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!

Angular Acceleration of a Fixed Rod

  1. Apr 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider a thin 14 m rod pivoted at one end. A uniform density spherical object (whose mass is 9 kg and radius is 3.2 m) is attached to the free end of the rod and the moment of inertia of the rod about an end is Irod =1/3mL^2 and the moment of inertia of the sphere about its center of mass is Isphere =2/5mr^2 . What is the angular acceleration of the rod immediately after it is released from its initial position of 35◦ from the vertical? The acceleration of gravity g = 9.8 m/s^2

    If needed a graphical representation of the question can be found at the following link:
    http://i1221.photobucket.com/albums/dd470/jokerg908/435.jpg

    2. Relevant equations
    τ = Iα
    I = (1/3)MR2
    τ = FRsin(Θ)



    3. The attempt at a solution
    Finding the position of the center of mass (x meters from pivot):
    (x-7) = (14+3.2-x)
    x=12.1 meters

    Using xcm = 12.1 m, we can solve for α:
    I = (1/3)(18 kg)(12.1 m)2 ≈ 878.46 kg·m2
    τ = (18 kg)(9.8 m/s2)(12.1 m)sin(35°) ≈ 1224.264489 N·m
    α = τ/I ≈ 1.39361 rad/s2

    This answer is taken as incorrect by my online program. Any suggestions or assumptions that I have made in my calculations would be helpful. Thanks. One thing that I was worried about was that this method didn't use the moments of inertia that I was given.
     
    Last edited: Apr 7, 2014
  2. jcsd
  3. Apr 7, 2014 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Did you try working out the torque due to each mass separately?
    How did you combine the moments of inertia for the two parts of the rig? I'd have expected a sum in there somewhere.
     
    Last edited: Apr 7, 2014
  4. Apr 7, 2014 #3
    Moment of Inertia

    Should I have used the parallel-axis theorem (I think it's called):

    I = Icm + Md2

    Then I wouldn't know what I should take as my value of M due to the mass distribution of the object?

    My attempt to find I:

    I = (Isphere + Md2) + (Irod + Md2)
    I = (2/5 (9 kg)(3.2 m)2) + (3.2 kg)((D?)2) + ((1/3)(9 kg)(14 m)2) + (9 kg)((12.1 m)2)

    I don't know how to find D? = the distance between the sphere and the center of mass, because D would be greater than the radius of the sphere. So I feel that D > r would cause me to make an assumption that the sphere is greater than it actually is?
     
  5. Apr 7, 2014 #4
    Individual torques

    I quite honestly have no clue how to find, individual torques on a system composed of multiple parts.
     
  6. Apr 7, 2014 #5

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes, you should have.
    I think you have misunderstood the formula though - check that.
    the term "d" refers to the distance between the center of mass of the object going in a circle and the rotation axis. The "Icm" refers to the moment of inertia for the same object rotating about it's center of mass.

    Gravity acts on the com of each individual component separately.
    But I think you should concentrate on your understanding of moment of inertia first.
     
  7. Apr 7, 2014 #6
    d is the distance between the two parallel axes of rotation (aka the distance between the axis of rotation Icm (of that object) and the moment of inertia I about a parallel axis through some other point).

    In this case, we choose that other point to be the center of mass of the two combined objects in order to find the sum of their moments of inertia.

    Now that I am thinking about the problem a bit more, I think that one should take (D?) to be 17.2 meters. Because this is the distance between the sphere's center of mass, and the pivot point. Therefore you would be finding the moment of inertia about the pivot point for the sphere.
     
  8. Apr 7, 2014 #7

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You need to rethink this.
    The axis of rotation of the composite object is not at it's center of mass.
    Now you've got it.
     
  9. Apr 7, 2014 #8
    Think i figured it out.

    Correct I:

    I = (Isphere + Md2) + (Irod) (Note: Irod needs no adjustment because we are rotating about a perpendicular axis through the end of the rod.)
    I = (2/5 (9 kg)(3.2 m)2) + (9 kg)((17.2 kg)2) + (1/3)(9 kg)(14 m)2
    I≈3287.424 kgm2

    Xcm= 12.1 m

    PEi = KEf, rotational
    (18kg)(9.8 m/s2)(12.1*sin(90-35) m) = 0.5(3287.424 kgm2)(w2)
    ωfinal2 = 1.063708782 (rad/s)2
    ωfinal≈1.0313626 (rad/s)


    Applying analog to kinematics equations to find the solution:
    2α(Δθ) = ωf2 - ωi2
    where Δθ = (0-55)/2π, ωi2 = 0 rad/s

    α≈0.0607589035 rad/s2 (Edit: sign should be positive, however, I tried this answer. It's not correct! I am about to rip my hair out) I don't know what I am doing wrong.

    Any suggestions would be helpful, thanks. Do I have a large gap in my understanding of the concepts?
     
    Last edited: Apr 7, 2014
  10. Apr 8, 2014 #9
    @Tanya Sharma

    @Tanya Sharma

    To find the center of mass, you take the individual masses as point masses, and then calculate the center of mass between these two point masses.

    Xcm = [(m1)(x1)) + (m2)(x2)]/Total mass of the two objects

    Xcm = [(9 kg)(7 m) + (9kg)(17.2 m)] / (18 kg)

    Xcm = 12.1 m, where this distance is along the length of the rod.
     
    Last edited: Apr 8, 2014
  11. Apr 8, 2014 #10
    Hello Schulze

    The mass of the rod is not given in the question and from the OP I got the impression that mass of the rod was 18 kg .

    Anyways,if the mass of rod is 9 kg then you have correctly calculated the location of CM.

    This equation is applicable when angular acceleration α is constant. ‘α’ is variable in this problem.

    You have correctly calculated both the net torque on the system about the pivot , and the MI of the system .

    Now just apply τnet = Iα and get the value of α.

    Hopefully you should get the correct answer .
     
    Last edited: Apr 8, 2014
  12. Apr 8, 2014 #11
    Thanks for helping me out to understand the pieces of information that could be used to arrive at this answer.
    I did find out how to do the problem, and got the right answer!

    Is there any way that I mark this thread as solved?
     
  13. Apr 8, 2014 #12

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    You just did. Well done.
     
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: Angular Acceleration of a Fixed Rod
Loading...