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!

Rotational Collision Door Problem

  1. Jan 11, 2014 #1
    1. The problem statement, all variables and given/known data
    A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 49.0 kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud of mass 0.500 kg, traveling perpendicular to the door at 15.0 m/s just before impact. Find the final angular speed of the door (in rad/sec).


    2. Relevant equations
    ΣLBefore=ΣLAfter
    L=Iω=mvrsinϕ
    IDoor swinging on hinges=(1/3)mr2
    IOf point=mr2


    3. The attempt at a solution
    ΣLBefore=ΣLAfter
    mvrsinϕ=(IDoor+IMud
    (.5)(15)(.5)sin90=[(1/3)(49)(.5)2+(.5)(.5)2
    w=.8911 rad/sec

    Sorry, I did do .8911 on the question (webassign), I just typed it in wrong
     
    Last edited: Jan 11, 2014
  2. jcsd
  3. Jan 11, 2014 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    That looks right...
    ... but that seems much too big.
     
  4. Jan 11, 2014 #3
    I fixed it in OP, I meant to put .8911
     
  5. Jan 12, 2014 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Looks good.
     
  6. Jan 12, 2014 #5
    It's wrong though... (Or at least I used exact same method for practice another version and got it wrong but can't check since I'll be locked out if I get it wrong again.
     
  7. Jan 12, 2014 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    OK, I see it. It's an easy confusion.
    The moment of inertia of a rod length 2L about its centre is mL2/3.
    The moment of inertia of a rod length 2L about one end is 4mL2/3.
    The moment of inertia of a rod length L about one end is mL2/3.
    You halved the door width, but only used 1/3 instead of 4/3.
     
  8. Jan 12, 2014 #7
    The item in red should be width of the door i.e 1.That should give you ω = .23 rad/sec .
     
    Last edited: Jan 12, 2014
  9. Jan 12, 2014 #8
    Okay, thank you very much, I ended up getting ω = .2278 rad/sec which was right.
     
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: Rotational Collision Door Problem
  1. Rotating Door Question (Replies: 3)

Loading...