Physics Help Forum Q24, specimen paper 1

 Periodic and Circular Motion Periodic and Circular Motion Physics Help Forum

 Mar 28th 2017, 11:33 AM #1 Junior Member   Join Date: Mar 2017 Posts: 9 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 1.58×104 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) Attached Thumbnails   Last edited by akra99; Mar 28th 2017 at 02:01 PM.
 Mar 28th 2017, 01:50 PM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,333 Hmmm... seems to me your answer is correct! I get 8.77 m/s^2 as well. akra99 likes this.
 Mar 28th 2017, 03:47 PM #3 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,009 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 akra99 likes this.
 Mar 28th 2017, 03:55 PM #4 Junior Member   Join Date: Mar 2017 Posts: 9 Thank you guys, so the specimen paper has an error.

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