1. Not finding help here? Sign up for a free 30min 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!

Mastering Physics Help

  1. Nov 28, 2009 #1
    1. The problem statement, all variables and given/known data
    An astronaut notices that a pendulum which took 2.46 seconds for a complete cycle of swing when the rocket was waiting on the launch pad takes 1.26 seconds for the same cycle of swing during liftoff. What is the acceleration of the rocket?(Hint: Inside the rocket, it appears that g has increased.)


    2. Relevant equations

    T=(2pi)sqrt(L/g)

    3. The attempt at a solution

    I know I am supposed to use the above equation, but I don't know how to use it to solve for acceleration. Using g=9.8m/s^2 and the initial period, I solved for L and got 0.621m. Do I even need to do that for this problem? The fact that the problem said that "g" increased within the rocket is what confuses me the most. Am I solving for a new/increased value of "g" and using that as acceleration? If so, when I solved the problem this way, I got g=15.4 m/s^2 (using the second period and 0.621m as L), but this was incorrect.

    I really need help with this problem as my teacher disabled the hints option. Thanks in advance.
     
  2. jcsd
  3. Nov 28, 2009 #2
    How did you get 0.621? I got something different. Did you forget to square both sides of the equation at some point? There is a square root sign.
     
  4. Nov 28, 2009 #3
    yeah you are right. i did do something qrong. i redid it and got 1.50
     
  5. Nov 29, 2009 #4

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    1.50 m is correct. What do you get for the new g value using 1.50 m and the shorter period?

    Note: this new g value is not the acceleration of the rocket.
     
  6. Nov 29, 2009 #5
    Thanks. Sorry it took so long to respond I have been at work all day. Anyways, with using 1.50m as the length and the shorter period of 1.26 seconds, I get a new value of "g" as 37.3 m/s^2
     
  7. Nov 30, 2009 #6

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Sounds good. So, what must the rocket's acceleration be, to produce an apparent "g" of 37.3 m/s^2?
     
  8. Nov 30, 2009 #7
    This is where I get stuck. I know that in order to accelerate, the rocket must overcome this new "g" value, but I don't know what equation I would use to figure out the rocket's acceleration.
     
  9. Nov 30, 2009 #8

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Okay. Well, for example:

    If the rocket were not accelerating at all, the apparent g would just be the usual 9.8 m/s^2.

    If the rocket accelerates at 1.0 m/s^2, the apparent "g" becomes 10.8 m/s^2.

    etc.
     
  10. Nov 30, 2009 #9
    I got it! Thank you so much! Wow I didn't think it could have been that easy!! Thanks again!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook