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Angular Acceleration 
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#1
Oct505, 12:24 AM

P: 1

An interesting question a prof posed in class to us one day:
A bicycle wheel is upsidedown and is spinning. It has radius R. As it is raining, water droplets fly off tangentially from the bicycle. You measure the heights of the droplets flying off vertically from the bicycle (at the point whee the tangent is vertical). You notice that the height of the first droplet is greater than that of the second droplet (h1>h2). The drop in height directly corresponds to a drop in angular acceleration. From this information, compute the average angular acceleration. Thats the question. he left it at that. He gave us a hint, and that was to remember conservation of energy, but that was all. Does anyone have any ideas how to solve this? desi 


#2
Oct505, 01:39 AM

Sci Advisor
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P: 1,204

So. How can you tell if a bicycle wheel is right side up or upside down. They look pretty much the same to me either way.
But getting on with your problem, the hint I would have given you instead is that you need to relate the angular velocity of the wheel with the height to which the drops go. A change in height makes a change in angular velocity and that's the definition of angular acceleration, when divided by time. But to do this, you're going to have to know how long between times at which those two drops are let go, I think. Carl 


#3
Oct505, 04:42 AM

P: 1,017

Assuming by angular acceleration you mean the instantaneous acceleration of a point on the wheel (not what I'd call angular acceleration, since it is translational in nature), this is simply, though I don't see why you need to recall conservation of energy.
The maximum height reached by the water droplet is given by the equation of displacement of a particle in 1dimensional nonuniform motion. This depends on v, which is the instantaneous velocity of a particle on the wheel (v = rw, where w = angular velocity of wheel), the acceleration on the particle once it has left the wheel (g) and the time taken to reach that height. The time taken to reach the height is the half the duration of a vertical projectile with an initial height of zero. Put those together and you get an expression for height in terms of (rw)^2. Since tangential acceleration is given by rw^2, you can express h in terms of a/r. Rearrange for a and take the average and you'll get an expression in terms of r, g and delta h. 


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