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Periodic and Circular Motion Periodic and Circular Motion Physics Help Forum

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Old Mar 28th 2017, 12:33 PM   #1
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Q24, specimen paper 1

24) The diagram shows the turbine of a wind generator. The tip of one blade moves in a circle of diameter 64.0 m.

The rotor blades make 300 revolutions per hour.

What is the centripetal acceleration of the tip of the rotor blades?

A 0.26 ms^–2
B 4.39 ms^–2 ---> correct answer
C 17.5 ms^–2
D 1.58104 ms^–2
__________________________________
My working out:
r = 32 m, T = (60*60)/300 = 12 secs, ω = 2π/12 = π/6

a = v^2/r = (rw)^2/32 = (16/3π^2)/32 = 8.77 ------> which is 2*(B)
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Q24, specimen paper 1-q24.png  

Last edited by akra99; Mar 28th 2017 at 03:01 PM.
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Old Mar 28th 2017, 02:50 PM   #2
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Hmmm... seems to me your answer is correct! I get 8.77 m/s^2 as well.
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Old Mar 28th 2017, 04:47 PM   #3
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Yes I agree the correct answer is 8.77.

300 cyclesph = 300/3600 cps = pi/6 rads/s = angular velocity

angular acceleration = r(w)^2 = 32 (pi/6)^2 = 8.77
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Old Mar 28th 2017, 04:55 PM   #4
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Thank you guys, so the specimen paper has an error.
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