Transcript rotational equilibrium

Rotational Equilibrium
and Dynamics
Chapter 8
Magnitude of a Torque
• What is a torque?
– A quantity that measures the ability of a force to
rotate an object around some axis.
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Example…
Imagine opening a door around a hinge.
The door is free to rotate about a fixed axis.
This axis is the door’s axis of rotation
When an individual pulls the door open at the
handle with a force that is perpendicular to the
door, the door opens.
– The measurement of this force is the Torque of the
motion.
Torque depends on a force
and a lever arm
• If someone opens the door from the previous
example, but at a point closer to the hinge,
the door would be much more difficult to
rotate.
• How easily an object rotates depends on how
much force is applied as well as WHERE the
force is applied.
Torque depends on a force
and a lever arm
• The farther the force is from the axis of
rotation, the easier it is to rotate the object
and the more torque is produced.
• The PERPENDICULAR distance from the axis of
rotation to a line drawn along the direction of
the force is called the lever arm, or moment
arm.
Torque and Lever Arm
Angular speed
d
F
axis of rotation
Force applied to an extended
object can produce a torque.
This torque in turn, causes this
object to rotate.
TORQUE Equation
• Torque also depends of the angle between a
force and a lever arm
Torque = Force (lever arm)(sin theta)
T = Fd(sin 0)
**SI unit of torque: Newton meter or (N x m)
Calculating Torque at an angle
F(sin )
F
T = d x (sin )
d
Conceptual Question…
• 1: A student pushes with minimal force of
50.0 N on the middle of a door to open it.
– What minimum force must be applied at the edge
of the door in order for the door to open?
– What minimum force must be applied to the
dinged side of the door in order for the door to
open?
Conceptual Question…
• 1: A student pushes with minimal force of 50.0 N
on the middle of a door to open it.
– What minimum force must be applied at the edge of
the door in order for the door to open?
– 25 Newtons
– (half a much because the distance in increased by 2)
– T=d*F
– What minimum force must be applied to the hinged
side of the door in order for the door to open?
– The door cannot be opened by a force at the hinge
location. It can be broken but not opened normally.
Torque can be (+) or (-)
• Torque is a vector quantity (i.e. has direction)
• To find net torque, simply add up the
individual torques.
Tnet = Sum of Torques = T1 + T2 = (F1d1) + (F2d2)
The sign of the net torque value (+ or -) will tell you the
direction an object rotates.
Clockwise rotation = - torque
Counterclockwise rotation = + torque
Example: Torque
• A basketball is being pushed by two players
during tip-off. One player exerts a downward
force of 11 N at a distance of 0.07 m from the
axis of rotation. The second player applies an
upward force of 15 N at a perpendicular
distance of 0.014 m from the axis of rotation.
• Find the net torque acting on the basketball.
Example: Torque
• A basketball is being pushed by two players during tip-off. One
player exerts a downward force of 11 N at a distance of 0.07 m
from the axis of rotation. The second player applies an upward
force of 15 N at a perpendicular distance of 0.14 m from the
axis of rotation.
• Find the net torque acting on the basketball.
Givens:
F1 = 15 N d1 = 0.14 m
F2 = 11 N d2 = 0.07 m
Formula:
Tnet = T1 + T2
Example: Torque
• A basketball is being pushed by two players during tip-off. One player exerts
a downward force of 11 N at a distance of 0.07 m from the axis of rotation.
The second player applies an upward force of 15 N at a perpendicular
distance of 0.14 m from the axis of rotation.
• Find the net torque acting on the basketball.
Givens:
F1 = 15 N d1 = 0.14 m
F2 = 11 N d2 = 0.07 m
F2 = 11 N
Formula:
Tnet = T1 + T2
d1 = 0.14 m
F1 = 15 N
d2 = 0.07 m
• A basketball is being pushed by two players during tip-off. One player exerts a
downward force of 11 N at a distance of 0.07 m from the axis of rotation. The
second player applies an upward force of 15 N at a perpendicular distance of 0.14
m from the axis of rotation.
• Find the net torque acting on the basketball.
Givens:
F1 = 15 N d1 = 0.14 m
F2 = 11 N d2 = 0.07 m
T1 = F1 d1 = -(15)(0.14) = -2.1 Nm
T2 = F2 d2 = -(11) (0.07)= -0.77Nm
Formula:
Tnet = T1 + T2
F2 = 11 N
d1 = 0.14 m
(both forces produce a clockwise rotation
Therefore both torques are negative)
F1 = 15 N
d2 = 0.07 m
• A basketball is being pushed by two players during tip-off. One player exerts a
downward force of 11 N at a distance of 0.07 m from the axis of rotation. The
second player applies an upward force of 15 N at a perpendicular distance of 0.14
m from the axis of rotation.
• Find the net torque acting on the basketball.
T1 = F1 d1 = -(15)(0.14) = -2.1 Nm
T2 = F2 d2 = -(11) (0.07)= -0.77Nm
Tnet = T1 + T2 = -2.1 – 0.77
Tnet = -2.9 N m
F2 = 11 N
d1 = 0.14 m
Negative torque thus
Ball rotates in a clockwise direction.
F1 = 15 N
d2 = 0.07 m
Individual Practice
• Take 5 minutes to work the following on your
own:
*problem 1 (practice 8A) on page 282
ROTATION and INERTIA
• Identify Center of Mass
• Distinguish between mass and moment of
Inertia
• Define the second condition of equilibrium
• Solve problems involving the first and second
conditions of equilibrium
Center of Mass
• Center of Mass – the point at which all the
mass of the body can be considered to be
concentrated when analyzing translational
motion.
• Consider a stick thrown in the air while playing
fetch with a puppy…
• Since the center of mass is considered to be the
point at which all the mass of an object is
concentrated…
• The stick rotates in the air around its center of mass
• The complete motion of the stick in
the air is both rotational and
translational.
The center of mass moves as if the stick were a point
mass, with all of its mass concentrated
at that point for purposes of analyzing its
translational motion.
Moment of Inertia
• Moment of inertia: the tendency of a body
rotating about a fixed axis to resist a change in
rotational motion.
• The moment of inertia is a measure of the
object’s resistance to a change in its rotational
motion about some axis.
Calculating Moment of Inertia
• Depends of the shape of the object.
• SI units: kg x m2
See table 8-1 in textbook for the different formulas.
Rotational Equilibrium
• Rotational Equilibrium is different from
translational equilibrium
– If the net force on an object is zero, then the object is
in translational equilibrium
– If the net torque on an object is zero, then the object
is in rotational equilibrium
– THUS, for an object to be completely in equilibrium,
the net force and the net torque must be zero
– The dependence of equilibrium on the absence of net
torque is called the second condition for equilibrium.
ROTATIONAL DYNAMICS
• Describe Newton’s 2nd law for rotation
• Calculate the angular momentum for various
rotating objects.
Newton’s 2nd Law for Rotation
 net = 
Net torque = moment of Inertia x angular
acceleration
Remember…for translational motion, F=ma
Ex: Newton’s 2nd law for rotation
• A student tosses a dart using only the rotation
of her forearm to accelerate the dart. The
forearm rotates in a vertical plane about an
axis at the elbow joint. The forearm and dart
have a combined moment of Inertia of 0.075
kgm2 about the axis, and the length of the
forearm is 0.26m. If the dart has a tangential
acceleration of 45 m/s2 just before released,
what is the net torque on the arm and dart?
A student tosses a dart using only the rotation of her forearm to
accelerate the dart. The forearm rotates in a vertical plane about
an axis at the elbow joint. The forearm and dart have a combined
moment of Inertia of 0.075 kgm2 about the axis, and the length of
the forearm is 0.26m. If the dart has a tangential acceleration of 45
m/s2 just before released, what is the net torque on the arm and
dart?
GIVEN:
= 0.075 kg x m2
a = 45 m/s2
d = 0.26 m
=?
= 
=  (where = a/d)
= (a/d)
= (0.075)(45/0.26)
= 13 Nm
MOMENTUM
• Angular Momentum: the product of a
rotating object’s moment of inertia and
angular speed about the same axis
L = ( )
Angular momentum = moment of Inertia x angular speed
Conservation of Angular Momentum
• When the net external torque acting on an
object or objects is zero, the angular
momentum of the object(s) does not change.
This is the:
Law of Conservation of Angular Momentum.
• Assuming the friction between skates and the ice
is negligible, there is no torque acting on an ice
skater.
When she brings her hands in and feet closer to her
body, more of her mass on average is nearer to
her axis of rotation. Thus, the moment of inertia
is constant and her angular speed increases to
compensate for her smaller moment of inertia.
Rotational Kinetic Energy
• Rotational KE = ½ x moment of Inertia x angular speed 2
KErot = ½ 
2
Rotational KE = energy of an object due to
its rotational motion
Rotational KE is conserved in the absence of any external force
moving an object.
Be ready for a vocab quiz on Friday
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Torque
Work
Lever Arm
Energy
Center of Mass
Momentum
Moment of Inertia
Angular Speed
Angular Momentum
Angular Acceleration
Rotational Kinetic Energy Rotational Motion
6 types of simple machines “SAWLIP”