Transcript Circular.Rotary Motion

Rotational Motion
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Circular Motion
An object moving in a circle at a constant
speed is accelerated
Centripetal acceleration depends upon
the object’s speed and the radius of the
circle
Centripetal force causes centripetal
acceleration.
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Ch 6 Circular Motion Problems: 12,14-20,49,61-62
ac
V
Fc
Definitions
• Centripetal Force - a center seeking force
that appears to pull an object toward the
center of the circle along which it is
moving
• Centripetal Acceleration - acceleration
toward the center ac = v2/r
Newton’s Law enters in…
• Since f = ma
• The centripetal force is determined by
substituting
• ac = v2/r
a = f/m
• fc = mv2/r
Velocity of an object moving in a
circle can be shown as:
•
•
•
•
Revolutions/sec
Cycles/sec
m/s
Degrees/sec
• AKA: angular velocity
Converting rev/s to m/s
1 rev = 2 p r
So, if an object completes 3 rev/min in a
circle with r = 0.5 m the velocity in m/s is
3 rev
2 p (0.5m)
1 min
min
rev
60 sec
V = 0.157 = 1.57 x 10-1 m/s
Sample problem
Describing Rotary Motion
• A fraction of a revolution can be measured
in degrees, grads or radians
• A degree is 1/360 of a revolution
• One revolution is 2p radians
• One radian is ½ p of a revolution
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•
Ch8 Assigned problems
8/1,6a,11,16,30,34,75,76,82
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Angular Displacement
•The Greek letter q is used to
represent the angle of revolution
•The counterclockwise rotation is
considered positive while clockwise is
negative
•As the object rotates, the change in
angle Is called angular displacement
•For rotation through an angle, q, a
point at a distance, r, from the center
moves a distance given by d = r q
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Angular Velocity, w
• The angular displacement divided by the
time required to make the displacement.
w=Dq
w, omega q, theta
Dt
Angular velocity can be measured in rad/s,
rev/s, degrees/s
For Earth, ωE = (2π rad)/(24.0 h)(3600 s/h) =
7.27×10─5 rad/s.
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• If an object’s angular velocity is ω, then
the linear velocity of a point a distance, r,
from the axis of rotation is given by v = rω.
• The speed at which an object on Earth’s
equator moves as a result of Earth’s
rotation is given by v = r ω = (6.38×106 m)
(7.27×10─5 rad/s) = 464 m/s.
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Angular Acceleration, a
• This is the change in angular velocity
divided by the time required to make that
change.
a=Dw
a, alpha
Dt
Units: rad/s2, rev/s2, degrees/s2
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Chart found on p. 199
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Problem
• When a machine is switched on, the angular
velocity of the motor increases by 10 rad/s for
the first 10 seconds before it starts rotating with
full speed. What is the angular acceleration of
the machine in the first 10 seconds?
a. π rad/s2
b. 1 rad/s2
c. 100π rad/s2
d. 100 rad/s2
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Answer: B
• Reason: Angular acceleration is equal to
the change in angular velocity divided by
the time required to make that change.
a = D w = 10 rad/s = 1 rad/s2
Dt
10 s
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Rotational Dynamics
• Torque is a measure of how effectively a force
causes rotation.
• The magnitude of torque is the product of the
force and the lever arm. Because force is
measured in newtons, and distance is measured
in meters, torque is measured in newton-meters
(N·m).
• Torque is represented by the Greek letter tau, t.
t = Fr sin q
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Torque problem
A bolt on a car engine needs to be
tightened with a torque of 35 N·m. You
use a 25-cm-long wrench and pull on
the end of the wrench at an angle of
60.0° from the perpendicular. How
long is the lever arm, and how much
force do you have to exert?
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Sketch the situation
Find the lever arm by extending the force vector backwards until a
line that is perpendicular to it intersects the axis of rotation.
Label your diagram indicating your known
Values
Known:
r = 0.25m
q= 60.0o
t = 35 Nm
Unknown:
L=?
F=?
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Solve for the length of the torque arm
L = r sinq = (0.25m) (sin 60.0o) = 0.22 m
Solve for the force
t = F r sin q
F= t
=
35 Nm
r sin q
(0.25m)(sin 60.0o)
F = 1.6 x 102 N
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Finding Net Force
• Translational equilibrium: all of the upward
forces must equal the downward forces
• Rotational equilibrium: counterclockwise
torque must equal clockwise torque
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Problem sample…
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Solution to prior problem
•
•
1.75m
6.00kg
4.25 kg
F = F so (4.25kg + 6.00kg)(9.8m/s2) = 100N
Let left end = x and right end = 100N - x
Choose right end as the axis of rotation (no
torque there)
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Solution continued…
1.75m
58.8N
41.6N
cwt = ccwt
F1D1 = F2D2 + F3D3
F1 = F2D2 + F3D3 =41.6(.875)+58.8(.5)
D1
1.75
F1 = 37.6N F2 = 100-37.6 = 62.4N
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Moment of Inertia, I - the resistance
to rotation
For a point mass:
I = mr2
Newton’s 2nd
Law for Rotational
Motion:
a=t/I
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The Center of Mass
The center of mass of an object is the point
on the object that moves in the same way
that a point particle would move.
The path of center of mass of the
object is a straight line.
An object is stable against rollover
if its center of mass is above its
base.
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• An object is in equilibrium if there are no
net forces exerted on it and if there are no
net torques acting on it.
• Centrifugal “force” and Coriolis “force” are
two apparent forces that appear when a
rotating object is analyzed from a
coordinate system that rotates with it.
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Sources
• All slides are copyrighted and are found in
• Glencoe Physics Principals and Problems
• 2005 Edition.
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