Download Rotational Equilibrium and Dynamics

Torque
• What is the best way to loosen a bolt using a wrench?
• Notice that where you exert the force matters!
• Which of the following forces is most likely to loosen
the bolt (assuming they all have the same
magnitude):
Bolt
Answer: Force C, because the tendency of a force
to cause a rotation about some point depends on
its magnitude and the perpendicular distance l
between the line of action of the force and that
lC point
Point about which bolt rotates
Line of action of force A
Line of action of force B
A
(lA = 0)
Line of action of force C
lB
B
C
Torque
• The distances lA, lB, and lC are called the lever arms
(or moment arms) of forces A, B, and C (respectively)
about the point of rotation
• Torque (or moment) is a “twist” or a “turn” associated
with a force and has a magnitude defined by: t = FlF
– It is always defined with respect to a specific reference
point
– Note the similarity to force (a “push” or a “pull”)
– Torque is the rotational analog of force
• The torques associated with forces A, B, and C (see
previous slide) with respect to the rotation point of the
bolt are: tA = 0 , tB = BlB , and tC = ClC
• Since the magnitudes of forces B and C are the same
and lC > lB , then tC > tB
Torque
• Forces B and C both tend to cause a
counterclockwise rotation, which we can define to be
the positive sense of rotation:
– Thus tB and tC are both positive
+
• However, a force directed in the opposite direction
as B or C (tightening the bolt) would tend to cause a
clockwise rotation, and the torque associated with
this force would be negative
• When a force acts in some arbitrary direction:
r f
F
f
lF
Line of action of force F
lF = lever arm of force F = rsinf
Torque
• The magnitude of the torque associated with force F
is: t = FlF = Frsinf
• Torque is a vector quantity  it has a magnitude and
a direction

– Direction
is perpendicular to both the position ( r ) and force

( F ) vectors
• The direction of torque is given by the Right-Hand
Rule:
– Point fingers of right hand in direction of position
vector

(begins at reference point and ends where F acts)
– Curl fingers toward direction of the force
– Thumb points in direction of the torque
• In the last example (see previous slide), torque
points into of the page (or screen) – tends to cause
clockwise rotation
CQ1: If all the forces below have equal
magnitude, which one creates the most
torque?
A)
B)
C)
D)
Figure A
Figure B
Figure C
Figure D
CQ2: A carpenter who is having a difficult
time loosening a screw puts away his
screwdriver and chooses another with a
handle with a larger diameter. He does this
because:
A)
B)
C)
D)
increasing force increases torque.
decreasing force decreases torque.
increasing lever arm increases torque.
decreasing lever arm decreases torque.
Center of Mass
• Note that weight doesn’t act at a single point – it is
distributed over the entire body
• However, we can always calculate the torque due
to a body’s weight by assuming that the entire force
of gravity (weight) is concentrated at a single point
called the center of gravity
• If the acceleration of gravity has the same value at
all points on a body, its center of gravity is identical
to its center of mass (where all mass of object is
considered concentrated at a single point)
• A good assumption, considering that g at the
bottom of the Sears Tower in Chicago is only about
0.014% greater than at the top
– Center of gravity only about 2 cm below center of mass
Center of Mass
m2
• For a system of point particles: m1
• The center of mass of the system is the point in
space having coordinates (xCM, yCM) defined by:
mx

m x  m x  m x 


m  m  m 
m
i i
xCM
1 1
2 2
1
2
3 3
i
3
i
i
my

m y  m y  m y 


m  m  m 
m
i
yCM
1 1
2
1
2
2
3 3
i
i
3
i
i
– Average position of set of particles, weighted by their
masses
– Coordinates where all the mass of entire system can be
considered to be concentrated
m3
Center of Mass
• Sometimes can use symmetry to deduce position of
center of mass
• Where would center of mass be for the previous
system of 3 equally-spaced point masses if m1 = m2 =
m3 ?
m1
m2
m3
center of mass
• For solid bodies (which are a continuous distribution
of matter), the sums given earlier would be impractical
• Center of mass can usually be deduced from
symmetry
Homogeneous
solid sphere:
CM at center of sphere
Donut:
CM at center of donut hole
CQ3: The system below consists of three
spheres of equal mass m. The center of
mass of the system is located at point:
A)
B)
C)
D)
3
4
5
6
Equilibrium of a Rigid Body
• We can apply Newton’s 1st law two different ways to
a rigid body when it is in equilibrium

• For translational equilibrium:  F  0
– Linear acceleration

a 0
• For rotational equilibrium:
– Angular acceleration

 0

t  0
Example Problem #8.17
Consider the model at right of a
person bending forward to lift a 200-N
object. The spine and upper body are
represented as a uniform horizontal
rod of weight 350 N, pivoted at the
base of the spine. The erector spinalis
muscle, attached at a point 2/3 of the
way up the spine, maintains the
position of the back. The angle
between the spine and this muscle is
12.0°. Find the tension in the back
muscle and the compressional force in
the spine.
Solution (details given in class):
l
2l/3
l/2
y
+
T = 2.71  103 N = 2.71 kN
Rx = 2.65 kN
(Note that when object is in equilibrium, you have freedom to choose the
point about which to calculate the torques.)
x
CQ4: A sign hangs by a rope attached at 30° to the
middle of its upper edge. It rests against a frictionless
wall. If the weight of the sign were doubled, what would
happen to the tension in the string? (Note: sin30° = 0.5;
cos30° = 0.87)
A)
B)
C)
D)
It would remain the same.
It would increase by a factor of 1.5.
It would increase by a factor of 2.
It would increase by a factor of 4.
CQ5: A telephone pole stands as shown below. Line A
is 4 m off the ground and line B is 3 m off the ground.
The tensions in line A and line B are 200 N and 400 N
respectively. What is the net torque on the pole?
A)
B)
C)
D)
0 Nm
400 Nm
800 Nm
2000 Nm
Example Problem #8.23
An 8.00-m, 200-N uniform ladder rests against a
smooth wall. The coefficient of static friction
between the ladder and the ground is 0.600,
and the ladder makes a 50.0° angle with the
ground. How far up the ladder can an 800-N
person climb before the ladder begins to slip?
Solution (details in class):
6.15 m
Energy in Rotational Motion
• Any matter in motion has kinetic energy
• What is the kinetic energy of a rotating rigid body?
• Representing the rigid body as the sum of individual
point particles, the total kinetic energy of the body is
the sum of the individual kinetic energies:
K tot 
1
1
1
m1v12  m2v22     mi vi2
2
2
i 2
• Each point particle has the same value of w, so:
Ki 
1
1
mi vi2  mi ri 2w 2
2
2
• Therefore, the kinetic energy of the rigid body is:
1
1
1 

K tot   mi vi2   mi ri 2w 2  w 2   mi ri 2 
2  i
i 2
i 2

Moment of Inertia
2
m
r
• Quantity  i i = moment of inertia I of the body
i
•
•
•
•
about a specific rotation axis (ri is distance from
object i to the rotation axis)
1 2
Rotational kinetic energy: K R  2 Iw
Note similarity to translational (linear) kinetic energy
(KT = ½ mv2)
Moment of inertia is the rotational analog of mass,
and has to do with how mass is distributed
Consider 2 dumbbells:
Small moment of inertia 
easy to start or stop rotation
Large moment of inertia 
difficult to start or stop rotation
Moment of Inertia
• For continuous distributions
of matter (spheres, rods,
cylinders, etc.), one would
need methods of calculus to
calculate moments of inertia
• Moments of inertia of some
common objects given in
Table 8.1, p. 241 in textbook
• Note that moment of inertia depends on how the
mass is distributed and on the location of the axis of
rotation
Total Mechanical Energy of Rigid Bodies
• Rigid bodies can have rotational and/or translational
(linear) kinetic energy
1 2 (v = velocity of center of mass)
K T  mvcm
cm
2
1 2
K R  Iw
2
• The rigid body could also have gravitational
potential energy
Ug = Mgycm (ycm = vertical position of center of mass)
• In general, total mechanical energy of rigid body is
thus: K1T + K1R + U1 = K2T + K2R + U2
• If work is done by non-conservative forces (like
friction): K1T + K1R + U1 + Wnc = K2T + K2R + U2
Newton’s 2nd Law for Rotation: Dynamics

• Newton’s
Law for linear motion:  F  ma
• Rotational analog for rigid bodies:


2nd
– Notice the strong similarity to the
linear version of Newton’s 2nd Law!
2nd
• Newton’s
Law for translation
of a rigid body:
t  I


 F  macm
– Depends on acceleration of the center of mass
• Remember also the strong similarity between
translational (KT) and rotational (KR) kinetic energy:
1 2
K T  mvcm
2
(For a rolling rigid body, I = Icm)
1 2
K R  Iw
2
Example Problem: Ball Rolling Down an Incline
A ball of mass m and radius R starts from rest at a
height of 2.00 m and
q rolls without slipping down a
30.0° slope as shown. Determine the linear
acceleration of the ball down the incline.
Solution (details given in class):
3.50 m/s2
CQ6: In a lecture demonstration, Dr. Kaye “races” a
solid sphere, a solid disk, and a thin cylindrical shell
(hoop) by releasing them from rest at the top of an
inclined plane of height h. In what order will the objects
reach the bottom of the incline?
A)
B)
C)
D)
E)
F)
Sphere, disk, hoop
Sphere, hoop, disk
Disk, sphere, hoop
Disk, hoop, sphere
Hoop, sphere, disk
Hoop, disk, sphere
Objects Rolling Down Incline Interactive
Example Problem #8.40
An Atwood’s machine consists of
blocks of masses m1 = 10.0 kg and m2 =
20.0 kg attached by a cord running over
a pulley as shown. The pulley is a solid
cylinder with mass M = 8.00 kg and
radius r = 0.200 m. The block of mass
T1
T
2
m2 is allowed to drop, and the cord
turns the pulley without slipping. (a)
Why must the tension T2 be greater
than the tension T1? (b) What is the
acceleration of the system, assuming
the pulley axis is frictionless? (c) Find
the tensions T1 and T2.
Solution (details given in class):
(b) 2.88 m/s2
(c) T1 = 127 N, T2 = 138 N
CQ7: Interactive Example Problem: The
Flywheel Elevator
What is the student’s linear speed after
falling 5.0 m, just before she reaches
the ground?
A) 1.0 m/s
B) 5.0 m/s
C) 10 m/s
D) 25 m/s
(ActivPhysics Online Problem #7.10/7.12, Pearson/Addison Wesley)
Angular Momentum
• The rotational analog of linear momentum is
angular momentum, defined by:
L  Iw
– Note similarity to linear relation p = mv
– Angular momentum is a vector quantity
– Direction is the same as w
 p
• Remembering Newton’s 2nd law written as: F 
 t
we have the following rotational analog:  L
t 
 t
• If the net torque acting on a body is t  L  0
zero, then:
t
– Angular momentum is constant in time, or conserved
– Statement of conservation of angular momentum
– Note similarity to conservation of linear momentum
Example Problem #8.60
A playground merry-go-round of radius 2.00 m has a
moment of inertia I = 275 kgm2 and is rotating about a
frictionless vertical axle. As a child of mass 25.0 kg
stands at a distance of 1.00 m from the axle, the
system (merry-go-round and child) rotates at the rate
of 14.0 rev/min. The child then proceeds to walk
toward the edge of the merry-go-round. What is the
angular speed of the system when the child reaches
the edge?
Solution (details given in class):
1.17 rad/s