Rotational Equilibrium and...
Transcript of Rotational Equilibrium and...
Chapter 8
Rotational Equilibrium and Dynamics
• The Torque, τ
• Elements of Statics. Center of Mass
• Rotational Dynamics. Moment of Inertia, I
• Rotational Energy
• Angular Momentum, L
What causes rotations?
• To rotate an object which is initially at
rest, a force is needed
• Beside the magnitude of the force, the
rotational effect of the force depends on
the position of the force with respect to
the axis of rotation, and on its direction
• So, in order to study how one can
produce rotations, one must take into
consideration all these quantities, not
only the simple force as in the purely
translational case
• Let’s first define some geometrical
characteristics:
a) the strength (magnitude) of the force
b) the position where the force is applied with
respect to the hinges (axis of rotation)
Ex: Consider a door acted upon by a force F.
The rotational effect depends on
c) the direction of the force (or line of action)
zero effect
maximum effect
larger
effect
smaller
effect F F
F
F
Def 1: A line along the vector force is
called the line of action
Def 2: The perpendicular distance from
the axis of rotation to the line of action
is called the lever arm
Ex: A longer lever arm is very helpful in rotating objects:
this is why a wrench can be used to loosen a bolt
• Formally, this is due to the fact that a longer r increases
the torque for the same F and θ
• Also, applying the force perpendicular on the arm will
maximize the torque for the same F and r, since sinθ is
maximum when θ = 90°
• What about the direction of the torque? The vector torque is
perpendicular on the plane of the rotation that it attempts to produce
• We can describe the direction using a sign convention:
sinFr
Torque – Definition
Def: The physical quantity that models the rotational
effect of a force is called torque, τ, with magnitude
defined by taking into account all the influences upon
the rotation:
Magnitude of force F Distance r to the axis Angle between F and r
+ – τ > 0 if it tries to rotate the
object counterclockwise
τ < 0 if it tries to rotate the
object clockwise
Axis of rotation
r
F
θ
SIm N
Torque – Calculation
The magnitude of the torque can be viewed in two ways:
1. Note that that the force is perpendicular on the lever
arm ℓ, so
Torque = (Force Magnitude) × (Lever Arm)
sinrF F
sinFr rF
Axis of rotation
r
F
θ
ℓ = rsinθ
Axis of rotation
r
F
θ
θ
2. Note that that the component of the force along r does not produce
rotation, so we remain only with the perpendicular component:
Torque = (Distance to Axis) × (Perpendicular Component of Force)
sinF F
F
Problem:
1. Net torque on a wheel: Two forces, F1 = 7.50 N and F2 = 5.30 N,
are applied to a wheel with radius r = 0.330 m, as in the figure.
What is the net torque on the wheel due to these forces for an axis
perpendicular on the wheel and passing through its center?
Exercise 1: Net torque on a dumbbell: Two balls of masses m1 = m2 = m are attached to the
ends of a massless rod which pivots, at distances L1 and L2 from the pivot, as shown in the
figure. Initially the rod is held in the horizontal position and then released.
Axis of rotation
r
F1
F2 ½r
L1 L2 m1
a) Sketch the forces action on the
dumbbell.
b) Calculate the magnitude and
direction of the net torque acting
on the system relative to the pivot.
m2
Statics – Elements
• Based on our discussion about different types motion, we see that the motion is
controlled by forces in the case of translations and torques in the case of rotations
• Therefore, we can introduce the conditions for an object to be completely at rest (or
moving with constant velocity)
• A solid body that is at rest is said to be in motional equilibrium
• To be in complete equilibrium, a solid body must satisfy both translational and a
rotational conditions of equilibrium:
00
0
x
y
FF
F
0
2. if the net torque is zero with respect with any arbitrary
center of rotation (pivot), there is no rotation
Ex: Rotation without
translation: Even when the
net force is zero, the object
can still rotate. So, a zero
balance of force does not
grant complete equilibrium
1. if the net force is zero there is no translation
• Since the pivot is arbitrary, we can choose its such
that the equation won’t contain unknown forces
Calculating the position of the Center of Mass
• If the mass of an object is uniformly distributed, and the system is geometrically
symmetric, the center of mass (CM) is located either in the center of symmetry or on
the axis of symmetry
1 1 2 2CM CM CM
1 2
...1
...iMg m g i i i i
i i i
m x m xMx m x x m x x
M m m
Ex: In the center of a sphere, or along the height of a cone.
Finding the center of mass in a one dimensional case: consider a number of
particles with masses mi (adding to a total mass M) and positions along x-axis xi (or
larger objects with the centers of mass located at xi) then the position of the center
of mass is given by:
• However, if an object is not symmetrically shaped, the position of the CM must be
calculated by using the fact that its weight is acted in the CM
• Hence, the torque of the net weight is equal to the sum of the torques of the weights
of each part of the system, and we can derive the following procedure:
Ex: Five particles of equal masses m are aligned at positions
given on the figure below. Find their center of mass
–2 –1 0 1 2 3 4 x (m)
1 2 3 4 5CM
15
5
2 0 1 3 4 m 1.2 m
mx mx mx mx mxx
m
CM
3. Translational equilibrium. A chandelier of mass m = 200 kg
is suspended from the ceiling by a rope making an angle θ = 60°
with the ceiling, and from a lateral wall by a horizontal rope, as
in the figure. Calculate the tension forces T1 and T2 necessary to
hold it in equilibrium.
4. Rotational equilibrium: A uniform, 256 kg beam is
supported using a cable connected to the ceiling, as shown in
the figure. The lower end of the beam rests on the floor. What
is the tension in the cable?
Problems:
2. CM of a two objects with size: A 2-kg sphere and a 1-kg
square plate are aligned as on the figure. Knowing that the
distance between their CMs is d = 2 m, find the center of mass
of this combination with respect to the center of the sphere.
256 kg
m1 m2
d
m
θ
Rotational Dynamics – Torque and Rotational Inertia
• If we consider the particle as a rigid body, what is
the corresponding torque?
2 tFr ma r mr I
2 i im r
• Let’s first consider a particle performing circular
motion of radius r
• Now we know that, if its speed varies, there must
be a tangential force accelerating it, such that, by
Newton’s 2nd law
t tF ma mr
vr path
tF
• This is for a single point mass; what about a rigid body seen as an extended object
containing many particles of mass mi at positions ri from the center of rotation?
• Then, since the angular acceleration is the same for the whole object, we can write:
2I mrDef: moment of inertia of a particle
2
i iI m rDef: moment of inertia of a system of particles
2
SI kg mI
• Notice that the same object can have different
moments of inertia with respect to different
axes of rotation
• The rotational inertia increases the further the
mass is distributed from the axis of rotation
• In this class, the moments of inertia of rigid
bodies with continuously distributed mass will
be given (see the adjacent table)
• The net moments of inertia of a system of
objects with known individual moments Ii is
2
i iI m r
1 2 3 ...netI I I I
• So, the rotational inertia of an object made
of many particles is given by its moment of
inertia
Quiz: The two cylinders
shown have the same
mass. Which has a
larger rotational inertia?
Moment of Inertia – Concept and formulas
Newton’s 2nd Law for Rotations
Comments:
• This law works only for rigid bodies (such that α is the same for all parts)
• Like forces that can modify the motion of an object only they are external forces,
the angular acceleration is determined only by external torques: the internal torques
cannot modify the rotation of the system
• Notice that the angular acceleration has the same vector direction as the net torque
(perpendicular on the circle of rotation)
I
A net torque Στ acting on a rigid body with moment of inertia I will determine an
angular acceleration α proportional to the torque as given by:
Ex: The weight of a rigid body will play a role in the
rotation in most cases. Since the weight can be
considered as acting in CM, the weight will have a
torque and rotate the body only if the if the center of
rotation doesn’t pass through the CM.
CM
mgPivot
m1 m2
Problems:
5. Net moment of inertia: A helicopter rotor blade can be
considered a long thin rod, as shown in the figure.
a) If each of the three rotor helicopter blades has length L
= 3.75 m and has a mass m =160 kg, calculate the moment
of inertia of the three rotor blades about the axis of
rotation.
b) How much torque must be applied to bring the blades
up to a speed ω in a time t?
6. Dynamics of Atwood machine revisited : Two boxes
with masses m1 and m2 are connected by a massless cord
passing over a pulley, as in the figure. Consider the pulley
as a disk of mass M and radius R. Use the dynamics of the
system to calculate
a) the acceleration a of the masses and
b) the angular acceleration α of the pulley vv
M
R
• The rotational kinetic energy of a rigid body
formed of particles indexed by i (as the wheel
on the right) is the sum of the kinetic energies
of all its parts
2 2 21 1 1rot 1 1 2 22 2 2
... i iKE m v m v m v
Physical situation:
• Consider some rigidly connected
particles rotating about their center of
mass with angular speed ω, and in the
same time translating with speed vcm
• Then, an arbitrary particle of mass mi
located at distance ri from the center of
rotation, moves with speed vi
2 2 2 21 1rot 2 2
212i i i i IKE m r m r
1 212
2
2 cne cmmtKE m Iv
Rotation Translation
• By the argument on the left, the net
kinetic energy is the sum of the kinetic
energies associated with its two motions
• By substituting the rotational quantities, we
find that the rotational kinetic energy can be
written
• Therefore, a rigid body that has both
translational motion (motion of its cm) and
rotational motion (about its cm) has both
translational and rotational kinetic energies:
Net moment of inertia I of the
system of particles
Rotational
velocity, ω
ri
vi
Translational velocity
vcm
Energy in Rotational Motion – Kinetic energy
Energy in Rotational Motion – Conservation
• When evaluating the conservation of energy for rotating rigid bodies, the only
change from our previous approach is that, beside the kinetic energy associated with
the translation of the CM, the total kinetic energy contains a rotational term
• So the expression for the mechanical energy becomes
• The conservation of energy can still be written
21rot CM
2 21 1transl M 2C2 2
E KE PE mKE v kx gI m y
nonconservativeE W
Exercise 2: Rolling motion. Assume that all the
objects on the figure have the same mass m and
are all released down the frictionless incline from
rest, from the same initial height. If the radius r
of each rolling object is the same, which object
will move faster at the bottom of the incline?
m1 m2
m1
m2
h
h
Problem:
7. Atwood machine treated using energy: Two boxes with masses m1 and m2 are
connected by a massless cord passing over a pulley, as in the figure. Consider the pulley as
a disk of mass M and radius R. If the system is released from rest and the masses travel a
distance h, use conservation of energy to calculate:
a) the final speed of the masses
b) the final angular velocity of the pulley
c) the corresponding angular displacement of the pulley
A B
M R
Angular Momentum – Conservation
• In analogy with linear momentum, we can define angular momentum:
L I2 1
SIkg m sL
after before i i i iafter beforeL L I I
L
t
pF
t
Def: A rigid body rotating with an angular speed ω about an axis of rotation with
respect to which its moment of inertia is I, carries a “quantity of rational motion”
given by the angular momentum:
• Hence, the net torque is the rate of change of angular momentum as the force is the
rate of change in linear momentum
• This expression leads us to a condition for the conservation of momentum: if the net
external torque acting on an object is zero, the total angular momentum is constant
• Thus, if the parts of an isolated rotating system redistribute due to some internal
torques, the net angular momentum stays constant, even though its structure changes:
…compare to
Therefore, systems that can change their rotational inertia through
internal torques will also have to change their rate of rotation.
Ex 1: No net rotations but rotating body parts
• A falling cat twisting its body
• An astronaut in space trying to rotate his body
• A helicopter must have two rotors
Ex 3: A ballerina redistributing the mass of her body
The total angular momentum stays the same, but the angular
velocity changes when the mass is redistributed changing the
rotational inertia:
large small small largeI I
Ex 2: Kepler’s 2nd law: planets move faster when they are closer to the sun
Angular Momentum – Examples
Problems:
8. Conservation of angular momentum: A person of mass m stands at the center of a
rotating merry-go-round platform of radius r, moment of inertia I and rotating without
friction with angular speed ω. The person walks radially to the edge of the platform.
Calculate the angular speed when the person reaches the edge.
9. Conservation of angular momentum of a turntable: Consider a turntable to be a
circular disk of moment of inertia It rotating freely at a constant angular velocity ωi. The
axis of the disk is vertical and the disk is supported by frictionless bearings. A record of
moment of inertia Ir is dropped onto the turntable. There is friction between the two disks.
After this "rotational collision," the disks will eventually rotate with the same angular
velocity. What is the final angular velocity, ωf, of the two disks?