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Rotational Kinematics Problem

  • Thread starter Moxin
  • Start date
24
0
A wheel has eight spokes and a radius of 38.0 cm. It is mounted on a fixed axle and is spinning at 2.80 rev/s. You want to shoot a 27.0 cm long arrow through the wheel, parallel to this axle, without hitting any of the spokes. Assume that the arrow and the spokes are very thin and evenly spaced. What minimum speed must the arrow have?

Ok.. This is how I tackled it:

First I converted 2.8 rev/s to rad/s using the equation 1 rev = 2*pi rad and got 17.593 rad/s

Then for the distance of a section of the wheel between the spokes I used the equation s = R*angle where R is the radius, so:

s = (0.38 m)((2*pi rad)/8) = 0.298 rad*m

Then I wasnt sure what I could do to get the minimum speed but I figured I'd find how long it'd take for s to "go by" using the angular velocity.. so I divided s by 17.593 and got 0.0170 seconds.. then I divided the length of the arrow (0.27 m) by 0.0170 seconds and got 15.88 m/s

Apparently that's not the answer. Anyone know whats up ?
 
17
0
I think you're making this harder than it has to be. One thing I see is that you lost a unit, when you divided s by w (lowercase omega, the angular velocity) you lost the meters, where'd the meters go? s is a distance along the curve, in meters and dividing it by the angular velocity, well I'm not sure if that is an actual representation of anything, if you divide the circumference (your s*8) by the angular velcoty I believe you get the tangential velcoity.

But anyways I would approach the problem like this, you know how fast the wheel is spinning and you have to get an arrow to pass just after one spoke and just before the next so you need to know how long it takes from when one spoke passes a position to when the next spoke reaches the same position. Now the distance between spokes is 1/8 of a revolution. so take .125rev / 2.8rev/s = .045s Now an arrow is .27m long so .27m / .045s = 6m/s See if that answer is correct.
 
24
0
thanks Zimm, I knew the problem couldn't be as hard as I was makin it (mainly because it was at the beginning of the problem set I was given and those are typically easier), and yeh, your answer worked
 

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