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Rotational Kinematics and Energy Problem.

  1. Apr 1, 2004 #1
    Please forgive me if this is sloppy. And sorry for not posting this thread here in the first place (I didn't know about this section of the forums). Thanks in advance for any help you can provide.

    A 2.0kg cylinder (radius=.1m and length=.5m) is released from rest at the top of a ramp, and is allowed to roll without slipping. The ramp is .75 m high and 5.0m long. When the cylinder reaches the bottom of the ramp, what is
    (a) it's total KE (b) its rotational KE and (c) its translational KE?

    Here's what I've got so far. Let me know if I'm off course.
    I decided to use the Total Mechanical Energy Formula, and the conservation of energy:
    E(total) = 1/2*m*v^2 + 1/2*I*ω^2 + m*g*h
    E(total) = TranslationalKE + RotationalKE + GravitationalPE

    Since the cylinder is at rest initially:
    E(initial)=0+0+ 2.0kg*9.8*.75m ; E=14.7 J
    From here I assume that at the end of the ramp, when the PE becomes 0, the total KE will equal 14.7 J as well. So there's (a) solved.

    Next I plug in the knowns into the E(final) equation, and come up with:
    14.7 J = (1/2)*2kg*v^2 + (1/2)*(1/2*2kg*.1^2)*ω^2 +0
    which simplifies to:
    14.7 J = V^2 + .005*ω^2

    Here's my problem. First the equation for the E(final) seems off to me. And second, I'm not sure how to go about solving for ω^2 AND v^2 (I can't think of a third formula to throw in). Any advice? Thank You!
     
  2. jcsd
  3. Apr 1, 2004 #2
    As I already posted in your original thread:
    v = ωr.
     
  4. Apr 1, 2004 #3
    Sorry I missed that. Thank you!
     
  5. Apr 1, 2004 #4
    Allright, great I got the right answer. But will someone take me through the steps to show how to solve this equation algerbraiclly? Exponents (and consequently square roots) are my weak suit, and I somehow always mess things up when multiple exponents/square roots are involved (I used Mathematica to solve it for me...I know I'm a cheater but I'd like be able to do it myself). The final equation ends up being:

    14.7 J = (.1*ω)^2 + .005*ω^2

    Thank You!!!

    And any tips for solving equations with multiple exponents involved? I'm pretty sure it's just a reverse of the order of operations. But I'm not positive...
     
  6. Apr 1, 2004 #5
    14.7 = (.1w)^2 + .005w^2
    14.7 = (.1)^2*(w^2) + .005w^2
    14.7 = w^2*(.1^2 + .005)
    w^2 = 14.7/(.01 + .005) = 14.7/(.015)

    You can take it from there, right?

    cookiemonster
     
  7. Apr 1, 2004 #6
    Thanks Cookie Monster! Hahaha, And yes I think I can handle the last of it (just barely). I thought that was pretty funny. I get it now, Much Obliged. :biggrin:
     
  8. Apr 2, 2004 #7
    You can also let [tex]t = \omega ^2[/tex] and solve for t, if it makes your life easier. :smile:
     
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