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: – Hollow shell:

– Ship:

Center of Mass

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

• Mid-jump: Dunk:

• 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

Center of Mass

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

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

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

• 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:

• Here p is the momentum of everything that was within the system at time t – including the mass that was ejected during dt

• So:

ext,net

d

dt= p

F

( ) ( ) ( )d d d

M

M M M

== + + + −

i

f

p v

p v v u

Velocity of ejected mass

Note the sign: If rocket is ejecting mass, dm is a negative number!

• So, our original equation becomes:

• vrel is the velocity of the ejected mass with respect to the rocket

d d d d d dM M M M M M= − = + + + − −f ip p p v v v v u v

Product of two small numbers – can be ignored!

( )ext,net

d d

d d

d d d d d

d d d

M M M MM

tM

t

t

Mt

t

+ −= =

−

+ −

= rel

v v u v

vv

F v u

• Consider the case where no external forces act on the rocket:

• 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

d d0

d dd d

d d

MM

t tM

Mt t

− =

= +

rel

rel

vv

vv

Thrust of rocket

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?

• The force diagram looks like: T

mg

• The mass of the fuel ejected per second is:

• The net force is the thrust minus the weight of the rocket, or:

• So the rocket does lift off, with initial acceleration of:

( ) ( )3 4rel

7

d2.7 10 m/s 1.3 10 kg/s

d

3.5 10 N

MT v

t= − = − − × ×

= ×

52

4

d 1 d 11.3 10 N/s

d d 9.8m/s

1.3 10 kg/s

M W

t g t= = ⋅ ×

= ×

7 6 2net

6

3.5 10 N 2.8 10 kg 9.8m/s

7.6 10 N

F T mg= − = × − × ⋅

= ×

62net

6

7.6 10 N2.7m/s

2.8 10 kg

Fa

M

×= = =×