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1. I cannot for the life of me figure out why the answer to the following question is 2π^2mf^2D...

"A disc of diameter D is turning at a steady angular speed at frequency f about an axis through its centre.

What is the centripetal force on a small object O of mass m on the perimeter of the disc?"

Why would the answer not just be m2πf^2D/2?

F=mω^2r,
ω= 2πf
so F=m2πf^2D/2?

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Just quoting in Puddles the Monkey so she can move the thread if needed
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(Original post by Puddles the Monkey)
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3. Hi Swifty 21,

When you substitute ω= 2πf into your equation for F you get

F=mω^2r = m(2πf)^2r = m(4π^2f^2)r = 4π^2mf^2r = 4π^2mf^2(D/2) = 2π^2mf^2D

Hope this helps.
4. (Original post by swifty21)
I cannot for the life of me figure out why the answer to the following question is 2π^2mf^2D...

"A disc of diameter D is turning at a steady angular speed at frequency f about an axis through its centre.

What is the centripetal force on a small object O of mass m on the perimeter of the disc?"

Why would the answer not just be m2πf^2D/2?

F=mω^2r,
ω= 2πf
so F=m2πf^2D/2?

Basically, you've forgotten to square the 2π as well as square the f

F = (mv^2)/r
= m(ω^2)r
= m(ω^2)D/2
= m((2πf)^2)D/2
= 4m(π^2)(f^2)D/2
= 2m(π^2)(f^2)D
5. (Original post by Teenie2)
Hi Swifty 21,

When you substitute ω= 2πf into your equation for F you get

F=mω^2r = m(2πf)^2r = m(4π^2f^2)r = 4π^2mf^2r = 4π^2mf^2(D/2) = 2π^2mf^2D

Hope this helps.
Yeah thank you very much, I understand now!

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