Transcript Chapter 8x - HCC Learning Web

Chapter 8 Rotational Equilibrium
and Rotational Dynamics
Ying Yi PhD
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Outline
 Torque
 Two conditions for equilibrium
 Center of Gravity
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Torque
 There are three factors that determine the effectiveness of
the force in opening the door:
 The magnitude of the force
 The position of the application of the force
 The angle at which the force is applied
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Torque, cont
 Torque,  , is the tendency of
object about some axis
 = r F
a force to rotate an
 is the torque
Symbol is the Greek tau
 F is the force
 r is the length of the position vector

 SI unit is N.m
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Direction of Torque
 Torque is a vector quantity
 The direction is perpendicular to the plane determined
by the position vector and the force
 If the turning tendency of the force is
counterclockwise, the torque will be positive
 If the turning tendency is clockwise, the torque will be
negative
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Multiple Torques
 When two or more torques are acting on an object,
the torques are added
 As vectors
 If the net torque is zero, the object’s rate of rotation
doesn’t change
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Example 8.1 Battle of the revolving Door
Two disgruntled businesspeople are trying to use a
revolving door, as in Figure 8.3. The woman on the left
exerts a force of 625 N perpendicular to the door and
1.20 m from the hub’s center, while the man on the right
exerts a force of 8.5×102 N perpendicular to the door
and 0.800 m from the hub’s center. Find the net torque
on the revolving door.
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Group problem: Revolving door
A businessman enters the same revolving door on the
right, pushing with 576 N of force directed
perpendicular to the door and 0.700 m from the hub,
while a boy exerts a force of 365 N perpendicular to the
door, 1.25 m to the left of the hub. Find (a) the torques
exerted by each person and (b) the net torque on the
door.
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General Definition of Torque
 The applied force is not always perpendicular to the
position vector
 The component of the force perpendicular to the
object will cause it to rotate
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General Definition of Torque, cont
 When the force is
parallel to the position
vector, no rotation
occurs
 When the force is at
some angle, the
perpendicular
component causes the
rotation
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General Definition of Torque, final
 Taking the angle into account leads to a more general
definition of torque:

 = r F sinƟ
 F is the force
 r is the position vector
 Ɵ is the angle between the force and the position vector
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Lever Arm
 The lever arm, d, is the perpendicular distance from the
axis of rotation to a line drawn along the direction of the
force
 d = r sin Ɵ
 This also gives t = rF sin Ɵ
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Right Hand Rule
 Point the fingers in the
direction of the
position vector
 Curl the fingers
toward the force
vector
 The thumb points in
the direction of the
torque
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Net Torque
 The net torque is the sum of all the torques produced
by all the forces
 Remember to account for the direction of the tendency
for rotation
 Counterclockwise torques are positive
 Clockwise torques are negative
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Example 8.2: The swinging door
(a) A man applies a force of F=3.00×102 N at an angle
of 60.0º to the door of Figure 8.7a, 2.00 m from the
hinges. Find the torque on the door, choosing the
position of the hinges as the axis of rotation. (b)
suppose a wedge is placed 1.50 m from the hinges on
the other side of the door. What minimum force must
the wedge exert so that the force applied in part (a)
won’t open the door?
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Torque and Equilibrium
 First Condition of Equilibrium
 The net external force must be zero
F  0 or
Fx  0 and Fy  0
 This is a necessary, but not sufficient, condition to
ensure that an object is in complete mechanical
equilibrium
 This is a statement of translational equilibrium
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Torque and Equilibrium, cont
 To ensure mechanical equilibrium, you need to ensure
rotational equilibrium as well as translational
 The Second Condition of Equilibrium states
 The net external torque must be zero
  0
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Selecting an Axis
 The value of  depends on the axis of
rotation
 You can choose any location for calculating torques
 It’s usually best to choose an axis that will make at
least one torque equal to zero
 This will simplify the torque equation
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Equilibrium Example
 The woman, mass m, sits
on the left end of the seesaw
 The man, mass M, sits
where the see-saw will be
balanced
 Apply the Second
Condition of Equilibrium
and solve for the unknown
distance, x
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Axis of Rotation
 If the object is in equilibrium, it does not matter
where you put the axis of rotation for calculating
the net torque
 The location of the axis of rotation is completely
arbitrary
 Often the nature of the problem will suggest a
convenient location for the axis
 When solving a problem, you must specify an axis of
rotation
 Once you have chosen an axis, you must maintain that choice
consistently throughout the problem
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Example 8.3 Balancing act
A woman of mass m=55.0 kg sits on the left end of a
seesaw a plank of length L=4.00 m, pivoted in the
middle as in figure 8.8. (a) First compute the torques on
the seesaw about an axis that passes through the pivot
point. Where should a man of mass M=75.0 kg sit if
the system (seesaw plus man and woman) is to be
balanced? (b) Find the normal force exerted by the pivot
if the plank has a mass of mpl=12.0 kg. (c) repeat part
(a), but this time compute the torques about an axis
through the left end of the plank.
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Center of Gravity
 The force of gravity acting on an object must be
considered
 In finding the torque produced by the force of
gravity, all of the weight of the object can be
considered to be concentrated at a single point
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Calculating the Center of Gravity
 The object is divided up
into a large number of
very small particles of
weight (mg)
 Each particle will have a
set of coordinates
indicating its location
(x,y)
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Calculating the Center of Gravity, cont.
 We assume the object is free to rotate about its center
 The torque produced by each particle about the axis
of rotation is equal to its weight times its lever arm
 For example,1 = m1 g x1
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Calculating the Center of Gravity, cont.
 We wish to locate the point of application of the
single force whose magnitude is equal to the weight of
the object, and whose effect on the rotation is the
same as all the individual particles.
 This point is called the center of gravity of the object
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Coordinates of the Center of Gravity
 The coordinates of the center of gravity can be found
from the sum of the torques acting on the individual
particles being set equal to the torque produced by
the weight of the object
xcg
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mi xi
mi y i

and ycg 
mi
mi
Center of Gravity of a Uniform Object
 The center of gravity of a homogenous, symmetric
body must lie on the axis of symmetry
 Often, the center of gravity of such an object is the
geometric center of the object
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Experimentally Determining the
Center of Gravity
 The wrench is hung freely
from two different pivots
 The intersection of the lines
indicates the center of gravity
 A rigid object can be balanced
by a single force equal in
magnitude to its weight as
long as the force is acting
upward through the object’s
center of gravity
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Notes About Equilibrium
 A zero net torque does not mean the absence of
rotational motion
 An object that rotates at uniform angular velocity can be
under the influence of a zero net torque
 This is analogous to the translational situation where a zero net
force does not mean the object is not in motion
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Solving Equilibrium Problems
 Diagram the system
 Include coordinates and choose a rotation axis
 Isolate the object being analyzed and draw a free
body diagram showing all the external forces
acting on the object
 For systems containing more than one object, draw a
separate free body diagram for each object
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Problem Solving, cont.
 Apply the Second Condition of Equilibrium
 This will yield a single equation, often with one
unknown which can be solved immediately
 Apply the First Condition of Equilibrium
 This will give you two more equations
 Solve the resulting simultaneous equations for all
of the unknowns
 Solving by substitution is generally easiest
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Example 8.6 A weighted forearm
A 50.0 N bowling ball is held in a person’s hand with
the forearm horizontal, as in Figure 8.13a. The biceps
muscle is attached 0.0300 m from the joint, and the ball
is 0.350 m from the joint. Find the upward force F
exerted by the biceps on the forearm (the ulna) and the
downward force R exerted by the humerus on the
forearm, acting at the joint. Neglect the weight of the
forearm and slight deviation from the vertical of the
biceps.
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Group Problem: Walking a horizontal beam
A uniform horizontal beam 5.00 m long and weighing
3.00×102N 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 53.0º with the horizontal.
If a person weighing 6.00×102 N stands 1.50 m from
the wall, find the magnitude of the tension T in the
cable and the force R exerted by the wall on the beam.
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Homework
1, 2, 4, 7, 10, 20, 21
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