201: Physics Lecture 15
Reprise on collisions and rocket motion Equilibrium Center of mass Dynamics of center of mass
3/27/08
Physics 201, UW-Madison
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Question
Two balls--one heavy (basketball) and the other light (tennis ball)--are stacked on each other, lighter ball on the top of the heavier one, and dropped to the ground. The speed of the recoiling tennis ball will be: Smaller than Same as CORRECT Larger than the speed attained by it if dropped by itself at the same height
3/27/08
Physics 201, UW-Madison
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Rocket Motion
"Professor Goddard does not know the relation between action and reaction and the need to have something better than a vacuum against which to react. He seems to lack the basic knowledge ladled out daily in high schools." -1921 New York Times editorial about Robert Goddard.
"Correction: It is now definitely established that a rocket can function in a vacuum. The 'Times' regrets the error." 1969 New York Times
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Physics 201, UW-Madison
Rocket Propulsion
The operation of a rocket depends on the law of conservation of momentum as applied to a system, where the system is the rocket plus its ejected fuel This is different than propulsion on the earth where two objects exert forces on each other road on car train on track The rocket is accelerated as a result of the thrust of the exhaust gases This represents the inverse of an inelastic collision Momentum is conserved Kinetic Energy is increased (at the expense of the stored energy of the rocket fuel)
3/27/08
Physics 201, UW-Madison
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Rocket Propulsion
p = Impulse
Ve
Initially M v 0 after t M-M v+v (v-ve)M
The rocket's mass The rocket's speed v Momentum of fuel
(The fuel, M, is ejected at ve with respect to rocket -- v+(-ve) in our coordinate system.)
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Rocket p = Impulse ( v ! ve ) "M + ( M ! "M )( v + "v ) ! Mv = F"t
momentum after "t - initial p = impulse
M dv = ve dM
Integrate
v!M " ve !M + Mv " v!M + M !v = F!t M !v " ve !M = F!t dv dM M " ve =F dt dt In the special case in which F=0 (no gravity, air resistance); dM
!
dv = ve
M
Mi v f " vi = ve ln Mf
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3/27/08
Question
Suppose rain falls vertically into an open cart rolling with negligible friction along a straight horizontal track. As a result of the accumulating water, the speed of the cart A. increases. B. does not change. C. decreases.
Mass is increasing Momentum is conserved So the velocity must decrease!
3/27/08
Physics 201, UW-Madison
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Question
Suppose rain falls vertically into an open cart rolling along a straight horizontal track with negligible friction. As a result of the accumulating water, the kinetic energy of the cart A. increases. B. does not change. CORRECT C. decreases.
Momentum is constant. Mass increases velocity decreases and so
does kinetic energy (p2/2M)
1 2 1 2 2 1 p2 KE = mv = m v = 2 2 m 2m
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Center of Mass
How do we describe a system made up of many parts? Define the Center of Mass (average position). Define the Moment of Inertia (mass distribution about CM) For a collection of N individual pointlike particles whose masses and positions we know:
! RCM =
! ! mi ri
N i =1 N
m1 r1 m4 r 4
y
m2 r2 RCM
x
! mi
i =1
r3
m3
(In this case, N = 4)
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3/27/08
Center of Mass
Ruler
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
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Physics 201, UW-Madison
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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
+
+
+
+
+
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Physics 201, UW-Madison
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Center of Mass
For a continuous solid, we have to do an integral.
dm
y
! RCM =
! ! rdm
r
x
! dm
=
! ! rdm M
where dm is an infinitesimal mass element.
3/27/08
Physics 201, UW-Madison
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Center of Mass
We find that the Center of Mass is at the "center" of the object. The location of the center of is mass an intrinsic property of the object!! (it does not depend on where you choose the origin or coordinates when calculating it).
y
RCM
x
3/27/08
Physics 201, UW-Madison
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Experimentally Determining the Center of Gravity
The wrench is hung freely from two different pivots The intersection of the lines indicates the center of gravity A rigid object can be balanced by a single force equal in magnitude to its weight as long as the force is acting upward through the object's center of gravity
3/27/08
Physics 201, UW-Madison
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Center of Mass
The center of mass for a combination of objects is the average center of mass location of the objects:
m2
! RCM =
! ! mi Ri
N i =1 N
+ R2 - R1
+
+ m1 R1
RCM
R2
!m
i =1
i
so if we have two objects:
y x
! RCM
! ! m R + m2 R2 = 1 1 m1 + m2
! m2 ! ! = R1 + R2 ! R1 M
(
)
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Physics 201, UW-Madison
Problem 1: Center of Mass
The disk shown below (1) clearly has its CM at the center. Suppose the disk is cut in half and the pieces arranged as shown in (2): Where is the CM of (2) as compared to (1)?
(a) higher
(b) lower
(c) same
X CM (2)
(1)
Problem 1: Solution
The CM of each half-disk will be closer to the fat end than to
the thin end (think of where it would balance).
The CM of the compound object will be halfway between the CMs of the two halves. This is higher than the CM of the disk
X X CM X X
(1)
(2)
Question 2
Where is the center of gravity of a "yummy" donut?
It is at the origin of the circular ring, half way from the bottom of the donut - where there is no dough.
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Physics 201, UW-Madison
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A moment later......
M& R # New c.g. = old c.g.+ % !m " ! ( $ 8'M R shift = 8
C G has shifted along the line of symmetry away from the bite.
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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.
So:
And:
! 1 N ! RCM = ! mi ri M i =1 ! ! ! dRCM 1 N dri VCM = = ! mi dt dt M i =1 ! ! N ! dV 1 dv ACM = CM = mi i ! dt dt M i =1
N " % $ M = ! mi ' # & i =1
1 = M
1 = M
! ! mi vi
N i =1
N
! mi ai !
i =1
The velocity and acceleration of the CM is just the weighted average velocity and acceleration of all the particles.
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Linear Momentum:
Recall the linear momentum
! ! p = mv
So px = mvx etc.
Newton's 2nd Law:
! ! Fext = m a ! dv d ! =m = (m v) dt dt
! ! dp F= dt
Units of linear momentum are kg m/s.
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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.
! ! ! ! ! Fi, NET = F13 + F12 + F1, EXT
i
! = F1, EXT
( ! + (F ! + (F
)
F13
! 21 + F 23 ! 31 + F32
) )
F31
m3 F32 F23 F21 m2
(since the other forces cancel in pairs...Newton's 3rd Law)
m1 F12 F1,EXT
All of the "internal" forces cancel !! Only the "external" force matters !!
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Problem : Center of Mass Motion
A man weighs exactly as much as his 20 foot long canoe. Initially he stands in the center of the motionless canoe, a distance of 20 feet from shore. Next he walks toward the shore until he gets to the end of the canoe. What is his new distance from the shore. (There no horizontal force on the canoe by the water).
20 ft before 20 ft ? ft after
(a) 10 ft (b) 15 ft (c) 20 ft
Solution
Since the man and the canoe have the same mass, the CM of the man-canoe system will be halfway between the CM of the man and the CM of the canoe. Initially the CM of the system is 20 ft from shore.
X X 20 ft CM of system x
Solution
Since there is no external force acting on the system in the x-direction, the location of the CM of the system can't change! Therefore, the man ends up 5 ft to the left of the system CM, and the center of the canoe ends up 5 ft to the right. He ends up moving 5 ft toward the shore (he is 15 ft away).
15 ft
X
10 ft X x
20 ft
5 ft CM of system