Distance Traveled by Point on Wheel: Rotational Motion

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SUMMARY

A 40cm diameter wheel accelerates uniformly from 240rpm to 360rpm over 6.5 seconds. To calculate the angular displacement, the equation θ = 0.5(ω₀ + ω)t is utilized, where ω₀ is the initial angular speed and ω is the final angular speed. The conversion from revolutions to radians is necessary for precise calculations, as 1 revolution equals 2π radians. The diameter of the wheel is only relevant if the problem specifically asks for linear distance traveled.

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A 40cm diameter wheel accelerates uniformly from 240rpm to 360rpm in 6.5s. How far will a point on the edge of the wheel have traveled in this time?


I think we have to use w=2pi*f?


I used that equation and plugged in w=2pi*(120rpm/0.108min)=1107.69 revolutions. I don't know where to go from there.
 
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I'm not sure what you mean by "w," if that's your omega, you'll need to use a different approach. You're given the initial angular speed, and the final angular speed in revolutions per minute. The object has a constant angular acceleration, so you can use the equations for rotational kinematics. You'll probably want to use \theta = .5(\omega_0 + \omega) t. Does that help?
 
ohh, yea, i think i get it. so the eq. would be:

theta = .5 (240 + 360)(0.108)?
 
You've got it. Just remember to convert units if you must.
 
Thanks!
 
so for distance, how should i find that? my answer is in radians.
 
Ok, so your theta is going to be in revolutions. If you want radians, you convert using the fact that 1 revolution equals 2\pi radians. Does that help?
 
yes, it does. so the 40cm is irrelevant?
 
sw3etazngyrl said:
yes, it does. so the 40cm is irrelevant?

Well, if we were asked the linear distance traveled by the point, then it would be necessary. But if we're simply considering its angular displacement over some period of time, it's not necessary. Make sure the problem doesn't ask for linear distance.
 
  • #10
it's asking for the distance the edge of wheel will be at or have traveled
 
  • #11
As in a wheel rolling across the ground?
 
  • #12
i think so
 
  • #13
i got it now, THANKS!
 
  • #14
You're welcome. I'd just ask that you be 100% sure it's not asking for the linear distance. If it is, you'll have to use the "rolling constraints" v_{linear} = \omega r, a_{linear} = \alpha r. If it's asking for the answer in radians or revolutions, then you're good to go, I wouldn't want you getting the wrong answer :)
 

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