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Constant angular acceleration problem

  • Thread starter saturn67
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  • #26
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Thank you i hope he good XD, tell me right anyway after u done
 
  • #27
16
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ur equation is correct.... i dont kno why

4pi = 1/2*a*(0.480)^2

mastering physics accepteed his answer but mine.......
 
  • #28
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werid

it don;t accept mine
 
  • #29
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what his time?

0.339 sec?
 
  • #30
rl.bhat
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Angular displacement = wo + 1/2*a*t^2. Since you are talking about first and second rotation , it must be starting from rest. Therefore wo = 0.
Let it take t second to complete first rotation and t + 0.48 s for two rotation. So 2pi = 1/2*a*(t)^2 and 4pi = 1/2*a*(t + 0.48)^2 . Divide these two equations. a gets cancelled out. Solve for t. Put this value in the first equation. You will get the value of a.
 
  • #31
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how to Divide these two equations to get a cancel ?
 
Last edited:
  • #32
rl.bhat
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2pi/4pi = 1/2*a*t^2/1/2a*(t+0.48)^2
1/2 = t^2/(t+0.48)^2
solve for t
 
  • #33
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that give t=0.48
which it like not solving anything :(
 
  • #34
rl.bhat
Homework Helper
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Cross multiplying the above expression we get (t + 0.48 )^2 = 2t^2
Taking sqrt. on both side we get t + 0.48 = 1.41t or 0.41t = 0.48 or t = 0.48/0.41
 
  • #35
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woot u were right
LOL stupid me
thank you
 

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