1. Lecture 17: More on Center of Mass,
and Variable-Mass Systems
• A Note on Center of Mass Location:
– The center of mass is of a solid object is not required to be
within the volume of the material
• Examples: – Ship:
– Hollow shell:
Center of Mass

2. Applications of Center of Mass Motion
• Some basketball players are said to “hang” in the air
• How can that be, given the their center of mass must move
as a projectile – that is, parabolically?
• Consider how the player configures his body as he flies
through the air

3. • Mid-jump: Dunk:
Center of Mass
• The center-of-mass moves parabolically, but the distance
between the center-of-mass and the ball varies throughout
the jump (less in the middle, greatest at the end)
– Ball appears to “hang”, or move in a straight line

4. Another Application: High Jump
• High-jumpers contort their bodies in a peculiar way when
going over the bar:
• This keeps the jumper’s center of mass below any part of
his body
– Means he might clear the bar even though his center of mass
goes below it

5. Variable-Mass Systems
• So far, we’ve considered the motion of systems of particles
with constant mass
• Not too much of a restriction, since we know that mass is
never created nor destroyed
• However, in some cases it’s more convenient to draw our
system boundary such that mass can leave (or enter) the
system
• A rocket is the best example
– It expels gas at high velocity – since the rocket applies a
force to the gas, the gas in turn applies a force to the rocket
(Newton’s Third Law again!); this force propels the rocket
forward
– While we care about the motion of the rocket, we don’t care
about how the gas moves after it’s exhausted

6. • In other words, we want to draw our system boundary as:

7. • At some time t, our system has mass M and is moving at
velocity v
• At a later time t + dt both the mass and velocity of the
system have changed
• Newton’s Second Law tells us that:
dp
Fext,net =
dt
• Here p is the momentum of everything that was within the
system at time t – including the mass that was ejected
during dt
Velocity of ejected mass
• So:
p i = Mv
pf = ( M + dM )( v + dv ) + u ( −dM )
Note the sign: If rocket is ejecting mass, dm is a negative number!

8. dp = pf − p i = Mv + Mdv + vdM + dvdM − udM − Mv
Product of two small numbers – can be ignored!
• So, our original equation becomes:
Mdv + vdM − udM dv dM
Fext,net = =M + ( v − u)
dt dt dt
dv dM
=M − v rel
dt dt
• vrel is the velocity of the ejected mass with respect to the
rocket

9. • Consider the case where no external forces act on the
rocket:
dv dM
M − v rel =0
dt dt
dv dM Thrust of rocket
M = + v rel
dt dt
• So the rocket accelerates even though no external forces
act on it
– However, momentum is conserved for the rocket + gas
system as a whole
• Our equation works equally well for cases in which a
system is gaining mass
– Sand being poured into a moving rail car, for example

10. Example: Saturn V Rocket
• The first stage of a Saturn V rocket (used to launch
astronauts to the moon) burns 15 tons of fuel per second,
and ejects the gasses at a velocity of 2700m/s. The rocket,
when fully loaded, has a mass of 2.8 x 106 kg.
• Can the rocket lift off the pad, and if so, what is its initial
acceleration?
T
• The force diagram looks like:
mg

11. • The mass of the fuel ejected per second is:
dM 1 dW 1
= = ⋅ 1.3 × 105 N/s
dt g dt 9.8m/s2
= 1.3 × 104 kg/s
= − ( −2.7 × 103 m/s )(1.3 × 104 kg/s )
dM
T = − vrel
dt
= 3.5 × 107 N
• The net force is the thrust minus the weight of the rocket, or:
Fnet = T − mg = 3.5 × 107 N − 2.8 × 106 kg ⋅ 9.8m/s 2
= 7.6 × 106 N
• So the rocket does lift off, with initial acceleration of:
Fnet 7.6 × 106 N
a= = = 2.7m/s2
M 2.8 × 106 kg