Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Pebble on rolling wheel

  1. Jan 15, 2013 #1
    1. The problem statement, all variables and given/known data

    only concerned with part (b)

    2. Relevant equations
    F = ma

    3. The attempt at a solution
    So first, just to visualize the problem a bit better, I transformed to the frame of an observer co - moving (but not rotating) with the wheel. Since the wheel is moving at constant speed this would just be a regular galilean boost and the equations of motion of the pebble will remain invariant under this transformation. In this frame, the wheel is simply rotating with angular velocity ω = V / R and not translating. Up until the pebble actually starts slipping relative to the wheel, it is constrained to move as one with the wheel so in this frame the pebble starts at the top of the wheel and sweeps out an angle θ right before sliding, simply by rotating about the center of the wheel. At this point on the wheel and in this frame, the equations of motion should be [itex]f - mgsin\theta = ma_{\theta } = 0[/itex] (since there is no tangential acceleration of the pebble as viewed in this frame) and [itex]N - mgcos\theta = ma_{r} = m\frac{V^{2}}{R}[/itex]. Right before sliding we have that [itex]f = \mu N = N[/itex] so we can solve the two and get [itex]sin\theta - cos\theta = \frac{V^{2}}{Rg}[/itex]. This is not correct as you can see by what the book gives as the answer in the problem statement. I'm really not sure where the 1/√2 factors come in before taking any cosines of angles. The only way I could see them coming in is if the problem involved an isosceles triangle in some way. thanks
  2. jcsd
  3. Jan 15, 2013 #2


    User Avatar
    Science Advisor

    Based on the way you wrote down your laws of motion, I assume you took the outward radial direction to be positive. Therefore you seem to have made a small typo as it should be [itex]N - mgcos\theta = -m\frac{V^{2}}{R}[/itex] which results in [itex]cos\theta - sin\theta = \frac{V^{2}}{Rg}[/itex] in actuality. That being said, try to derive the following trig relationship: [itex]cos\theta - sin\theta = \sqrt{2}sin(\frac{\pi }{4} - \theta ) = \sqrt{2}cos(\frac{\pi}{2} - (\frac{\pi }{4} - \theta)) = \sqrt{2}cos(\frac{\pi }{4} + \theta) [/itex]. If you can derive that then you have gotten the form the book wants.
  4. Jan 15, 2013 #3
    Thank you
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook