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!

Acceleration and velocity of a mass attached to a sphere

  1. Mar 29, 2016 #1
    1. The problem statement, all variables and given/known data

    Hi everybody! Still checking my comprehension of physics, and hopefully those posts will help other users!

    A homogeneous wooden sphere with mass M and radius R rotates about a perpendicular axis going through its center of mass. At the top of this axis is a homogeneous metal bar of length 2R and of same mass M. A massless string is rolled around the ball (see attached picture) and on its other end goes through a wheel and has a suspended mass attached to it. There is no friction.
    a) calculate the moment of inertia of the system wooden ball-metal bar regarding its rotation axis. The bar should be considered as infinitely thin. The moment of inertia of a sphere is JK = (2/5)MR2.
    b) what is the acceleration of the mass m in the field of gravity of Earth?
    c) what is the velocity of the mass m when it has gone down a distance h?

    2. Relevant equations

    Moment of inertia, torque, Newton 2nd and 3rd laws, conservation of energy

    3. The attempt at a solution

    Though it is not the hardest problem, I find the calculations messy and the risk of missing something high. Here is what I've done:

    a) First I calculate the moment of inertia of the bar:

    JS = (1/12)M(2R)2 = ⅓⋅MR2
    ⇒ Jtotal = (2/5 + ⅓)MR2 = (11/15)MR2

    Im pretty sure I can add the moments of inertia that way since they have the same axis of rotation, right?

    b) By Newton's 3rd law I know that the force of tension of the string on the mass is equal to the force of tension of the string on the sphere. I try to find an expression for T using the torque and the angular acceleration:

    Στ = T⋅R
    ⇒ α = Στ/J = 15T/11M

    The angular acceleration is also equal to tangential acceleration/radius aT/R, and in that case aT = a (the acceleration of the suspended mass):

    ⇒15T/11M = a/R ⇒ T = 11aR/15M

    I plug that into the Newton's 2nd law:

    a = ΣF/m = g - 11a⋅m⋅R/15M
    a = g/(1 + 11m⋅R/15M)

    c) For that part I set up an equation of conservation of energy. I assume that the system starts at rest:

    m⋅g⋅h = ½⋅m⋅v2 + ½⋅J⋅ω2

    Is that correct? I have a little doubt about the rotational energy.

    Then I substitute ω by v/R. Here again I have a little doubt: I allow myself this substitution because I think the velocity of the string at the sphere should be the same as the velocity at the mass. Anyway that gives me for the velocity:

    v = √(2⋅m⋅g⋅h/(m + 11M/15))

    What do you guys think? It's always a bit tricky not to misjudge such a situation..


    Thank you in advance for your answers!


    Julien.
     

    Attached Files:

  2. jcsd
  3. Mar 29, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Yes. Jtotal looks good.

    If you mean M to be in the denominator, then you should use parentheses: 15T/(11M). However, this expression is not correct as you can tell by the dimensions. Did you get all the factors of R taken care of?

    Again, you can check the dimensions of the last term in parentheses and see that this cannot be correct. Is little m equal to big M? If so, you can simplify.

    Your work for (c) looks good to me. Again, you can simplify if m = M. You can also check whether or not the answer for (c) is consistent with (b) by using the acceleration to find the final v.
     
  4. Mar 29, 2016 #3
    @TSny Thank you for your answer. I've reviewed the calculations and I indeed forgot a factor R. I now get T = (11/15)a⋅M and therefore a = g/(1 + 11M/(15m)). I believe the units are consistent this time.

    Unfortunately as far as I know m ≠ M, so I cannot simplify any further, but the R I forgot already simplified the expression a little bit since it does not depend on it anymore.


    Julien.
     
  5. Mar 29, 2016 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    OK. Your answers look right to me. You can check some limits to see if it makes sense. If M goes to zero, what is a? If M goes to infinity, what is a? If m goes to zero, what is a? If m goes to infinity, what is a?
     
  6. Mar 29, 2016 #5
    @TSny thanks I'm always fond of ways to check the consistency of an answer!

    - if M goes to 0, then the acceleration is g. Makes sense since the mass m would not be held back by anything then.
    - if M goes to infinity, the acceleration goes to 0. Makes also sense since m would then be held back by a much heavier mass.
    - if m goes to 0, the acceleration also goes to 0. There is no more mass on which to apply the force of gravity,
    - if m goes to infinity, a also goes to infinity. The force of gravity being applied onto an infinitely heavy mass, that also makes sense.

    That's great! I will make sure I use those instinctive methods of verification during my exams!


    Julien.
     
  7. Mar 29, 2016 #6

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Great!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted