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!

Double Pendulum With Acceleration

  1. Jun 24, 2014 #1
    1. The problem statement, all variables and given/known data

    You are holding a special pendulum with two masses [itex] m_1 [/itex] and [itex] m_2 [/itex], instead of one, connected by a rope as shown in Fig. You lower the pendulum such that the tension in the rope between the two masses is half the weight of the bottom mass. Find the acceleration with which you lower the pendulum.

    2. Relevant equations

    [itex]\sum \hat{\text{F}} = m\hat{a}[/itex]

    [itex]\text{w} = m\hat{g}[/itex]

    3. The attempt at a solution

    I thought that I simply needed to consider the bottom mass, [itex]m_2[/itex] and so since the forces acting on it are gravity and the tension between the two masses, I set up my equations as such:

    [itex]\sum \hat{\text{F}}_\text{y} = \hat{\text{T}}_{1,2} - m_2\hat{g} = -m_2\hat{a}[/itex], where [itex]\hat{\text{T}}_{1,2}=\frac{1}{2}m_2 \hat{g}[/itex]

    so I end up with [itex]\hat{a} = \frac{\hat{g}}{2}[/itex], however the book gives an answer of

    [itex]\hat{a} = \frac{m_2\hat{g}}{m_1 + m_2}[/itex] but I can only get that answer if I start with

    [itex]\sum \hat{\text{F}} =\hat{\text{T}}_{1,2}- m_2\hat{g} = -(m_1 + m_2)\hat{a}[/itex]

    Can anybody explain to me why they are using the sum of the masses as the net force acting on [itex]m_2[/itex]?
     

    Attached Files:

  2. jcsd
  3. Jun 24, 2014 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Perhaps the a you calculate is the desired acceleration of mass 1 ? And how does the acceleration of mass 1 come about ?

    I do have difficulty believing the book answer too, since ##m_1\rightarrow 0## begets g which I would say is a 100 % overestimate.
     
  4. Jun 24, 2014 #3
    What exactly do you mean? Also, if there is a single mass, m, attached to a string with a downward acceleration of g/2, then then tension in the rope is mg/2; that, as well as your insight, leads me to believe their answer is wrong. However it may not be..
     
  5. Jun 24, 2014 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I agree with your answer a = g/2.
     
  6. Jun 24, 2014 #5
    Thanks for the confirmation. I am growing very displeased with this textbook due to the number of mistakes.. Take this for example; a question asking how fast you can spin a disk with a mass sitting on it before the mass slides off.

    Part b) Find an expression for the tension in the string.

    What string? Where is the string?? Not what i had in mind when I learned about "fictitious forces"
     
  7. Jun 24, 2014 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    :smile:
     
  8. Jun 25, 2014 #7

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Nothing so good for building self esteem as improving on a textbook..., but:

    I might agree with you on the spinning disk question being mistaken -- if that is the full tekst of the question. Is it ? Anyway, re the latter part of your post:

    Fictitious forces are just that: ficitious. Most of the time. No need for them here. If we disregard snapping ropes (strings...), then m1 and m2 experience the same acceleration (it's a string, not a spring): ##\ y_1- y_2\ ## is a constant. Hence ##\ \dot y_1 = \dot y_2\ ## and ##\ \ddot y_1 = \ddot y_2 ##.

    Check it out: ##\ T_{12} = {m_2g\over 2}\ ## is a given. ##\ \ddot y_1 = \ddot y_2 = g/2 \ ## is what I claim. For ##m_1## the force balance is then ## m_1 g/2 = m_1 g + m_2 g/2- T_1\ ##, or ##T_1 = (m_1+m_2)\, g/2##. The two balls behave as a whole.

    Not exciting at all; nothing fictitious needs to be brought in!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Double Pendulum With Acceleration
  1. Double pendulum (Replies: 7)

  2. Pendulum acceleration (Replies: 3)

  3. Accelerating Pendulum (Replies: 3)

Loading...