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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:
321
332211
mmm
xmxmxmxcm
321
332211
mmm
ymymymycm
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.
01 x
kgm 0.101
mx 0.122
kgm 0.302
321
332211
mmm
xmxmxmxcm
kgkg
mkgkgxcm 0.300.10
0.120.3000.10
m00.9
(b) Find the center of mass for this system relative to the mass at the right.
mx 0.121
kgm 0.101
02 x
kgm 0.302
321
332211
mmm
xmxmxmxcm
kgkg
kgmkgxcm 0.300.10
00.300.120.10
m00.3
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.
Mass 1
Mass 2
2 cm
Center of Mass of BothSticks together.
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:
gcmcm
gm 40.200.6
0.10
00.41
0,1 cmx
cmy cm 00.3,1
4 cm object:
gcmcm
gm 60.100.4
0.10
00.42
cmx cm 00.2,2 0,2 cmy
Mass 12.40 g Mass 2
1.60 g
2 cm
Center of MassX =Y =
321
332211
mmm
xmxmxmxcm
g
cmggxcm 00.4
00.260.1040.2 cm800.0
321
332211
mmm
ymymymycm
g
gcmgycm 00.4
060.100.340.2 cm80.1
Mass 12.40 g Mass 2
1.60 g
2 cm
Center of MassX = .800 cmY = 1.80 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.
Mm 1
21olx
Mm 82
olx 22 Mm 273 olx 5.43
321
332211
mmm
xmxmxmxcm
M
lMlMl
Mx
ooo
cm 36
5.427282
ocm lx6
23
TorqueTorque 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.
Equation of Torque:
pivot point
r distance from pivot to
applied force
F
applied force
angle between direction of force and pivot distance.
sinrFtorque
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.
rFrFtorque sin
The torque is also the product of the force times the lever arm distance, d.
FdrFtorque sin
Example #4: Calculate the torque for the force shown below.
sinrF 0.60sin30000.2 Nm
Nm520
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.
sinrF
20sin100.460sin250.2 NmNmnet
Nmnet 6.29
60degrees
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.
90sin21 RFnet
90sin22 RF
90sin13 RF
0sin24 RF0
mNmNmNnet 050.00.212.00.412.00.6
Nmnet 14.0
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.
0net
If the net torque is zero, then all the counterclockwise (ccw) torques must balance all the clockwise (cw) torques.
cwccw
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?
?1 F ?2 F
gmms
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. 90sin10022 cmF
cm100
Force mmsg makes a cw torque.
cm50
90sin0.50 cmgmmsms
Balance the net ccw and cw torque:
cmF 1002 cmgmms 0.50
gmF ms21
2 280.9150.021
smkg
NF 735.02
The other unknown must also equal half the weight, so:
NF 735.01
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?
?1 F ?2 F
gmms
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. 90sin0.7522 cmF
cm75
Force mmsg makes a cw torque.
cm50
90sin0.50 cmgmmsms
Balance the net ccw and cw torque:
cmF 0.752 cmgmms 0.50
gmF ms32
2 280.9150.032
smkg
NF 980.02
If force F2 holds 2/3 the weight, then F1 must hold the remaining 1/3 of the weight.
NF 490.01
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?
gmmsgmadded The meterstick behaves as if all of its mass was concentrated at its center of mass.cmcm 0.102.39
cm2.29 cmcm 2.397.49
cm5.10Calculate 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.
90sin2.29 cmgmaddedadded
Force mmsg makes a cw torque. 90sin5.10 cmgmmsms
Balance the net ccw and cw torque:
cmgmcmgm addedms 2.295.10
cm
cmgramsmms 5.10
2.290.50
gmms 139
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?
N200
m50.1
N700
m00.11F 2F
Put the pivot point on the left end. The force F1 does not contribute torque. Solve for F2.
mNmNmF 50.120000.170000.32
NF 3332
N200
m50.1
N700
m00.11F 2F
Solve F1 from Newton’s laws:
NNFF 20070021
NF 5671
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.
0A
90sin0.20 mFBBccw
90sin0.25 mgmbeammgcw
Balance the net ccw and cw torque: mFB 0.20 mgmbeam 0.25
m
mkgF s
m
B 0.20
0.2580.9200,1 2
N700,14
NFA 940,2
Solve FA from Newton’s laws:
280.9200,1s
mBA kgFF
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.
= 11,760 N
Static Equilibrium:Day #2
Example #6: Calculate (a) the tension force FT in the wire that supports the 27.0 kg beam shown below.
Beam length is L.
LL2
1
Put the pivot point at the left end. The wall support does not contribute to torque.
sinLFTTccw
Note that and 40° are supplements, so it does not matter which is used in the sine function.
180sinsin
0.40sinLFTTccw
0.90sin2
Lgmbeambeamcw
Balance the net ccw and cw torque:
0.40sinLFT 2
Lgmbeam
0.40sin2
80.90.27 2sm
T
kgF
NFT 206
(b) Determine the x and y components to the force exerted by the wall.
yF
xF
Balance forces in each component direction.
0.40cosTx FF N158
gmFF beamTy 0.40sin
NFy 132
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.
0.35sin35.1 mFTTccw
signbeamcw
mgmm
gm signbeamcw 70.12
70.1
Balance the net ccw and cw torque:
0.35sin35.1 mFT
mNm
N 70.12452
70.1155
NFT 708
yF
xF
0.35cosTx FF N580
0.35sinTy FF
NFy 1.6
gmgm signbeam
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.
L
2
L
3
2L
Put the pivot point at the left end. The hip support does not contribute to torque.
LNL
NL
T 2002
3500.12sin3
2
0.12sin
2002
350
2
3N
N
TNT 2705
lb610
0.12cosTRxN2646 lb600
NNTRy 2003500.12sin
NRy 5.12
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
22
2 2
mg Rh hF
R 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.
cwccw
Minimum F to lift comes when the normal force becomes zero.Balance torques about contact point A:
Φ
Φ
Find the Clockwise Torque
2R-h
h
X
Ƭcw = 2F X sinΦ There are two hands at work here
Ƭcw = 2F X 2R-h X
Ƭcw = 2F (2R-h)
2F
A
Find the Counterclockwise Torque
R
h
R-h
d
Ƭccw = F X sinΦ
Ƭccw = mg d
A
22d R R h Pythagorean Theorem
22d Rh h
22Tccw mg Rh h
Ƭcw = 2F(2R-h)
22Tccw mg Rh h
22 (2 ) 2F R h mg Rh h
22
2 2
mg Rh hF
R h
F = 313 N = 64 pounds.Do you see why ramps are needed around town?
cw ccw
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?
N240
gmrod
m00.1
mN 250.0240
mgmrod 00.1
kgmrod 12.6
(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?
N240
gmrod
N20
mN 250.0240
cos00.1 mgmrod
cos00.220 mN
6.0cos
1.53
Inertia and Rotary MotionMoment of Inertia
Handout
HW #8
Final Schedule: Mosig’s Class 2013
Monday 12/9 NotesTuesday 12/10 Finish notesWednesday 12/11 Study / Tower practice dayThursday 12/12 Study / Tower practice dayFriday 12/13 Final Exam {covers current unit}
Monday 12/16 Towers p. 0, 1, & 6Tuesday 12/17 Towers p. 2 & 3Wednesday 12/18 Towers p. 4 & 5Thursday 12/19 Elf Dance / Non – Academic dayFriday 12/20 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: maFnet
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:
221 mvKE
Let the circle have a radius r, and let the angular speed of the mass be . Write the kinetic energy in terms of the angular speed:
2221 mrKE
Define the moment of inertia I of this point mass to be:
Then: This is now the definition.
2mrI
221 IKE
rvt
Now build up to a more complicated object:
The total kinetic energy becomes:
2112
1 vmKE 2
33212
2221 vmvm
Mass m1 travels in a circle of radius r1, etc. All masses have the same angular velocity, . Write the kinetic energy in terms of this:
22332
122222
122112
1 rmrmrmKE
22122
332
222
1121 IrmrmrmKE
In general, the moment of inertia is found as: 2iirmI
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.
233
222
211 rmrmrmI
222 32 dmdmmdI
214mdI
2500.000.214 mkg 200.7 mkg
(b) If this apparatus is rotating at 4.00 rad/s, what is its kinetic energy?
221 IKE
2221 00.400.7 s
radmkgKE
JKE 0.56
For more complicated objects, the moment of inertia is tabulated below. The calculations of these are complex and beyond the scope of the class.
2iirmI
Turns into:
2I r dmEeek!!! 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.
maFnet Inet
This can be combined with the kinematics equations from earlier in the unit to solve uniform motion problems;
to
221 tto
222o
to
2
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?
221 mrIdisk
Inet
90sinrF
2
2 mrIrF
2
2mrrF
mr
F2
mkg
N
60.1240
2252 217.1
srad
to ss
rad 0.2417.10 2
srad1.28
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?
T
T
212
1 rmIdisk
2m
gm2
The tangential acceleration of the disk is equal to the linear acceleration of the falling mass.
ra The net torque on the disk gives one equation:
r
armITrnet
212
1
amT 121
Next use Newton’s 2nd law on the falling mass:
Tgmam 22
Combine the equations: Tam 121
TTgmamam 2121
2
121
2
2
mm
gma
Inertia and Rotary MotionDay #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:
2
21
cmcm vmKE
The portion associated with rotating around the center of mass is:
221 cmIKE 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?
2
21
cmtot vmKE 221 cmI
For a rolling ball: &252 mrIcm
r
v
r
v cmt
2
252
212
21
r
vmrvmKE cm
cmtot
2
1072
51
21
cmcmtot vmvmKE
2
107 00.50.10 s
mtot kgKE
JKEtot 175
(b) What percentage of the total kinetic energy is rolling?
2107
221
cmtot
roll
vm
I
KE
KE 2
107
22
52
21
cm
cm
vm
rv
mr
7
2
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!
bottomtop EE
bottombottomtoptop PEKEPEKE
0 0
221 mvmgh ghvpoint 2
(b) Solve for the speed of the sphere (solid ball) at the bottom.
Energy conservation!
bottomtop EE
2212
21 Imvmgh
Note that there is a fixed starting energy, and this is split between linear motion and rotation. The rolling objects are slower!
252 mrIcm
r
vcm
22
52
212
21
r
vmrmvmgh 2
107 mv
ghv 710
(c) Solve for the speed of the hoop at the bottom.
Energy conservation!
bottomtop EE
2212
21 Imvmgh
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.
2mrIcm
r
vcm
22
212
21
r
vmrmvmgh 2mv ghv
hoopdiskballpoint tttt
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!
bottomtop EE
2212
221
2 Ivmghm
212
1 rmIcm
Speed of falling mass equals tangential speed of disk:
r
v
r
v objectt
2m
2212
221
2 Ivmghm
2
221
212
221
2
r
vrmvmghm disk
221
221
2 vmmghm disk
diskmm
ghmv
21
2
22
kgkg
mkgs
m
0.2000.6
00.480.900.62
21
2
smv 42.5
(b) Solve for the acceleration of the hanging mass:
221
221
2 vmmghm disk
hmm
gmv
disk21
2
22 2
Shortcut: Remember the kinematics equation… xavv o 222
diskmm
gma
21
2
2
Same result as before!
Momentum and Rotary Motion
Handout
HW #9
