Transcript Rotational Mechanics

Rotational Mechanics
Torque
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The ability of a force to rotate an object around
some axis
Produces rotation
Produced when a force is applied with leverage
Longer the lever = more torque
More force = more torque
Ex: Door
More Torque
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Lever arm: distance from the turning axis to the point of
contact
Force is always perpendicular
to lever arm.
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Torque = Force x lever arm
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τ = Fd
Units: N x m = Nm
Torque also depends on the angle
between force and lever arm
τ
= Fd (sin θ)
 Perpendicular
= most torque
 Increase angle = less torque
 Increase angle more = least torque
Sign of Torque
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Counterclockwise rotation = (+) torque
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Clockwise rotation = (-) torque
Problems
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A force of 50N is applied
perpendicular to the
door, .5m from the
hinges. How much
torque is produced?
What if the force were
applied .1 m from the
hinges?
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If you cannot exert
enough torque with a
wrench to turn a
stubborn bolt, would
more torque be
produced if you fastened
a length of rope to the
wrench handle?
Example Problem
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Ned tightens a bolt in his car engine by
exerting 12 N of force on his wrench at a
distance of 0.40 m from the fulcrum. How
much torque must Ned produce to turn the
bolt?
Sample Problem 8A
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A basketball is being pushed by two players
during tip-off. One player exerts a downward
force of 11 N at a distance of 7.0 cm from the
axis of rotation. The second player applies an
upward force of 15 N at a perpendicular
distance of 14 cm from the axis of rotation.
Find the net torque acting on the ball.
Balanced Torques
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Ex: on a seesaw, balance can be achieved if
clockwise torque = counter clockwise torque
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If a meter stick is balanced in the center.
A 20 N block is attached to the 80cm mark,
and another block is attached to the 10cm
mark. What is the weight of the second block?
Balancing Torque Problem
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Mabel and Maude are seesawing on the school
playground and decide to see if they can move
to the correct location to make the seesaw
balance. Mabel weighs 400. N and she sits
2.00 m from the fulcrum of the seesaw. Where
should 450. N Maude sit to balance the
seesaw?
Center of Mass
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Average position of the mass
of an object
Object will rotate around this
point is gravity is the only
force acting on it.
When working problems – all
mass can be considered to be
concentrated here
For regular shapes – CoM is
at the center of the object
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Center of Gravity – average
position of weight of an object
An object will fall over if the
center of gravity extends
beyond the area of support.
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Falling over
C o G outside the object
Usually, the center of mass
and center of gravity are at
the same point.
Rotational Inertia
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The resistance of an object to changes in its
rotational motion.
AKA: moment of inertia
An object rotating about an axis tends to keep
rotating about that axis.
Moment of Inertia
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Torque is required to
change the rotational
state of motion of an
object.
Rotational inertia
depends on mass and
on the distribution of the
mass.
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The greater the distance
between the axis and the
majority of the mass, the
greater the rotational
inertia.
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Ex: Bat
Formulas for Rotational Inertia
Rolling
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Which will roll down an
incline with greater
acceleration, a hollow
cylinder or a solid
cylinder if both have the
same mass and
diameter?
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Solid cylinder
Why?
Smaller cylinder has less
rotational inertia.
Rotational Equilibrium
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Equilibrium requires zero net force and zero
net torque
Translational Equilibrium
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ΣF = 0
Rotational Equilibrium
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ΣΤ = 0
Sample Problem 8B
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A uniform 5.00 m long horizontal beam that
weighs 315 N is attached to a wall by a pin
connection that allows the beam to rotate. Its
far end is supported by a cable that makes an
angle of 53o with the horizontal, and a 545 N
person is standing 1.50 m from the pin. Find
the force in the cable, FT, and the force exerted
on the beam by the wall, R, if the beam is in
equilibrium.
Newton’s 2nd Law of Rotation
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Tnet = Iα
Net torque = moment of
inertia x angular accel.
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A bicycle tire of radius
.330 m and mass 1.50
kg is rotating at 98.7
rad/s. What torque is
needed to stop the tire in
2.00 s?
Sample Problem 8C
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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 kg
x m2 about the axis, and the length of the
forearm is 0.26 m. If the dart has a tangential
acceleration of 45 m/s2 just before it is
released, what is the net torque on the arm and
dart?
Angular Momentum
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Angular momentum = inertia of rotation
Angular momentum = rotational inertia x
rotational velocity
Angular momentum L =  x 
Angular momentum always acts outward along
the axis
Angular momentum = mvr
Law of Conservation of
Angular Momentum
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If no unbalanced
external torque acts on a
rotating system, the
angular momentum of
that system is constant.
Example
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A merry-go-round rotates at the rate of 0.30
rad/s w/an 80. kg man standing at a point 2.0
m from the axis of rotation. If the man walks to
a point 1.0 m form the center, what will be the
new angular speed?
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Merry-go-round = solid cylinder, m = 6.5 x 102 kg,
r=2.0 m
Sample Problem 8D
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A 65 kg student is spinning on a merry-goround that has a mass of 5.25 x 102 kg and a
radius of 2.00 m. She walks from the edge of
the merry-go-round toward the center. If the
angular speed of the merry-go-round is initially
0.20 rad/s, what is the angular speed when the
student reaches a point 0.50 m from the
center.
Homework, yea!!!!
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Practice problems
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8B: 4
8C: 1 & 3
8D: 3, 4, & 5
Stop complaining, it’s only 6 problems
Rotational Kinetic Energy
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Just as with an object moving in a straight line,
a rotating object also possess kinetic energy
KErot = ½ I ω2
This causes the conservation of energy
equation to change to
(PE + KEtrans +KErot)i = (PE + KEtrans +KErot)f
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We call this Mechanical Energy
ME = KEtrans + KErot + Peg
ME = ½ mv2 + ½ Iω2 + mgh
Example
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A 1.5 kg bicycle tire of radius .33m starts from
rest and rolls down a hill 14.8m high. What is
the translational speed when it reaches the
bottom of the hill?
Sample Problem 8E
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A solid ball with a mass of 4.10 kg and a radius
of 0.050 m starts from rest at a height of 2.00
m and rolls down a 30.0o slope. What is the
translational speed of the ball when it leaves
the incline?
Homework
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Practice 8E (all)
Simple Machines
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Machine – any device that transmits or
modifies force, usually by changing the force
applied to an object.
Types of simple machines
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Lever
Wheel & Axle
Wedge
Inclined Plane
Pulley
Screw
Mechanical Advantage
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MA compares output force to input force
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MA = Fout/Fin
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Fout = resistance
Fin = effort
The MA can also be compared to the distance
moved – like in a lever or incline
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MA = din/dout
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din = effort
dout = resistance
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In the absence of
friction, energy is
conserved in a machine
The force applied may
be decreased, if the
distance it is applied is
increased.
The product of the 2
remains constant
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In reality, no machine
exists without friction
We can calculate the
efficiency of a machine
by..
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Eff = Wout/Win
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Wout = resistance
Win = effort
Example
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You are pushing a block up a ramp that is 3.0
m high and 6.0 m long. It is a 30.o ramp. The
block has a mass of 5.0 kg. It is takes you 37 N
of force to push it up the incline.
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Draw this picture.
What is the Effort Force?
What is the Resistance Force?
What is the Effort Distance?
What is the Resistance Distance?
What is the Mechanical Advantage?
What is the Efficiency of the machine?
Homework
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Section Review (all)
Let’s try these
Starting on page 306:
#12, 14, 18, 21, 26, 31, 35, & 45
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