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I have the question "A big wheel at a fairground has a radius of 23[m] and takes 5 mins to turn once. What speed does a point on the edge of the wheel have?."

I know that the speed of a point on the edge of the wheel is linear velocity and so to calculate this you use the equation

V = rw

Here is my working is the final answer correct ?

enter image description here

Dan
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    Yes, what you've done is correct. – hamam_Abdallah Nov 22 '16 at 18:53
  • Easiest way to check your answer is to take the velocity and multiply by 5 minutes (since that is the time to make one revolution) and see if the distance traveled is the circumference of the circle in question. – nurdyguy Nov 22 '16 at 18:53
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    You made a small mistake in the second last step. Your angular velocity is $\frac{\pi }{150}$ not $\frac{1}{\pi(150)}$ – Swapnil Rustagi Nov 22 '16 at 18:55
  • Is it not the same thing? – Dan Nov 22 '16 at 18:57
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    @dan If replying to me, no obviously not. One is pi divided by 150, (approx. value 0.02) and the other is well 1 divided by product of (pi and 150). (approx. Value is 0.0021) – Swapnil Rustagi Nov 22 '16 at 19:03
  • But when I put Pi divided by 150 into the calculator I get (1/150)Pi or 1 divided by 150 multiplied by Pi. It's not the product of Pi and 150 it is the product of (1/150) and Pi. – Dan Nov 22 '16 at 20:18
  • That is my fault I should have written it more clearly but that is what I meant. – Dan Nov 22 '16 at 20:19
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    The correct answer is 0.48 m/s. You just need to correct the mistake I told you about. Your working is correct. – Swapnil Rustagi Nov 23 '16 at 11:00
  • Thank you for the correction. – Dan Nov 23 '16 at 12:12

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You know that $V=r\omega$ and $\omega=2\pi v$. Now all you need to do is apply the given data in the formula's:

$v$=1/300 (as $T=5 minutes$)

$\omega=2\pi v=2\pi {1/300}$.

$V=r\omega={23}*2\pi {1/300}$ which roughly turns out to be $0.5m/s$

Vidyanshu Mishra
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