Transcript A Simple Model of a Rocket

A Simple Model of a Rocket
How it gets into the air
• The engine produces
thrust from burning of
powder
• The trust from the
burning of the powder
results in kinetic
energy.
First … Thrust
• Thrust is in
Newtons for a
given time
• To clarify use a
simplified curve
• 8 Newtons for
1.8 seconds
• This is like a finger pushing the rocket into the
air
Kinetic Energy and Work Done
• Kinetic Energy
𝐾=
1
𝑚𝑣 2
2
• Work by a constant applied force:
𝑊 = 𝐹𝑥 ∆𝑥
• So then the change in kinetic energy is:
∆𝐾 = 𝑊𝑛𝑒𝑡
• The result is that the work done by the rocket
engine, we are looking at it as a constant
applied force is then turned into kinetic
energy
First Part of the Model
• Launch, and powered ascent
• Launch occurs at time t=0
• End of powered
ascent is at 𝑇𝑓𝑜
• T=0, K=0
• Work done by
engine results in
𝑊𝑒𝑛𝑔𝑖𝑛𝑒 = 𝐾𝑚𝑎𝑥
at 𝑡 = 𝑇𝑓𝑜
Need to Find ∆𝑥
𝐹𝑡
• The engine burns for 1.8 seconds
• Free body diagram: 𝐹 = 𝐹𝑡 − 𝐹𝑔
𝐹𝑡 = 8𝑁
𝐹 = 𝑚𝑎 or 𝑎 =
𝑎=
𝐹𝑔 = 𝑚𝑔
𝐹
𝑚
(𝐹𝑡 −𝐹𝑔 )
𝑚
With the mass of the rocket at ~80g
But now we have a value for constant
acceleration …… we want distance ∆𝑥
𝐹𝑔
The result of this exercise …
• The distance an object travels is
∆𝑥 = 𝑣 ∗ 𝑡
𝑡ℎ𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑡𝑖𝑚𝑒𝑠 𝑡ℎ𝑒 𝑡𝑖𝑚𝑒
• The acceleration is related to the
velocity, and we know the time (1.8
seconds)
• From calculus and physics
𝑣=
𝑎 𝑑𝑡
What does this mean for us?
We will do this graphically
• At time zero, the rocket is standing still
• At time 𝑡 = 0.1𝑠𝑒𝑐 the rocket has been
accelerating for 0.1 sec and, in this case,
10 𝑚 𝑠2.
• Every 0.1
second the
speed
increases
by 1 𝑚 𝑠
So how far did we go?
• Remember, this is only a model up to 𝑇𝑓𝑜 , or
1.8 seconds
• For example (not the answer) the average
velocity is
𝑣𝑎𝑣𝑔 = 9 𝑚 𝑠
for 1.8 seconds
• The distance is
∆𝑥 = 𝑣𝑎𝑣𝑔 ∗ 𝑇𝑓𝑜
• Why did we
want this?
The work done is converted to kinetic
energy ….
• Work done
𝑊 = 𝐹 ∗ ∆𝑥
with:
𝐹 = 𝐹𝑡 − 𝐹𝑔
and:
∆𝑥 = 𝑣𝑎𝑣𝑔 ∗ 𝑇𝑓𝑜
And
𝑎=
(𝐹𝑡 −𝐹𝑔 )
𝑚