1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work and kinetic energy (dynamics)

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data

    See attachment.

    2. Relevant equations

    (1) F=ma
    (2.13) W=ΔT
    (2.20) (weight) x (decrease in altitude of mass center of body)


    3. The attempt at a solution

    I don't get where |v(t1)| and |v(t2)|^2 comes from. The book says |v(t1)| = 0, which I suppose is because initially there was no movement, so kinetic energy is 0? Does |v(t1)|=0 have anything to do with T=mgsinθ+mlθ'^2?

    Then it says |v(t2)|^2 = l^2θ'^2, and this totally confused me. First, why is v(t2) squared and not v(t1)? Second, obviously v(t2) is somehow related to T=mgsinθ+mlθ'^2, but I don't see how you get v(t2) from that. Also originally θ'^2 was already squared, so wouldn't |v(t2)|^2 = l^2θ'^4??
    Very confused on this.

    Any help would be appreciated thanks.
     

    Attached Files:

  2. jcsd
  3. Feb 5, 2012 #2

    rock.freak667

    User Avatar
    Homework Helper

    |v(t1)| = 0 as no movement at t=t1.


    |v(t2)|^2 = l^2θ'^2 because v(t2)= radius x angular velocity. So it is just squared. (I am ignoring the modulus signs now).

    For the second part, the change in PE = change in KE

    so the height went from lsin(30) to lsin(θ) so the change in PE is mg[lsinθ-lsin30]

    similarly change in KE is 1/2mv22-1/2mv12 = 1/2mv22 = 1/2m(lω)2 = ½ml2ω2

    Do you understand how they the first part for the tension T at least?
     
  4. Feb 5, 2012 #3
    I understand where T=mgsinθ+mlθ'^2 came from. But for |v(t2)|^2, why is l squared but theta dot not to the 4th power? I totally don't get this. And PE = ΔKE is the same as W=ΔKE right? Thanks.
     
  5. Feb 5, 2012 #4

    rock.freak667

    User Avatar
    Homework Helper

    Right, for the velocity, they multiply the radius (l) by the angular velocity (theta dot or ω)

    so v(t2)=lω so that v(t2)2=l2ω2


    Yep, same idea.
     
  6. Feb 5, 2012 #5
    so |v(t2)|^2 has absolutely nothing to do with T=mgsinθ+mlθ'^2? Meaning it wasn't derived from there or anything? It's just something you should intuitively know?
     
  7. Feb 6, 2012 #6

    rock.freak667

    User Avatar
    Homework Helper

    Well the kind of did it backwards in a way.

    Normally you'd start with ΔPE=ΔKE

    to get mg[lsinθ-lsin30]= 1/2mv22

    and it would be better to have everything in terms of θ so putting [itex]v_2= l \dot{\theta}[/itex] is easier.

    so you'd get

    [tex]mg[lsin\theta - lsin30]=\frac{1}{2}ml^2 \theta ^2 \Rightarrow mg[sin\theta -sin 30]=\frac{1}{2}ml \dot{\theta}^2[/tex]


    and if we look at the the [itex]ml \dot{\theta}^2[/itex] you'd see that is mass*radius*angular velocity squared i.e. centripetal force.

    The only way that relates to tension T is from the first line

    [tex]T+mgsin\theta = ml \dot{\theta}^2[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Work and kinetic energy (dynamics)
Loading...