Download Rotational Dynamics

Rotational
Dynamics
Physics
Montwood High School
R. Casao
• The rotational equivalent of a force is a torque.
The figure shows a small object of mass m
moving in a circle of radius r. A force F is
applied to the mass, tangential to the circle
(perpendicular to the radius). The acceleration
in the direction of this force is the tangential
acceleration at. Substituting into Newton’s
Second Law, F = m·at, and multiplying both
sides by r to get the torque:
T  F  r  m  at  r  m  r   r  m  r 
2
• Torque for a hoop or wheel spinning on an axis
through the center: T = m·r2·α
• The angular mass is given a special name,
moment of inertia, I, and therefore: T = I·,
which describes rotation about a fixed axis
in exactly the same manner that F = m·a.
Moment of inertia is also referred to as
rotational inertia.
• Moment of inertia (rotational inertia) is the
resistance of a rotating object to changes in
its angular velocity .
• An object rotating about an axis tends to
remain rotating about the same axis
unless interfered with by some external
influence. Bodies that are rotating tend to
remain rotating, while non-rotating bodies
tend to remain non-rotating.
• If we wish to change the rate of rotation about
an axis (change of angular velocity), we must
apply a torque about the axis.
• The angular acceleration  this torque produces
depends on the mass of the rotating object and
upon the distribution of its mass with respect to
the axis of rotation.
• If the mass remains fixed in position, torque and
angular acceleration are directly proportional.
• If the mass is closer to the axis of rotation, the
acceleration produced by the given torque is greater
than if the mass is farther from the axis of rotation.
• Moving the mass farther from the axis of rotation
changes the distribution of the mass and increases
the rotational inertia. The greater the distance
between the bulk of the mass of an object and the
axis about which rotation takes place, the greater the
rotational inertia.
 Industrial flywheels are constructed so that most of
their mass is concentrated along the rim. Once
rotating, they have a greater tendency to remain
rotating than they would if their mass were
concentrated nearer the axis of rotation.
 The greater the rotational inertia of an object, the
harder it is to change the rotational state of the object.
If rotating, it is difficult to stop; if at rest, it is difficult to
rotate.
 Equation: F·r = I·; the units for rotational inertia are
kg·m2/rad2.
 Rotational inertia and legs: short legs have less rotational
inertia than long legs. An animal with shorter legs has a
quicker stride than one with long legs (same is true for
pendulums). When running, we bend our legs to reduce the
rotational inertia so that we can rotate them back and forth
more quickly.
• The moment of inertia must be specified
with respect to a chosen axis of rotation.
• See the table containing the rotational
inertia equations for various shapes and
axes of rotation.
Rolling
• A solid cylinder will roll down an
incline faster than a hollow one,
whether or not they have the same
mass or diameter, as illustrated. The
object with the mass concentrated
farthest from the axis of rotation, the
hollow cylinder, has a greater
rotational inertia, therefore, less
acceleration.
• Any solid cylinder will roll down an
incline with greater acceleration than
any hollow cylinder, regardless of the
mass or radius. The hollow cylinder
has a greater rotational inertia.
• Objects of the same shape but different
sizes will accelerate equally when rolled
down an incline. The objects have the
same rotational inertia per mass.
• Just as the different masses of objects in
free fall do not influence the acceleration,
the different masses of the rolling objects
does not influence the acceleration.
• When rotation is introduced, the object
with the larger rotational inertia compared
to its own mass has the greater resistance
to a change in its motion.
Atwood’s Machine Revisited
 When we previously examined
Atwood’s machine, we ignored
friction in the bearings and the
mass and inertia of the pulley.
Not any more!
 When a cord passes over a
frictionless pulley of negligible
mass, there is a single value for
the tension throughout the cord.
 If the pulley is heavy, or if friction
is present in the pulley’s
bearings, the tension on one side
of the pulley is different from the
tension on the other side.
• The Atwood’s machine
problems will now involve
applying the F = m·a and
T = I· equations to both
masses as well as the pulley.
Substitute as required.
• m1 = upward moving mass;
m2 = downward moving mass
• For the side that accelerates
downward:
Fw 2  FT 2  m2  a
• For the side that accelerates upward:
FT1  Fw1  m1  a
• Replace linear acceleration a with r·.
• For the rotation of the pulley:
FT2  FT1   r  I  
• Solve the linear tension equations for the tension
and substitute into the torque equation for the
pulley.
• Solve the subsequent equation for the angular
acceleration. With the angular acceleration, the
linear acceleration and the tensions in the system
can be determined.
Work in Rotational Motion
• When you pedal a bicycle, you
apply forces to a rotating body
and do work on it; a rotating
motor in a drill turns the drill bit.
• In both cases, work can be
expressed in terms of torque and
angular displacement.
• When the kid applies a tangential
force on the merry-go-round, the
force acts through the radius of
the merry-go-round to produce a
torque which will cause the
merry-go-round to turn through
an angular displacement .
• Avoiding the calculus: the work done by the force
as the merry-go-round rotates is:
W = T or W = Fr
 has to be in radians
• If the torque remains constant while the angle
changes from 1 to 2, then:
W  T    T  2  1 
• If torque is in N·m and  is in radians, work is in
joules J.
• When an object is accelerating, the net torque is:
T  TFf  I  
• Maximum torque results
when the force is applied
tangentially to the radius.
• If the force is applied at
an angle to the radius,
only the tangential
component will do work.
• A radial component
would pass through the
pivot and the torque
would be 0 N·m.
Rotational Kinetic Energy
• When a torque does work on a rotating rigid
body, the kinetic energy changes by an
amount equal to the work done (workenergy theorem).
Work = ΔKlinear + ΔKrotary
• Rotating bodies often have two types of
kinetic energy (Work = ΔKrotary ):
– the kinetic energy due to the linear motion of
the object and,
– the kinetic energy due to the rotational motion of
the object.
Rotational Kinetic Energy
– Linear kinetic energy: Klinear = 0.5·m·v2
– Rotational kinetic energy: Krotational = 0.5·I·2
– Total kinetic energy: Ktotal = Klinear +Krotational
• Many problems ask for the translational
velocity of a rolling object, using v = ·r,
solve for  = v/r and substitute this into the
rotational kinetic energy equation.
• See the example at the end of the notes.
Conservation of Energy for
Rotational Motion
• Energy before = Energy after
• Ugi + Ki linear + Ki rotational = Ugf + Kf
linear + Kf rotational
2
2
m  g  hi  0.5  m  v i  0.5  I  ω i 
2
m  g  hf  0.5  m  v f  0.5  I  ω f
2
Power and Rotational Motion
• When a torque acts on a body that rotates
with angular velocity , the power (rate of
doing work) is: P = T·
• If torque is in N·m and  is in rad/s, power is
in watts W.
• Once an object is rotating at constant
angular velocity ω, the only torque needed to
keep the object rotating is the frictional
torque; P = TFf·ω
Translational Velocity of Solid
Ball at the Bottom of an Incline
• Potential energy at the top of an incline is
transformed into kinetic energy at the
bottom of the incline.
• Kinetic energy = 0.5·m·v 2 + 0.5·I·2
• Rotational inertia for a solid ball: 0.4·m·r 2
• Relationship between linear and angular
velocity: v = ·r;  = v/r
• Substituting:
2
2
2 v
K  0.5  m  v  0.5  0.4  m  r   
r
2
2
2 v
K  0.5  m  v  0.2  m  r  2
r
2
2
K  0.5  m  v  0.2  m  v
K  0.7  m  v
2
• Ug top = K bottom:
m  g  h  0.7  m  v 2
v
gh
10  g  h

0.7
7
Angular Momentum
• Rotating objects have angular momentum
L.
• Angular momentum is the rotational equivalent
of linear momentum.
• Equation: L = I·
 in rad/s
• Units: kg  m2
s
• Conservation of Angular Momentum:
Lbefore = Lafter
SI1·1 = SI2·2
• (for each object involved)