Transcript 11_LectureOutline

Lecture Outline
Chapter 11
Physics, 4th Edition
James S. Walker
Copyright © 2010 Pearson Education, Inc.
Chapter 11
Rotational Dynamics and
Static Equilibrium
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Units of Chapter 11
• Torque
• Torque and Angular Acceleration
• Zero Torque and Static Equilibrium
• Center of Mass and Balance
• Dynamic Applications of Torque
• Angular Momentum
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Units of Chapter 11
• Conservation of Angular Momentum
• Rotational Work and Power
• The Vector Nature of Rotational Motion
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11-1 Torque
From experience, we know that the same force
will be much more effective at rotating an
object such as a nut or a door if our hand is not
too close to the axis.
This is why we have
long-handled
wrenches, and why
doorknobs are not
next to hinges.
Applying a torque
(a) When a wrench is used to loosen a nut, less force is required if is
applied far from the nut. (b) Similarly, less force is required to open a
revolving door if it is applied far from the axis of rotation.
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11-1 Torque
We define a quantity called torque:
The torque increases as the force increases,
and also as the distance increases.
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11-1 Torque
Only the tangential component of force causes
a torque:
Only the tangential component of a force causes a torque.
(a). A radial force causes no rotation. In this case, the force F is opposed by an equal and
opposite force exerted by the axle of the merry-go-around. The merry-go-around does not
rotate. (b) A force applied at an angle Ɵ with respect to the radial direction. The radial
component of this force, F cos θ, cause no rotation; the tangential component, F sin θ, can
cause a rotation. Because it is the tangential component alone that causes rotation, we
define the torque to have a magnitude of r(F sin θ).
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11-1 Torque
This leads to a more general definition of torque:
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11-1 Torque
If the torque causes a counterclockwise angular
acceleration, it is positive; if it causes a
clockwise angular acceleration, it is negative.
Sign Convention for Torque
By convention, if a torque ‫ ד‬acts alone, then
r > 0 if the torque causes a counterclockwise angular acceleration
r < 0 if the torque causes a clockwise angular acceleration
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11-2 Torque and Angular Acceleration
Newton’s second law:
α = a/r; α = a/r = F/mr
If we consider a mass m rotating around an axis a distance r away, we
can reformat Newton’s second law to read:
Finally, multiplying both numerator and denominator by r gives
Or equivalently,
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11-2 Torque and Angular Acceleration
Once again, we have analogies between linear
and angular motion:
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11-2 Torque and Angular Acceleration
A light rope wrapped around a disk-shaped pulley is pulled tangentially with a force
of 0.53 N. Find the angular acceleration of the pulley, given that its mass is 1.3 kg
and its radius is 0.11 m.
The torque applied to the disk is ‫ = ד‬r x F = (0.11 m)(0.53 N) = 5.8 x 10-2 N.m
Since the pulley is a disk, its moment of inertia is given by I = ½ mr2
I = ½ mr2 = ½(1.3 kg)(0.11 m)2 = 7.9 x 10-3 kg.m2
Thus, the angular acceleration of the pulley is α = ‫ ד‬/ I = 5.8 x 10-2 N.m
7.9 x 10-3 kg.m2
α = 7.3 rad /s2
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11-3 Zero Torque and Static Equilibrium
Static equilibrium occurs when an object is at
rest – neither rotating nor translating.
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11-3 Zero Torque and Static Equilibrium
If the net torque is zero, it doesn’t matter which
axis we consider rotation to be around; we are
free to choose the one that makes our
calculations easiest.
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11-3 Zero Torque and Static Equilibrium
A child of mass m is supported on a light plank by his parent, who exert the forces
F1 and F2 as indicated. Find the forces required to keep the plank in static
equilibrium. Use the right end of the plank as the axis of rotation. .
Set the net force acting on the plank equal
to zero.
F1 + F2 – mg = 0
Set the net torque acting on the plank
equal to zero.
-F1(L) + mg(1/4 L) = 0
F1 = 1/4mg
F2 = mg – 1/4mg = 3/4mg
F2 = ¾ mg
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11-3 Zero Torque and Static Equilibrium
A 5.0 m long diving board of negligible mass is supported by two pillars. One pillar
is at the left end of the diving board, as shown below; the other is 1.50 m away.
Find the forces exerted by the pillars when a 90.0-kg diver stands at the far end of
the board.
Set the net force acting on the diving board equal to
zero.
∑Fy = F1,y + F2,y – mg = 0
Set the net torque acting on the diving board equal to
zero. Torque about the F1
-F2,y (d) - mg( L) = 0
F2,y = mg (L / d)
F2,y = (90.kg)(9.81 m/s2)(5.0 m/ 1.50 m)
F2,y = 2940 N
F1,y = mg – F2,y = (0 kg) (9.81 m/s2) – 2940 N
F1,y = -2060 N
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11-3 Zero Torque and Static Equilibrium
When forces have both vertical and horizontal
components, in order to be in equilibrium an
object must have no net torque, and no net force
in either the x- or y-direction.
∑‫ = ד‬T(V) – mg(H) = 0
Solve for the tension T,
T = mg(H / V)
∑Fy = fy – mg = 0 = => fy = mg
∑Fx = fx – T = 0 = => fx = T = mg(H / V)
A lamp in static equilibrium
A wall mounted lamp of mass m is
suspended from a light curved rod.
The bottom of the rod is bolted to the wall.
The rod is also connected to the wall by a
Horizontal wire a vertical distance V above
the bottom of the rod.
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11-3 Zero Torque and Static Equilibrium
An 85-kg person stands on a lightweight ladder as shown. The floor is
rough; hence, it exerts both a normal forces f1, and a frictional force, f2 on
the ladder. The wall, on the other hand, is frictionless; it exerts only a
normal force, f3. Using the dimensions given in the figure, find the
magnitudes of f1, f2 and f3.
Set the net torque acting on the ladder equal to
zero. Use the bottom of the ladder as the axis:
f3 (a) - mg (b) = 0  f3 = mg (b/a) = 150 N
∑Fx = 0 = f2 – f3 = 0  f2 = f3 = 150 N
∑Fy = 0  f1 – mg = 0  f1 = 830 N
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11-4 Center of Mass and Balance
If an extended object is to be balanced, it must
be supported through its center of mass.
Zero torque and balance
One section of a mobile. The rod is balanced when the net torque acting
on it is zero. This is equivalent to having the center of mass directly under
the suspension point
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11-4 Center of Mass and Balance
This fact can be used to find the center of mass
of an object – suspend it from different axes and
trace a vertical line. The center of mass is where
the lines meet.
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11-5 Dynamic Applications of Torque
When dealing with systems that have both
rotating parts and translating parts, we must be
careful to account for all forces and torques
correctly.
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11-6 Angular Momentum
Using a bit of algebra, we find for a
particle moving in a circle of radius r,
L = Iω = (mr2) (v/r) = rmv = Note that mv is the linear momentum p; we find that the
angular momentum of a point mass can be written in the following form :
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11-6 Angular Momentum
For more general motion,
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11-6 Angular Momentum
Looking at the rate at which angular momentum
changes,
∆L = Lf – Li = (∆ t) ‫ד‬
The final angular momentum of the object is Lf = Li + (∆ t) ‫ד‬
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11-7 Conservation of Angular Momentum
If the net external torque on a system is zero,
the angular momentum is conserved.
The most interesting consequences occur in
systems that are able to change shape:
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11-7 Conservation of Angular Momentum
As the moment of inertia decreases, the
angular speed increases, so the angular
momentum does not change.
Angular momentum is also conserved in
rotational collisions:
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11-8 Rotational Work and Power
A torque acting through an angular
displacement does work, just as a force
acting through a distance does.
The work-energy theorem applies as usual,
regardless whether the work is done by a
force or a torque.
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11-8 Rotational Work and Power
Power is the rate at which work is done, for
rotational motion as well as for translational
motion.
Again, note the analogy to the linear form:
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11-8 Rotational Work and Power
It takes a good deal of effort to make homemade ice cream. (a) If the
torque required to turn the handle on an ice cream maker is 5.7 N.m, how
much work is expended on each complete revolution of the handle? (b)
How much power is required to turn the handle if each revolution is
completed in 1.5 s?
a.
= (5.7 N.m) (2 π rad) = 36 J
(b). Power is the work per unit time; that is
P = W / ∆ t = (36 J) / (1.5 s) = 24 W
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11-9 The Vector Nature of Rotational
Motion
The direction of the angular velocity vector is
along the axis of rotation. A right-hand rule gives
the sign.
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11-9 The Vector Nature of Rotational
Motion
A similar right-hand rule gives the direction of
the torque.
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11-9 The Vector Nature of Rotational
Motion
Conservation of angular momentum means that
the total angular momentum around any axis
must be constant. This is why gyroscopes are
so stable.
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Summary of Chapter 11
• A force applied so as to cause an angular
acceleration is said to exert a torque.
• Torque due to a tangential force:
• Torque in general:
• Newton’s second law for rotation:
• In order for an object to be in static equilibrium,
the total force and the total torque acting on the
object must be zero.
• An object balances when it is supported at its
center of mass.
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Summary of Chapter 11
• In systems with both rotational and linear
motion, Newton’s second law must be applied
separately to each.
• Angular momentum:
• For tangential motion,
• In general,
• Newton’s second law:
• In systems with no external torque, angular
momentum is conserved.
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Summary of Chapter 11
• Work done by a torque:
• Power:
• Rotational quantities are vectors that point
along the axis of rotation, with the direction
given by the right-hand rule.
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