Physics 121,Lecture 17,Pg 1
Physics 121,Sections 9,10,11,and 12
Lecture 17
Today’s Topics:
? Homework 7,due Friday Nov,4 @ 6:00 PM.
? Chap,7,# 3,11,20,21,25,27,30,40,
46,47,52,and 68.
? Chapter 8:
? Torque
? Static equilibrium
? C.M,motion
? Rotation
? Moment of inertia
Physics 121,Lecture 17,Pg 2
Torque
? The lever arm is the perpendicular distance from the
axis of rotation to a line drawn along the direction of
the force
? = Fd = FL sin ?
d
?
L
F
?
L
F
?
F cos ?
F sin ?
The sign of the torque is positive if its turning tendency is counterclockwise
and negative if its turning tendency is clockwise (right-hand rule)
Physics 121,Lecture 17,Pg 3
Conditions for Equilibrium
? An object in mechanical equilibrium must satisfy:
1,The net external force must be Zero,SF = 0
2,The net external torque must be Zero,S? = 0
The first condition is translational equilibrium and the second
is rotational equilibrium
If the object is in equilibrium,then the choice of axis of
rotation does not influence the calculation of the net torque;
The location of the axis is arbitrary.
Physics 121,Lecture 17,Pg 4
Objects in Equilibrium
? Walking a Horizontal Beam
A Uniform horizontal 300 N beam,5.00 m long,is
attached to a wall by a pin connection that
allows the beam to rotate,Its far end is supported by
cable that makes an angle of 53.00,If a 600 N person
stand 1.5 m from the wall,find the tension in the cable
and the force exerted by the wall on the beam.
53.00
5.00 m
600 N
Physics 121,Lecture 17,Pg 5
Walking a Horizontal Beam
Fx = Rx - T cos 53.00 = 0
Fy = Ry - T sin 53.00 – 600 N – 300 N = 0
?0= (T sin 53.00)(5.00 m)
- (300 N)(2.5 m)
- (600 N)(1.5 m) = 0
? T = 413 N,Rx = 249 N,Ry = 570 N
53.00
300 N600 N
o
300 N600 N
o
Physics 121,Lecture 17,Pg 6
Statics:
? Objects are at rest (Static) when,
? =? 0F =? 0
When choosing axes about which to calculate torque,we
can be clever and make the problem easy....
No translation No rotation
AND
Physics 121,Lecture 17,Pg 7
Statics,Using Torque
? Now consider a plank of mass M suspended by two strings
as shown,We want to find the tension in each string:
L/2 L/4
Mx cm
T1 T2
Mg
y
x
F =? 0? First use
T1 + T2 = Mg
? This is no longer enough to
solve the problem !
?1 equation,2 unknowns.
? We need more information !!
Physics 121,Lecture 17,Pg 8
Using Torque...
? We do have more information,
?We know the plank is not rotating.
??TOT = 0
? The sum of all torques is zero.
? This is true about any axis
we choose !
L/2 L/4
Mx cm
T1 T2
Mg
y
x
? =? 0
Physics 121,Lecture 17,Pg 9
Using Torque...
? Choose the rotation axis
to be along the z direction
(out of the page) through
the CM:
L/2 L/4
Mx cm
T1 T2
Mg
y
x
? 2 2 4= T L
? The torque due to the string
on the right about this axis is:
? 1 1 2= ?T L
? The torque due to the string on
the left about this axis is:
Gravity exerts no
torque about CM
Physics 121,Lecture 17,Pg 10
Using Torque...
? Since the sum of all
torques must be 0:
L/2 L/4
Mx cm
T1 T2
T L T L2 14 2 0? =
Mg
y
x
T T2 12=
We already found that
T1 + T2 = Mg
T Mg1 13=
T Mg2 23=
Physics 121,Lecture 17,Pg 11
Lecture 17,Act 1
Statics
? A 1kg ball is hung at the end of a rod 1m long,The system
balances at a point on the rod 0.25m from the end holding the
mass,
?What is the mass of the rod?
(a) 0.5 kg (b) 1 kg (c) 2 kg
1kg
1m
Physics 121,Lecture 17,Pg 12
Example Problem,Hanging Lamp
? Your folks are making you help out on fixing up your house,
They have always been worried that the walk around back
is just too dark,so they want to hang a lamp,You go to the
hardware store and try to put together a decorative light
fixture,At the store you find a bunch of massless string
(kind of a surprising find?),a lamp of mass 2 kg,a plank of
mass 1 kg and length 2 m,and a hinge to hold the plank to
the wall,Your design is for the lamp to hang off one end of
the plank and the other to be held to a wall by a hinge,The
lamp end is supported by a massless string that makes an
angle of30o with the plank,(The hinge supplies a force to
hold the end of the plank in place.) How strong must the
string and the hinge be for this design to work?
Physics 121,Lecture 17,Pg 13
Example,Hanging Lamp
1,You need to solve for the forces on the string and the hinge
Use statics equations.
hinge
M
m
L
?
Physics 121,Lecture 17,Pg 14
Example,Hanging Lamp
2,You need to solve for T and components of FH.
Use SF = 0 in x and y.
Use S? = 0 in z,
m
?
T
FH
y
Mg mg
L/2 L/2
FHx
Physics 121,Lecture 17,Pg 15
Hanging Lamp...
3,First use the fact that in both x and y directions:
M
m
L/2
?
Fx
Fy
T
L/2
Mg
mg
y
x
F =? 0
x,T cos? - Fx = 0
y,T sin? + Fy - Mg - mg = 0
? =? 0? Now use in the z direction.
?If we choose the rotation axis to
be through the hinge then the
hinge forces Fx and Fy will not
enter into the torque equation:
0LT s in-mg2LLM g =??
Physics 121,Lecture 17,Pg 16
Hanging Lamp...
3 (Cont.) So we have three equations and three unknowns:
T cos ? + Fx = 0
T sin? + Fy - Mg - mg = 0
LMg + (L/2)mg – LTsin? = 0
Which we can solve to find,
M
m
L/2
?
Fx
Fy
T
L/2
Mg
mg
y
x
( )
?s in
g2mMT +=
( )
?t an
g2mMF
x
+=
F mgy = 12
Physics 121,Lecture 17,Pg 17
Hanging Lamp...
4,Put in numbers
M
m
L/2
?
Fx
Fy
T
L/2
Mg
mg
y
x
( ? ( ?( ? NsmkggmMT 50
)30s in (
/10~5.2
s in
2 2 ?=?=
?( ? ( ?( ?
Nsmkgg
mM
F x 43)30ta n ( /10~5.2ta n 2
2
?=?= ?
NsmkgmgF y 5)/10(~5.021 2 ?==
Physics 121,Lecture 17,Pg 18
Hanging Lamp...
5,Have we answered the question?
Well the string must be strong enough to exert a force of 50 N
without breaking.
We don’t yet have the total force the hinge must withstand.
M
m
L/2
?
Fx
Fy
T
L/2
Mg
mg
y
x
NNFFF yx 43435 2222 ??=?=
Buy a hinge that can take more than 43 N
Physics 121,Lecture 17,Pg 19
System of Particles:
? Until now,we have considered the behavior of very simple
systems (one or two masses).
? But real life is usually much more interesting !
? For example,consider a simple rotating disk.
? An extended solid object (like a disk) can be thought of as a
collection of parts,The motion of each little part depends on
where it is in the object!
Physics 121,Lecture 17,Pg 20
System of Particles,Center of Mass
? How do we describe the,position” of a system made
up of many parts?
? Define the Center of Mass (average position):
?For a collection of N individual pointlike particles
whose masses and positions we know:
R
r
CM
i i
i
N
i
i
N
m
m
= =
=
?
?
1
1
y
x
r1
m1
r3
r
2
r
4
m4
m2
m3RCM
(In this case,N = 4)
Physics 121,Lecture 17,Pg 21
System of Particles,Center of Mass
? If the system is made up of only two particles:
R
r
r r
CM
i i
i
N
i
i
N
m
m
m m
m m
= =
?
?
=
=
?
?
1
1
1 1 2 2
1 2
y
x
r2r
1
m1 m2RCM
r2 - r1
( ? ( ?( ?= ? ? ?
?
m m m
m m
1 2 1 2 2 1
1 2
r r r
( ?R r r rCM mM= ? ?1 2 2 1
where M = m1 + m2
So:
Physics 121,Lecture 17,Pg 22
System of Particles,Center of Mass
? If the system is made up of only two particles:
y
x
r2r
1
m1 m2RCM
r2 - r1
where M = m1 + m2
+
( ?R r r rCM mM= ? ?1 2 2 1
If m1 = m2
the CM is halfway between
the masses.
( ?R r r rCM = ? ?1 2 112
Physics 121,Lecture 17,Pg 23
System of Particles,Center of Mass
? If the system is made up of only two particles:
y
x
r2r
1
m1
m2
RCM
r2 - r1
where M = m1 + m2
+
If m1 = 3m2
the CM is now closer to
the heavy mass.
( ?R r r rCM mM= ? ?1 2 2 1
( ?R r r rCM = ? ?1 2 114
Physics 121,Lecture 17,Pg 24
System of Particles,Center of Mass
? The center of mass is where the system is balanced !
?Building a mobile is an exercise in finding centers of mass.
m1 m2+m1 m2+
Physics 121,Lecture 17,Pg 25
System of Particles,Center of Mass
? We can consider the components of RCM separately:
(,,),,X Y Z
m x
M
m y
M
m z
MCM CM CM
i ii i ii i ii= ?
?
??
?
?
??? ? ?
y
x
r1
m1
r3
r
2
r
4
m4
m2
m3RCM
(In this case,N = 4)
Physics 121,Lecture 17,Pg 26
Example Calculation:
? Consider the following mass distribution:
(24,0)
12m4 24m12)m2(0mM xmX i iiCM =??=?=
6m4 0m12)m2(0mM ymY i iiCM =??=?=
(0,0)
(12,12)
m
2m
m
RCM = (12,6)
Physics 121,Lecture 17,Pg 27
System of Particles,Center of Mass
The location of the center
of mass is an intrinsic
property of the object !!
(it does not depend on where
you choose the origin or
coordinates when
calculating it).
y
xR
CM
? We find that the Center of Mass is at the,center”
of the object.
Physics 121,Lecture 17,Pg 28
Center of Mass
? We can use intuition to find the location of the center of mass for
symmetric objects that have uniform density:
? It will simply be at the geometrical center !
+
CM
+ +
+ + +
Physics 121,Lecture 17,Pg 29
System of Particles,Center of Mass
? The center of mass for a combination of objects is the average
center of mass location of the objects:
+
+
y
x
R2
R1
RCM
m1
m2
+
21
2211
CM
N
1i
i
N
1i
ii
CM
mm
mm
m
m
?
?
=
?
?
=
=
=
RR
R
R
R
R2 - R1
so if we have two objects:
( ?1221 Mm RRR ??=
Physics 121,Lecture 17,Pg 30
System of Particles,Center of Mass
? The center of mass (CM) of an object is where we can freely
pivot that object.
? Gravity acts on the CM of an object (show later)
? If we pivot the object
somewhere else,it will
orient itself so that the
CM is directly below
the pivot.
? This fact can be used to find
the CM of odd-shaped objects.
+
CM
pivot
+ CM
pivotpivot
CM
mg
Physics 121,Lecture 17,Pg 31
System of Particles,Center of Mass
? Hang the object from several pivots and see where the
vertical lines through each pivot intersect !
pivot
pivot pivot
+
CM
? The intersection point must be at the CM.
Physics 121,Lecture 17,Pg 32
Velocity and Acceleration
of the Center of Mass
? If its particles are moving,the CM of a system can also move.
? Suppose we know the position ri of every particle in the system
as a function of time.
R rCM i i
i
N
M m= =?
1
1
??
==
=??=??= N
i
ii
N
i
iiCMCM mMtmMt
11
11 vrRVSo:
??
==
=??=??= N
i
ii
N
i
iiCMCM mMtmMt
11
11 avVAAnd:
? The velocity and acceleration of the CM is just the weighted
average velocity and acceleration of all the particles.
?????? ?= =N1i imM
Physics 121,Lecture 17,Pg 33
Linear Momentum:
P p v= =
= =? ?ii
N
i ii
N m
1 1
? For a system of particles the total
momentum P is the vector sum of
the individual particle momenta:
m Mi i
i
N
CMv V=? =1But we just showed that
P V= M CMSo
?????? ?= =N 1i iiCM mM1 vV
Physics 121,Lecture 17,Pg 34
Linear Momentum:
? So the total momentum of a system of particles
is just the total mass times the velocity of the
center of mass.
? Observe:
? We are interested in so we need to figure
out
?? ===??=??
i
n e ti
i
iiCMCM mMtMt,FaA
VP
t?
?P Fi net
i,?
P V= M CM
Physics 121,Lecture 17,Pg 35
Linear Momentum:
? Suppose we have a system of three particles as shown,Each
particle interacts with every other,and in addition there is an
external force pushing on particle 1.
m1
m3
m2
F13
F31 F
32
F23
F21F12
F1,EXT
( ?
( ?
( ?
F F F F
F F
F F
F
i net
i
EXT
EXT
,,
,
? = ? ?
? ?
? ?
=
13 12 1
21 23
31 32
1
(since the other forces cancel
in pairs...Newton’s 3rd Law)
All of the,internal” forces cancel !!
Only the,external” force matters !!
Physics 121,Lecture 17,Pg 36
Linear Momentum:
? Only the total external force matters !
m1
m3
m2
F1,EXT
E X TT O T
i
E X Tit,,FF
P ==
?
? ?
CME X TT O T Mt A
PF =
?
?=
,
Which is the same as:
Newton’s 2nd law applied to systems !
Physics 121,Lecture 17,Pg 37
Center of Mass Motion,Recap
? We have the following law for CM motion:
? This has several interesting implications:
? It tell us that the CM of an extended object behaves like a
simple point mass under the influence of external forces,
?We can use it to relate F and A like we are used to doing.
? It tells us that if FEXT = 0,the total momentum of the system
can not change.
?The total momentum of a system is conserved if there
are no external forces acting.
CME X T Mt A
PF =
?
?=
Physics 121,Lecture 17,Pg 38
Summary
(with comparison to 1-D kinematics)
Angular Linear
c o n s ta n t=?
?
? = ? 0 ??t
? ? ? ?= ? ?0 0 212t t
co n st a n ta =
v v at= ?0
x x v t at= ? ?0 0 212
And for a point at a distance R from the rotation axis:
x = R? v = ?R a = ?R
Physics 121,Lecture 17,Pg 39
Rotation & Kinetic Energy
? Consider the simple rotating system shown
below,(Assume the masses are attached to the
rotation axis by massless rigid rods).
? The kinetic energy of this system will be the sum
of the kinetic energy of each piece:
r1
r2r3
r4
m4
m1
m2
m3
?
Physics 121,Lecture 17,Pg 40
Rotation & Kinetic Energy...
? So,but vi = ?ri
r1
r2r3
r4
m4
m1
m2
m3
?v4
v1
v3
v2
?=
i
ii vmKE
2
2
1
( ? ?? ==
i
ii
i
ii rmrmKE
222
2
1
2
1 ??
which we write as:
2I
2
1 ?=KE
I = ? m ri i
i
2
Define the moment of inertia
about the rotation axis I has units of kg m2.
Physics 121,Lecture 17,Pg 41
Lecture 17,Act 2
Rotational Kinetic Energy
? I have two basketballs,BB#1 is attached to a 0.1m long rope,
I spin around with it at a rate of 2 revolutions per second,
BB#2 is on a 0.2m long rope,I then spin around with it at a
rate of 2 revolutions per second,What is the ratio of the
kinetic energy of BB#2 to that of BB#1?
A) 1/4 B) 1/2 C) 1 D) 2 E) 4
BB#1 BB#2
Physics 121,Lecture 17,Pg 42
Rotation & Kinetic Energy...
? The kinetic energy of a rotating system looks similar to that of a
point particle:
Point Particle Rotating System
2I
2
1 ?=KE
I = ? m ri i
i
2
2
2
1 mvKE =
v is,linear” velocity
m is the mass.
? is angular velocity
I is the moment of inertia
about the rotation axis.
Physics 121,Lecture 17,Pg 43
Recap of today’s lecture
? Homework 7,due Friday Nov,4 @ 6:00 PM.
?Chap,7,# 3,11,20,21,25,27,30,40,
46,47,52,and 68.
? Chapter 8:
?Torque
?Static equilibrium
?C.M,motion
?Rotation
?Moment of inertia