Transcript Torque - cloudfront.net

Torque and Center of Mass
Julius Sumner Miller
Center of Mass:
The center of mass (or mass center) is the mean
location of all the mass in a system.
The motion of an object can be characterized by this
point in space. All the mass of the object can be
thought of being concentrated at this location. The
motion of this point matches the motion of a point
particle.
Finding the Center of Mass:
Uniform geometric figures have the center of mass
located at the geometric center of the object.
Note that the center of
mass does not have to be
contained inside the
volume of the object.
Collections of Point Masses:
The center of mass for a collection of point masses
is the weighted average of the position of the
objects in space.
Each object will have a position in space. The
center of mass is found as:
xcm
m1 x1  m2 x2  m3 x3  

m1  m2  m3  
ycm
m1 y1  m2 y2  m3 y3  

m1  m2  m3  
Example #1: A 10.0 kg mass sits at the origin, and a 30.0 kg mass rests at
the 12.0 m mark on the x – axis. (a) Find the center of mass for this
system.
m1  10.0 kg
m2  30.0 kg
x1  0
x2  12.0 m
xcm
xcm
m1 x1  m2 x2  m3 x3  

m1  m2  m3  

10.0 kg 0  30.0 kg 12.0 m 

10.0 kg  30.0 kg
 9.00 m
(b) Find the center of mass for this system relative to the mass at the right.
m1  10.0 kg
m2  30.0 kg
x1  12.0 m
x2  0
xcm
xcm
m1 x1  m2 x2  m3 x3  

m1  m2  m3  

10.0 kg  12.0 m   30.0 kg 0

10.0 kg  30.0 kg
 3.00 m
Although numerically different, it is the same point in space relative to
the masses…
Example #2: A 10.0 cm long wire has a mass of 4.00 grams. This wire is
bent into an “L” shape that measures 6.00 cm by 4.00 cm, as shown below.
Determine the center of mass for this object.
Center of Mass of Both
Sticks together.
Mass 1
Mass 2
2 cm
Example #2: A 10.0 cm long wire has a mass of 4.00 grams. This wire is
bent into an “L” shape that measures 6.00 cm by 4.00 cm, as shown below.
Determine the center of mass for this object.
Treat as two objects:
6 cm object:
x1,cm  0
y1,cm  3.00 cm
 4.00 g 
6.00 cm  2.40 g
m1  
 10.0 cm 
4 cm object:
x2,cm  2.00cm
y2,cm  0
 4.00 g 
4.00 cm  1.60 g
m2  
 10.0 cm 
Center of Mass
X=
Y=
Mass 1
2.40 g
Mass 2
1.60 g
2 cm
xcm
xcm
m1 x1  m2 x2  m3 x3  

m1  m2  m3  

2.40 g 0  1.60 g 2.00 cm 

4.00 g
ycm
ycm
 0.800 cm
m1 y1  m2 y2  m3 y3  

m1  m2  m3  

2.40 g 3.00 cm   1.60 g 0 

4.00 g
 1.80 cm
Center of Mass
X = .800 cm
Y = 1.80 cm
Mass 1
2.40 g
Mass 2
1.60 g
2 cm
Example #3: Determine the center of mass of the following masses, as
measured from the left end. Assume the blocks are of the same density.
This is homework.
lo
x1 
2
m1  M
x2  2lo
m2  8M
m3  27 M
x3  4.5lo
xcm
xcm
m1 x1  m2 x2  m3 x3  

m1  m2  m3  
 lo 
M    8M 2lo   27M 4.5lo 
2


36M
xcm
23
 lo
6
Torque
Torque is the rotational equivalent of force. A
torque is the result of a force applied to an object
that tries to make the object rotate about some pivot
point.

F  applied force
Equation of Torque:
 distance from pivot to
r  applied force

pivot point

angle between direction
of force and pivot
distance.
  torque  rF sin 
Note that torque is maximum
when the angle  is 90º.
The units of torque are
Nm or newton · meter
The torque is also the product of the distance from
the pivot times the component of the force
perpendicular to the distance from the pivot.
  torque  rF sin   rF
The torque is also the product of the force times the
lever arm distance, d.
  torque  rF sin   Fd
Example #4: Calculate the torque for the force shown below.
  rF sin 
 2.00 m300 N sin 60.0
  520 Nm
Example #5: Calculate the total torque about point O on the figure below.
Take counterclockwise torques to be positive, and clockwise torques to be
negative.
60degrees
  rF sin 
 net  2.0 m25 N sin 60  4.0 m10 N sin 20
 net  29.6 Nm
Example #6: The forces applied to the cylinder below are F1 = 6.0 N, F2 =
4.0 N, F3 = 2.0 N, and F4 = 5.0 N. Also, R1 = 5.0 cm and R2 = 12 cm.
Determine the net torque on the cylinder.
 net  F1 R2 sin 90
 F2 R2 sin 90
 F3 R1 sin 90
 F4 R2 sin 0
0
 net  6.0 N 0.12 m  4.0 N 0.12 m  2.0 N 0.050 m
 net  0.14 Nm
Static Equilibrium:
Torque and Center of Mass
Static Equilibrium:
Static equilibrium was touched on in the
unit of forces. The condition for static
equilibrium is that the object is at rest.
Since the object is not moving, it is not
accelerating. Thus the net force is zero.
Shown at right is a typical example from
that unit: Find the force of tension in
each rope.
A new condition can now be added into
this type of problem: Since the object is
at rest, it must not be rotating, as that
would also require an acceleration.
If there is no rotation, there must not be a rotational acceleration. Thus the
net torque must be zero. This is an application of Newton’s 2nd law to
rotational motion.
 net  0
If the net torque is zero, then all the counterclockwise (ccw) torques must
balance all the clockwise (cw) torques.

ccw
  cw
If there is no rotation, where is the pivot point for calculating torque?
Answer: The pivot point can be put anyplace you want!
Hint: Put the pivot point at one of the unknowns. This
eliminates the unknown from the torque equation.
Example #1: A meter stick has a mass of 150 grams and has its center of
mass located at the 50.0 cm mark. If the meter stick is supported at each of
its ends, then what forces are needed to support it?
F1  ?
100 cm
50 cm
F2  ?
mms g
Show that the two forces are equal through torque. Put the pivot point at
the left end.
Force F1 does not contribute to torque. {force applied to pivot point!}
Force F2 makes a ccw torque.
 2  F2 100 cmsin 90
Force mmsg makes a cw torque.
 ms  mms g 50.0 cmsin 90
Balance the net ccw and cw torque:
F2 100 cm  mms g 50.0 cm
F2  12 mms g

1
2
0.150 kg9.80 m s 
F2  0.735 N
The other unknown must also equal half the weight, so:
F1  0.735 N
2
Example #2: Suppose the meter stick above were supported at the 0 cm
mark (on the left) and at the 75 cm mark (on the right). What are the forces
of support now?
F1  ?
75 cm
50 cm
F2  ?
mms g
Find the two unknown forces through torque. Put the pivot point at the left
end.
Force F1 does not contribute to torque. {force applied to pivot point!}
Force F2 makes a ccw torque.
 2  F2 75.0 cmsin 90
Force mmsg makes a cw torque.
 ms  mms g 50.0 cmsin 90
Balance the net ccw and cw torque:
F2 75.0 cm  mms g 50.0 cm
F2  23 mms g

2
3
0.150 kg9.80 m s 
F2  0.980 N
If force F2 holds 2/3 the weight, then F1 must hold the
remaining 1/3 of the weight.
F1  0.490 N
2
Example #3: A meterstick is found to balance at the 49.7-cm mark when
placed on a fulcrum. When a 50.0-gram mass is attached at the 10.0-cm
mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is
the mass of the meter stick?
madded g
39.2 cm  10.0 cm
29.2 cm
mms g
The meterstick behaves
as if all of its mass was
concentrated at its center of mass.
49.7 cm  39.2 cm
10.5 cm
Calculate the torque about the pivot point. The support force of the
fulcrum will not contribute to the torque in this case.
Force maddedg makes a ccw
torque.
 added  madded g 29.2 cmsin 90
Force mmsg makes a cw torque.
 ms  mms g 10.5 cmsin 90
Balance the net ccw and cw torque:
mms g 10.5 cm  madded g 29.2 cm
 29.2 cm 

mms  50.0 grams 
 10.5 cm 
mms  139 g
Example #4: A window washer is standing on a scaffold supported by a
vertical rope at each end. The scaffold weighs 200 N and is 3.00 m long.
What is the tension in each rope when the 700-N worker stands 1.00 m
from one end?
F1
1.00 m
1.50 m
F2
200 N
700 N
Put the pivot point on the left end. The force F1 does not
contribute torque. Solve for F2.
F2 3.00 m  700 N 1.00 m  200 N 1.50 m
F2  333 N
F1
1.00 m
1.50 m
200 N
700 N
Solve F1 from Newton’s laws:
F1  F2  700 N  200 N
F1  567 N
F2
Example #5: A cantilever is a beam that extends beyond its supports, as
shown below. Assume the beam has a mass of 1,200 kg and that its center
of mass is located at its geometric center. (a) Determine the support forces.
Put the pivot point at the left
end and balance the torques.
A  0
 ccw   B  FB 20.0 msin 90
 cw   mg  mbeam g 25.0 m sin 90
Balance the net ccw
and cw torque:
FB 20.0 m  mbeam g 25.0 m
1,200 kg9.80 m s 25.0 m
FB 
20.0 m
2
 14,700 N
Solve FA from Newton’s laws:
FA  FB  1,200 kg 9.80 m s 2 
= 11,760 N
FA  2,940 N
The fact FA is negative means that the
force really points downwards.
When the wrong direction is chosen for a force, it just
comes out negative at the end.
Static Equilibrium:
Day #2
Example #6: Calculate (a) the tension force FT in the wire that supports
the 27.0 kg beam shown below.
Put the pivot point at the left end. The wall
support does not contribute to torque.
 ccw   T  FT L sin  

Note that  and 40° are supplements,
so it does not matter which is used in
the sine function.
sin    sin 180   
L
1
2
L
 ccw   T  FT L sin 40.0
Beam length is L.
 cw
L
  beam  mbeam g sin 90.0
2
Balance the net ccw and cw torque:
L
FT L sin 40.0  mbeam g
2
FT
27.0 kg 9.80 m s 

2 sin 40.0
2
FT  206 N
(b) Determine the x and y components to the force exerted by the wall.
Balance forces in each component direction.
Fx  FT cos40.0
Fy
 158 N
Fy  FT sin 40.0  mbeam g
Fx
Fy  132 N
Example #7: A shop sign weighing 245 N is supported by a uniform 155 N
beam as shown below. Find the tension in the guy wire and the horizontal
and vertical forces exerted by the hinge on the beam.
Put the pivot point at the left end. The wall
support does not contribute to torque.
 ccw   T  FT 1.35 msin 35.0
 cw   beam   sign
 cw
 1.70 m 
 mbeam g 
  msign g 1.70 m 
 2 
Balance the net ccw and cw torque:
FT 1.35 msin 35.0
 1.70 m 
 155 N 
  245 N 1.70 m 
 2 
FT  708 N
Fx  FT cos35.0
Fy
Fx
 580 N
Fy  FT sin 35.0
 mbeam g  msign g
Fy  6.1 N
The fact Fy is negative means that the
force really points downwards.
Example #8: A person bending forward to lift a load “with his back” (see
figure below) rather than “with his knees” can be injured by large forces
exerted on the muscles and vertebrae. The spine pivots mainly at the fifth
lumbar vertebra, with the principal supporting force provided by the erector
spinalis muscle in the back. To see the magnitude of the forces involved, and
to understand why back problems are common among humans, consider the
model shown in the figure below 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 two-thirds of the way up the spine, maintains the
position of the back. Let the distance from the hinge point to the weight be
distance, L. 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.
Put the pivot point at the left end. The
hip support does not contribute to
torque.
2L
3
L
L
2
 2L 
L
T   sin 12.0  350 N    200 N L 
 3 
2
 350 N 

  200 N 
3 2 
T  
sin 12.0
2
Rx  T cos12.0
 2646 N
T  2705 N
 610 lb
 600 lb
Ry  T sin 12.0  350 N  200 N
Ry  12.5 N
Example #9: A person in a wheelchair wishes to roll up over a sidewalk
curb by exerting a horizontal force to the top of each of the wheelchair’s
main wheels (Fig. P8.81a). The main wheels have radius r and come in
contact with a curb of height h (Fig. P8.81b). (a) Assume that each main
wheel supports half of the total load, and show that the magnitude of the
minimum force necessary to raise the wheelchair from the street is given by
mg 2 Rh  h 2
F
2  2R  h 
where mg is the combined weight of the wheelchair and person. (b) Estimate
the value of F, taking mg = 1 400 N, R = 30 cm, and h = 10 cm.

ccw
  cw
Minimum F to lift comes when
the normal force becomes zero.
Balance torques about contact
point A:
Find the Clockwise Torque
2F
Φ
X
2R-h
Φ
A
h
Ƭcw = 2F X sinΦ There are two hands at work here
Ƭcw = 2F X 2R-h
X
Ƭcw = 2F (2R-h)
Find the Counterclockwise Torque
R
R-h
A
d
h
Ƭccw = F X sinΦ
d  R   R  h
Ƭccw = mg d
d  2Rh  h
2
Tccw  mg 2Rh  h2
2
2
Pythagorean Theorem

Ƭcw = 2F(2R-h)
cw
  ccw
Tccw  mg 2Rh  h2
2F (2R  h)  mg 2Rh  h
mg 2 Rh  h
F
2  2R  h 
2
2
F = 313 N = 64 pounds.
Do you see why ramps are
needed around town?
Example #10: A circular disk 0.500 m in diameter, pivoted about a
horizontal axis through its center, has a cord wrapped around its rim. The
cord passes over a frictionless pulley P and is attached to an object that
weighs 240 N. A uniform rod 2.00 m long is fastened to the disk, with one
end at the center of the disk. The apparatus is in equilibrium, with the rod
horizontal. (a) What is the weight of the rod?
240 N
1.00 m
240 N 0.250 m
 mrod g 1.00 m
mrod  6.12 kg
mrod g
(b) What is the new equilibrium direction of the rod when a second object
weighing 20.0 N is suspended from the other end of the rod, as shown by the
broken line in the image below? That is, what angle does the rod then make
with the horizontal?
240 N 0.250 m
 mrod g 1.00 mcos
 20 N 2.00 mcos
240 N
mrod g
20 N
cos  0.6
  53.1
Inertia and Rotary Motion
Moment of Inertia
Handout
HW #8
Final Schedule: Mosig’s Class 2013
Monday
Tuesday
Wednesday
Thursday
Friday
12/9
12/10
12/11
12/12
12/13
Notes
Finish notes
Study / Tower practice day
Study / Tower practice day
Final Exam {covers current unit}
Monday
Tuesday
Wednesday
Thursday
Friday
12/16
12/17
12/18
12/19
12/20
Towers p. 0, 1, & 6
Towers p. 2 & 3
Towers p. 4 & 5
Elf Dance / Non – Academic day
No School
Definition: Inertia is the ability of an object to resist a change in its motion.
Inertia was introduced earlier in the Force unit. For straight line motion, the
inertia of an object is measured through mass: The more massive an object,
the more it is able to resist changes to its (straight line) motion.
Force and acceleration were related through inertia:
Fnet  ma
There is an equivalent to inertia in rotary motion. Here, inertia would try to
resist a change to the angular motion. This form of inertia will depend on
mass, just as before, but it will also depend on the distribution of the mass.
Demonstration:
When the mass is distributed close to the center of rotation, the object is
relatively easy to turn. When the mass is held much further away, it is more
difficult to rotate the object.
For example, if you had a ring (hoop) and a disk with the same radius and
same mass, the ring would show more resistance to rotation than would the
disk. The disk has mass uniformly distributed across its body, so some of
the mass is near the center of rotation. The ring has more mass concentrated
at the outside edge of the body than does the disk, so it will show more
inertia (even though the mass and radius are the same).
The dependence of the inertia on mass and distribution can be built up
through the kinetic energy of an object moving through a circular path.
The kinetic energy of a mass m moving at a
speed v around the circle is:
KE  mv
1
2
2
Let the circle have a radius r, and let the angular
speed of the mass be w. Write the kinetic
energy in terms of the angular speed:
KE  12 mr w
2
2
Define the moment of inertia I of this point mass to be:
Then:
KE  Iw
1
2
2
vt  rw
I  mr 2
This is now the definition.
Now build up to a more complicated object:
The total kinetic energy becomes:
KE  m v
2
1 1
1
2
 m2v2  m3v3
2
1
2
1
2
2
Mass m1 travels in a circle of radius r1, etc. All masses have the same
angular velocity, w. Write the kinetic energy in terms of this:
KE  m r w  m2 r2 w  m3r3 w
1
2
KE 
1
2
2
11
m r
2
11
2
2
1
2
2
1
2

2
2
 m2r2  m3r3 w 2  12 Iw 2
2
2
In general, the moment of inertia is found as:
I   mi ri
2
The farther the mass is from the center of rotation, the higher the moment of
inertia.
Example #1: (a) If m = 2.00 kg and d = 0.500 m for the image below,
determine the moment of inertia of this group of objects.
I  m r  m2 r2  m3r3
2
11
2
2
I  md  m2d   m3d 
2
2
I  14md
2
2
2



 14 2.00 kg 0.500 m
 7.00 kg  m2
(b) If this apparatus is rotating at 4.00 rad/s, what is its kinetic energy?
KE  12 Iw 2
KE 
1
2
7.00 kg  m 4.00 
2
KE  56.0 J
rad 2
s
For more complicated objects, the
moment of inertia is tabulated
below. The calculations of these
are complex and beyond the scope
of the class.
I   mi ri
2
Turns into:
I   r 2 dm
Eeek!!! Calculus! Run for
your lives!
Example of calculating the rotational inertia of a solid ball. You are not
responsible for knowing these calculations…
Newton’s 2nd Law and Rotational Motion:
There is an equivalent to Newton’s 2nd law for rotational motion.
 net  I
Fnet  ma
This can be combined with the kinematics equations from earlier in the unit
to solve uniform motion problems;
w  wo   t
  wot   t
1
2
w  wo  2
2
2
2
w  wo
2


t
Example #2: A force of 225.0 N is applied to the edge of a disk that can
spin about its center. The disk has a mass of 240 kg and a diameter of
3.20 m. If the force is applied for
24.0 s, how fast will the disk be
turning if it starts from rest?
I disk  12 mr 2
 net  I
  rF sin 90
mr 
rF    I 
2
2
mr 
rF 
2
2
2F

mr
2225 N 

240 kg 1.60 m 
w  wo   t

 1.17 rad
s2

24.0 s
 0  1.17 rad
s2
w  28.1 rad
s
Example #3: Variation of the Atwood’s machine. A 6.00 kg mass is tied
via a cord to a heavy wheel (solid disk) with mass 20.0 kg and radius
20.0 cm. What is the acceleration of the hanging mass downwards?
I disk  12 m1r 2
T
The tangential acceleration of the disk
is equal to the linear acceleration of the
falling mass.
a  r
T
m2
m2 g
The net torque on the disk gives one
equation:
 net
a
 Tr  I  m1r  
r
T  12 m1a
1
2
2
Next use Newton’s 2nd law on the falling mass:
m2a  m2 g  T
Combine the equations:

1
2
m1a  T
m2 a  12 m1a  m2 g  T  T
 m2 g
a
m2  12 m1
Inertia and Rotary Motion
Day #2
Handout
HW #8
Rolling and Energy Conservation.
When an object rolls along the ground, the tangential speed of the outside
edge of the object is the same as the speed of the center of mass of the object
relative to the ground.
The rolling object has two parts to its motion. First is the motion of the
center of mass, and second is the rotation around the center of mass. The
total kinetic energy is the sum of the kinetic energy associated with each
part.
The kinetic energy associated with the center of mass moving in a straight
line is given by the term:
KEcm  mvcm 
1
2
2
The portion associated with rotating around the center of mass is:
KE  12 I cmw 2
Icm is the moment of inertia of the object about its
center of mass. Refer to the given table for values.
Example #4: A solid ball of radius 10.0 cm and mass 10.0 kg rolls at a given
speed vo of 5.00 m/s. (a) What is the total kinetic energy of this rolling ball?
KEtot  12 mvcm 
2
For a rolling ball:
I cm  mr
2
5
2
KEtot  mvcm  
2
1
2
 12 I cmw 2
&
1 2
2 5
 vcm 
mr 

 r 
KEtot    mvcm  
1
2
1
5
vt vcm
w 
r
r
2
2
2
7
10
mvcm 
2
KEtot 
7
10
10.0 kg5.00 
m 2
s
KEtot  175 J
(b) What percentage of the total kinetic energy is rolling?
KEroll

KEtot
Iw 2
2
7
10 mvcm 
1
2

 vcm 
mr 

 r 
2
7
10 mvcm 
1 2
2 5
2
2
2

7
Example #5: Four different objects are placed at the top on an incline, as
shown below. A point particle can slide down without friction. The other
three objects will roll down the incline. In what order will the objects reach
the bottom, from fastest to slowest? (a) What is the speed of the sliding
point particle when it reaches the
bottom?
Energy conservation!
Etop  Ebottom
KEtop  PEtop  KEbottom  PEbottom
0
mgh  12 mv 2
0
v point  2 gh
(b) Solve for the speed of the sphere (solid ball) at the bottom.
I cm  52 mr 2
Energy conservation!
Etop  Ebottom
vcm
w
r
mgh  12 mv 2  12 Iw 2
Note that there is a fixed starting energy, and this is split between linear
motion and rotation. The rolling objects are slower!
2
mgh  mv 
1
2
2
v
10
7
1 2
2 5
gh
 v   7 mv 2
mr  
10
r
2
(c) Solve for the speed of the hoop at the bottom.
I cm  mr 2
Energy conservation!
Etop  Ebottom
vcm
w
r
mgh  12 mv 2  12 Iw 2
2
v
2

mv
mgh  mv  mr  
r
1
2
2
1
2
2
v  gh
The hoop has the slowest speed, and thus takes the longest to reach the
bottom. The disk will be between the ball and the hoop.
t point  tball  t disk  t hoop
Example #6: Variation of the Atwood’s machine. A 6.00 kg mass is tied via
a cord to a heavy wheel (solid disk) with mass 20.0 kg and radius 20.0 cm.
(a) How fast will the mass be traveling after it falls a distance of h = 4.00 m
downwards?
Energy conservation!
Etop  Ebottom
m2 gh  12 m2v 2  12 Iw 2
m2
I cm  12 m1r 2
Speed of falling mass equals
tangential speed of disk:
vt vobject
w 
r
r
m2 gh  12 m2v 2  12 Iw 2
m2 gh  m2 v 
2
1
2
m2 gh 
2m2 gh
v
m2  12 mdisk
1
2

1 1
2 2

v
mdisk r  
r
2
2
m2  12 mdisk v 2
26.00 kg 9.80 m s 2 4.00 m

6.00 kg  12 20.0 kg 
v  5.42 ms
(b) Solve for the acceleration of the hanging mass:
m2 gh 
1
2
m2  12 mdisk v 2
m2 g
v 2
h
1
m2  2 mdisk 
2
Shortcut: Remember the
kinematics equation…
m2 g
a
m2  12 mdisk
v  vo  2ax
2
2
Same result as before!
Momentum and Rotary
Motion
Handout
HW #9