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Homework Help: Accelerating frame + energy

  1. Aug 5, 2010 #1
    1. The problem statement, all variables and given/known data

    A simple pendulum that consists of a mass [tex] m [/tex] and a massless cord of length [tex] L [/tex] is hanged on the ceiling of an elevator, which is intially at rest. The pendulum is performing a simple harmonic motion. The energy of the mass (Kinetic energy + potential energy with respect to the lowest point of the mass' trajectory) is [tex] E_{0} [/tex].
    The following experiment if performed:
    * at [tex] t=0 [/tex], when the mass is in its lowest point along the trajectory, the elevator's acceleration jumps from [tex] a=0 [/tex] to [tex] a=0.75g [/tex], upwards.

    * at [tex] t=\frac{T}{4} [/tex] (T being the cycle time), when the mass is in an extremum (highest point), the elevator's acceleration jumps from [tex] a=0.75g [/tex] to [tex] a=1.5g [/tex], upwards.

    What is the total energy of the mass at the ened of the experiment, when the acceleration is [tex] a=1.5g [/tex] ?

    2. The attempt at a solution

    First of all, the upward acceleration defines a "new gravitational acceleration", which is [tex] g_{eff} = a+g [/tex].
    At the lowest point the total energy of the mass in Kinetic since the reference point is defined to be there, therefore the "new gravity" doesn't affect the energy.
    At the highest point, however, the total energy is Potential because the velocity is zero over there, theresore the corresponding "new gravity" has some effect there.

    Now, how can I relate all of these stuff together?

    for the 1st experiment, the total energy is: [tex] E_{1} = \frac{1}{2}mv^{2} [/tex]
    for the 2nd experiment, the total energy is: [tex] E_{2} = m(g+a)h [/tex]

    But how can I make these stuff useful?!

    I don't think that I can use the conservation of energy principle since a fictitious force is regarded as an external force for all energy consideration purposes.
  2. jcsd
  3. Aug 5, 2010 #2
  4. Aug 5, 2010 #3


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    Not necessarily - even regular gravity is a fictitious force, technically, but that doesn't automatically make it external.

    Here's a little thought experiment that might help make your problem a little clearer. Imagine another pendulum, identical to the one in your problem (mass m suspended from a massless cord of length L), hanging from the roof of another elevator. At the beginning of the experiment, the pendulum is at rest, hanging straight down, and the elevator is also at rest.

    Q) Choosing the reference point to be the roof of the elevator, what is the potential energy of the pendulum? What is its total energy?

    Then the elevator's acceleration suddenly changes from 0 to 0.75g upward.

    Q) In the reference frame of the elevator, again choosing the reference point to be the roof of the elevator, what is the potential energy of the pendulum? What is its total energy?

    I think answering those questions may tell you something useful about the problem you've been given.
  5. Aug 6, 2010 #4
    I know that when the pendulum is released from a certain height [tex] h_{0} [/tex], it has a total energy of [tex] E_{0} [/tex] (GIVEN), now when it reaches it's lowest position (i.e. the suspension angle is 0 degrees ) the elevator accelerates upwards, but this doesn't affect the total energy at that position since it is all kinetic, however, the mass will reach a lower height since the "new" gravity now is [tex] g+a [/tex] and not [tex] g [/tex].
    I have the physical understanding of the problem, but i'm failing to realte the appropriate equations in order to get somewhere. HELP! =/
  6. Aug 6, 2010 #5


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    Really? I don't see how the mass is able to reach a lower height.

    Did you try thinking about the situation I described in my last post?
  7. Aug 7, 2010 #6
    Actually, I did ... but couldn't go with it any further.

    Can you please expand your explanation? I'm seriously having troubles with this one =/
  8. Aug 7, 2010 #7


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    Right :wink: Trust me, we're getting there.

    So is energy conserved when the value of g changes?
  9. Aug 7, 2010 #8
    energy conservation essentially means that we're gonna get the same number for the total energy "before" and "after". In this case, however, we don't so energy is not conserved when the value of g changes. (right?)
  10. Aug 8, 2010 #9
    I think I got it ...

    Let us consider the lowest point in the pendulum's trajectory to be our reference. So, the total initial energy, i.e. [tex] E_{0} [/tex], is conserved up until [tex] t=0 [/tex] because at that state the total energy is kinetic solely and the acceleration applied doesn't affect anything. Now, in the interval [tex] 0<t<\frac{T}{4} [/tex], energy is conserved but with a "new g" which is [tex] g_{1}=g+0.75g=1.75g [/tex], so we can find the height in which the mass is located at its final state in the following manner:

    [tex] E_{0}=mg_{1}h \to h=\frac{4}{7} \frac{E_{0}}{mg} [/tex]

    Finally, at [tex] t=\frac{T}{4} [/tex], the total energy is potential solely but with [tex] g_{2}=1.75g+0.75g=2.5g [/tex], So:

    [tex] E_{tot} = mg_{2}h=m(2.5g)(\frac{4}{7} \frac{E_{0}}{mg})=\frac{10}{7} E_{0} [/tex]

    Seems reasonable?!
  11. Aug 8, 2010 #10


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    Yep, seems reasonable. I just posed that thought experiment to help you realize that energy isn't conserved when g changes, and once you figure that out, it's not so hard :wink:
  12. Aug 8, 2010 #11
    Thanks buddy :smile:
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