How to find power of an engine in UCM?

  • Thread starter chrishans
  • Start date
  • #1
chrishans
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Homework Statement



A plane of mass 1930kg, driven by a pilot of mass 70kg moves at a constant speed of 200 metres/sec describing a vertical circular trajectory, so that in the top of the circle the pilot is 'upside down'. Find the minimun power of the engine so that they can move from the bottom to the top of the trajectory.
mPLANE=1930 kg
mPILOT=70 kg
radius=1000 mtr
speed=200m/sec
I know the formulas of P(power) and L(work) but I really don't know how can the data fit into any of these. Please, help me! Thanks a lot!


Homework Equations


P=L/T
L=F·cos α·Δx (for constant forces)
L(A,B)=ΔMechanicalE=PotEB+KinEB+ElasticEB-(PotEA+KinEA+ElasticEA)
PotEb=m·g·h
KinEb=1/2·m·speed^2

The Attempt at a Solution


I took potential energy 0 at the bottom of the circle and there is no elastic energy, so:
*g=10m/s
L(Bottom, Top)=PotEb+KinEb-KinEa-Pota
L(Bottom, Top)=2000kg·10m/s^2·2000m+2000kg·1/2·(200m/s)^2-2000kg·1/2·(200m/s)^2-2000kg·10m/s·0m
L(Bottom, Top)=40000000J+40000000J-40000000J-0J
L(Bottom, Top)=40000000J

The semicircle I'm asked about has a length of 1000m*π=3140m
At 200m/s the plane should take 3140/200=15.7 seconds.

Then:
P=L/T -> P=40000000J/15.7sec=2547770.7 Watts

Is this correct? I was given this excercise in an exam and did it wrong, and now I am trying to solve it at home, but I have no one to ask whether it is OK or not.
Thanks!
 

Answers and Replies

  • #2
kushan
256
0
looks good to me
 

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