure of the resistance of an object to changes in its rotational motion, similar to mass being a measure of an object ’ s resistance to changes in its translational motion.
The moment of inertia depends on the mass and how the mass is distributed around the rotational axis.
Figure 10.11
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Example
A uniform rod of length L and mass M is attached to one end of a frictionless pivot and is free to rotate about the pivot in the vertical plane. The rod is released from rest in the horizontal position. What are the initial angular acceleration of the rod and the initial translational acceleration of its right end? Figure 10.12
Falling Smokestack Example
When a tall smokestack falls over, it often breaks somewhere along its length before it hits the ground.
Each higher portion of the smokestack has a larger tangential acceleration than the points below it.
The shear force due to the tangential acceleration is greater than the smokestack can withstand.
The smokestack breaks.
Section 10.7
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Torque and Angular Acceleration, Extended
Consider the object consists of an infinite number of mass elements
dm
of infinitesimal size.
Each mass element rotates in a circle about the origin,
O.
Each mass element has a tangential acceleration.
From Newton ’ s Second Law
dF t
= (
dm
)
a t
The torque associated with the force and using the angular acceleration gives
d
t ext =
r dF t
=
a t r dm
= a
r
2
dm
Section 10.7
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Torque and Angular Acceleration, Extended cont.
Finding the net torque
å
t
ext
=
ò
a This becomes S = a
ò
t = Ia This is the same relationship that applied to a particle.
This is the mathematic representation of the analysis model of a
rigid body under a net torque.
The result also applies when the forces have radial components.
The line of action of the radial component must pass through the axis of rotation.
These components will produce zero torque about the axis.
Example
A wheel of radius, R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axis as shown to the right. A light cord wrapped around the wheel supports an object of mass m. When the wheel is released, the object accelerates downward, the cord unwraps off the wheel, and the wheel rotates with an angular acceleration. Find the expressions for the angular acceleration of the wheel, the translational acceleration of the object, and the tension in the cord.
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Moment of Inertia, cont
The moment of inertia of a system of discrete particles can be calculated by applying the definition for I.
For a continuous rigid object, imagine the object to be divided into many small elements, each having a mass of
Δm i.
We can rewrite the expression for
I
in terms of D
m.
I
= lim
m i
0 º
i r i
2 D
m i
= º With the small volume segment assumption,
I
= ò r If r is constant, the integral can be evaluated with known geometry, otherwise its variation with position must be known.
Section 10.5
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Notes on Various Densities
Volumetric Mass Density → mass per unit volume: r =
m
/
V
Surface Mass Density → mass per unit thickness of a sheet of uniform thickness,
t
: s = r
t
Linear Mass Density → mass per unit length of a rod of uniform cross-sectional area: l =
m
/
L
= r A
Section 10.5
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Moment of Inertia of a Uniform Rigid Rod
The shaded area has a mass
dm
= l
dx
Then the moment of inertia is
I y
= ò
I
= 1 12
ML
2 = ò -
L L
/ 2 / 2
x
2
M L dx
Section 10.5
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Moment of Inertia of a Uniform Solid Cylinder
A uniform solid cylinder has a radius R, mass M, and length L. Calculate its moment of inertia about its central axis.
Section 10.5
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Parallel-Axis Theorem
In the previous examples, the axis of rotation coincided with the axis of symmetry of the object.
For an arbitrary axis, the parallel-axis theorem often simplifies calculations.
The theorem states
I
=
I
CM +
MD
2
I
is about any axis parallel to the axis through the center of mass of the object.
I
CM
D
is about the axis through the center of mass.
is the distance from the center of mass axis to the arbitrary axis.
Figure 10.17
Section 10.5
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Rotational Kinetic Energy
An object rotating about some axis with an angular speed,
ω
, has rotational kinetic energy even though it may not have any translational kinetic energy.
Each particle has a kinetic energy of
K i
= ½
m i v i
2 Since the tangential velocity depends on the distance,
r
, from the axis of rotation, we can substitute
v i
= w
i r.
The total rotational kinetic energy of the rigid object is the sum of the energies of all its particles.
K R
=
i
å
K i
=
i
å 1
m r i i
2 2 w 2
K R
= 1 2
i
å
m r i i
2 w 2 = 1 2
I
w 2
There is an analogy between the kinetic energies associated with linear motion (K = ½ mv 2 ) and the kinetic energy associated with rotational motion (K
R
= ½ I
w
2 ).
Work and Power in Rotational Motion
Find the work done by
F
on the object as it rotates through an infinitesimal distance
ds
=
r d
q
dW
= (
F
=
F
sin f i
d
s
)
r d
q The radial component of the force does no work because it is perpendicular to the displacement.
The rate at which work is being done in a time interval
dt
is Power =
P
=
dW dt
= t
d
q
dt
= tw
Section 10.8
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Work-Kinetic Energy Theorem in Rotational Motion
The work-kinetic energy theorem for rotational motion states that
the net work done by external forces in rotating a symmetrical rigid object about a fixed axis equals the change in the object
’
s rotational kinetic energy.
The rotational form can be combined with the linear form which indicates
the net work done by external forces on an object is the change in its total kinetic energy, which is the sum of the translational and rotational kinetic energies.
Section 10.8
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Summary of Useful Equations
Section 10.8
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Example
A uniform rod of length L and mass M is free to rotate on a frictionless pin passing through one end. The rod is released from rest in the horizontal position. What is its angular speed when the rod reaches its lowest position? Determine the tangential speed of the center of mass and the tangential speed of the lowest point on the rod when it is in the vertical position. Figure 10.21
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Energy in an Atwood Machine, Example
Two blocks having different masses m1 and m2 are connected by a string pulley system. The pulley has a radius R and moment of inertia I about its axis of rotation. The string does not slip on the pulley, and the system is released from rest. Find the translational speeds of the blocks after block 2 descends through a distance h and find the angular speed of the pulley at this time.
Section 10.8
Rolling Object
The red curve shows the path moved by a point on the rim of the object.
This path is called a
cycloid.
The green line shows the path of the center of mass of the object.
In pure rolling motion, an object rolls without slipping.
In such a case, there is a simple relationship between its rotational and translational motions.
Section 10.9
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Pure Rolling Motion, Object
’
s Center of Mass
The translational speed of the center of mass is
v
CM =
ds dt
=
R d
q
dt
=
R
w The linear acceleration of the center of mass is
a
CM =
dv CM dt
=
R d
w
dt
=
R
a
Section 10.9
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Rolling Motion Cont.
Rolling motion can be modeled as a combination of pure translational motion and pure rotational motion.
The contact point between the surface and the cylinder has a translational speed of zero (c).
Section 10.9
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Total Kinetic Energy of a Rolling Object
The total kinetic energy of a rolling object is the sum of the translational energy of its center of mass and the rotational kinetic energy about its center of mass.
K
= ½
I
CM w 2 + ½
Mv
CM 2 The ½
I CM
w
2
represents the rotational kinetic energy of the cylinder about its center of mass.
The ½
Mv 2
represents the translational kinetic energy of the cylinder about its center of mass.
Section 10.9
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Total Kinetic Energy, Example
Accelerated rolling motion is possible only if friction is present between the sphere and the incline.
The friction produces the net torque required for rotation.
No loss of mechanical energy occurs because the contact point is at rest relative to the surface at any instant.
In reality, rolling friction causes mechanical energy to transform to internal energy.
Rolling friction is due to deformations of the surface and the rolling object.
Section 10.9
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Total Kinetic Energy, Example cont.
Apply Conservation of Mechanical Energy: Let
U
= 0 at the bottom of the plane
K f K f
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U f
=
K i
= ½ (
I
CM +
U i
/
R
2 )
v
CM 2 + ½
Mv
CM 2 = 1 2
U i
=
Mgh U f
=
K i
= 0
I CM R
2 +
M v
2
CM
Solving for v
v cm
= ê ê é ê 1 æ è 2
gh I CM MR
2 ø ù ú ú ú 1 2
Section 10.9
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Sphere Rolling Down an Incline, Example
For the solid sphere shown to the right, calculate the translational speed of the center of mass at the bottom of the incline and the magnitude of the translational acceleration of the center of mass.
Section 10.9