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Physics final tomorrow; any help would be appreciated!

  1. Dec 12, 2012 #1
    As the title suggests, I have a physics final tomorrow and have been doing questions from past finals to prepare. I've tried to solve these independently since that is what will be expected of me on the test, but there are a couple that have me stumped.

    1:

    The problem statement: http://imageshack.us/a/img688/6753/screenshot20121212at619.png [Broken]

    Here's what I've tried for this one. I broke up the situation into 3 stages, each of which is pictured in the diagram.

    Stage 1 - the wheel has kinetic energy KE = 1/2 Iω^2 + 1/2 mv^2. Using the substitution v = rω, the expression came out to be 3/4 mr^2 ω^2.

    Stage 2 - Next, since this was an elastic collision, KE was conserved, so I said that 1/2 Iω^2 - 1/2 mv^2 = 1/2 Iω^2 + 1/2 mv^2 because ω is still going in the same direction, but v is going in the opposite direction. But this gave me 0 = mv^2, which doesn't make much sense to me. I also think I should draw a free-body diagram showing the effects of static friction here, but I don't know how to start.

    EDIT: Apparently forgot the carat (^) sign to denote exponents... oops.
    2:

    The problem statement: http://imageshack.us/a/img641/6753/screenshot20121212at619.png [Broken]

    I tried solving this as a static equilibrium problem. I had T = (2.5)(-0.732 * 9.8) + (5)(25 * 9.8), but didn't manage to get much farther than that.

    For the second part, I used the fundamental frequency expression f1 = v/2L, and the wave speed v = sqrt(T/μ). Obviously I'll have to use the value for T I attempted to find above, but I don't know whether that value is correct to begin with.

    Any tips would be appreciated - thanks a ton!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 12, 2012 #2
    This is wrong! (-v)2 = v2, not -(v)2
     
  4. Dec 12, 2012 #3
    About the first problem, consevation of KE gives you nothing! Since v ang ω are the same before and after colision, conservation of KE formula gives a trivial equation. Try to evaluate the torque (τ = r Ffriction)applied to the cylinder and the consequent angular momentum rate of change. Suppose that after Δt time, CM velocity is v' and angular velocity is ω'. Evaluate the angular momentum after time Δt (L' = L - τΔt = Iω-FfrictionΔt=Iω') and CM momentum after Δt (P' = mv - FfrictionΔt= mv'). Since right after the colision v = ωr and since after Δt, v'= ω'r, you can substitute ω and ω' as an expression of v and v', respectively. Now you have two equations for two unknown quantities (v' and Δt). Solve to find v' (and ω' since ω'=v'/r). Now, about the traveled distance, since you known Δt yoy could try to evaluate it directly or you can do it another way: calculate the KE after regular rolling is re-established; it's difference from the initial KE will be the work of frictional force.
     
  5. Dec 12, 2012 #4
    About the second, draw a diagramm showing the applied forces. Using static quilibrium arguments you can find the tension. After you do it, you can calculate waves velocity (v=Sqrt(T/μ), where μ is the linear mass density μ = m/L and L is the lenght of the wire). This is the same for every oscillation frequency. About the natural one, it's wavelenght should be L/2 and since you know the velocity, you can calculate the frequency.
     
  6. Dec 12, 2012 #5
    @ clamtrox: That was a pretty bone-headed error. Thanks!

    @ cosmic dust: Wow, thanks for the help (especially for #1)! Could you look over my equation for T in the 2nd question and see if I'm on the right track?
     
  7. Dec 12, 2012 #6
    You are welcome! About the 2nd question, post it and I will try to verify it soon...
     
  8. Dec 12, 2012 #7
    This is what I had for #2:

     
    Last edited by a moderator: May 6, 2017
  9. Dec 12, 2012 #8
    Suppose M and L is the mass and the lenght of the post and m the mass of the wire. Try to find all the torques applied to each post:
    -there is one because of its weight: τ1 = Mgcosθ L/2
    -one because of tension: τ2 = L T sinθ
    -and one because of the wire's weight (I'm not so sure about this, what is your opinion?): since the posts balancing symmetrically, the weight of the wire distributes equally on the posts. So, τ3 = L (mgcosθ)/2.
    The post balances so: net torque = zero i.e.:
    τ2= τ13 (L disappears!)
    Solve for T...
     
    Last edited: Dec 12, 2012
  10. Dec 12, 2012 #9
    Thank you! That makes sense.
     
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