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!

Linear acceleration, Angular acceleration, tension.

  1. Feb 6, 2017 #1
    1. The problem statement, all variables and given/known data
    A mass of 0.5 kg is suspended from a flywheel as shown in FIGURE 2.
    If the mass is released from rest and falls a distance of 0.5 m in 1.5 s.
    Mass of wheel: 3kg
    Outside rad. of gyration of wheel: 300mm
    Radius of gyration: 212mm

    calculate:

    (a) The linear acceleration of the mass.
    (b) The angular acceleration of the wheel.
    (c) The tension in the rope.
    (d) The frictional torque, resisting motion

    2. Relevant equations

    a = u t + 2s / t^2
    alpha = a / r

    3. The attempt at a solution

    a. a = u t + 2s / t^2
    = 0.444 m s^-2

    b. alpha = a r
    = 0.444 / 0.3
    = 1.48 rad s^-2

    c. Now I have found online that this is T = m (g - a) However neither my learning materials or the online answers I have found offer an explanation of this, other than it being derived from the second law F = m a. Any explanation on this would be appreciated, as I am not willing to use an answer I do not understand.

    d. I am yet to try this one so am not yet looking for help, no doubt I will need it soon.

    I am aware these have been answered elsewhere before, but as I said I am looking to understand c. specifically.
     
  2. jcsd
  3. Feb 6, 2017 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hello Tom, :welcome:

    I wonder what u is in your step a) and in your first relevant equation. Must be something from SUVAT and must be zero; am I right ?
    Can't find fault with your a) and b) answers, so if we aren't both wrong, all is OK there :smile:

    For c) the conceptual thing to do is draw a free body diagram for the 0.5 kg mass. Can you post it ?
    The complementary FBD is for the rope revving up the wheel. That'll come in handy for part d), too.
     
  4. Feb 6, 2017 #3
    Hello and thank-you!

    The U in my step is is the initial velocity value, which in this case is 0 so becomes irrelevant.

    I have this picture from the question:
    Untitled.jpg

    I understand how the acceleration of the mass due to gravity acts upon the 0.5kg mass and causes the F = m g to act upon the rope, however its the subtraction of the mass times the acceleration I do not understand, I thought if anything that would also be causing tension on the rope rather than negating it as the acceleration is in the same x direction!

    I'm sure there is something glaringly obvious I am missing!

    Edit: I am not sure if it is to do with the mass of the flywheel oposing the freefall of the object perhaps?
     
  5. Feb 6, 2017 #4

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Not correct. ##mg## acts on the mass, not on the rope.

    THE free body diagram I am waiting for (:smile:) shows two forces acting on the 0.5 kg mass: the force of gravity and the tension in the rope.
    The equation of motion for this mass is simply ##F_{\rm net} = ma ##
     
  6. Feb 6, 2017 #5
    Ohh!
    Gravity is acting on the mass pulling it downwards, however as we calculated it is only accelerating at 0.44 m s ^-2 not 9.81 m s ^-2 therefore the rope is subject to the tensional force of the difference between the two once plugged into the second law?
    I have drawn a FBD (which is how I arrived at this conclusion) Not sure if it's quite right but it worked (assuming I am on the right track!)

    Untitled.jpg

    Thank-you so much for the help so far (again assuming I am thinking along the right lines) it seems very obvious now!
     
  7. Feb 6, 2017 #6

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I really like this kind of positive feedback: yes, a free-body diagram can really help establish insight into what's happening.
    If the rope weren't there, the acceleration would be ##g## indeed, so the tension in the rope must be responsible for the difference.
    As you see from writing out ##\ F_{\rm net} = ma \ ## : $$T - mg = ma $$ (up = positive, ##g## = 9.81 m/s2) leading to ##a = -9.81 ## m/s2 when T = 0 (as 'desired') and to ##a = - 0.444## m/s2 with the value of T the exercise wants as answer to c).
     
  8. Feb 6, 2017 #7
    Thank-you so much, I haven't done anything educational like this for a good few years now and am doing a distance learning course. Its very rewarding and I am enjoying it, however the lack of a lecturer is often a restriction as discussion is a brilliant way to aid understanding!

    So I have had a go at part d. now that I have a figure for the tension.

    2. Relevant equations

    I = m k^2
    T = I alpha
    total torque = (m g - m a)r

    3. The attempt at a solution

    I = m k^2
    = 3 x 0.212^2
    = 0.1348

    The torque needed to accelerate the flywheel:
    t = I x alpha
    = 0.1348 x 1.488
    = 0.2001 N

    So frictional torque is the total tourque minus the torque needed for acceleration:
    Total= (m g - m a) x r
    = 4.683 x 0.3
    =1.405 N

    Frictional torque = 1.405 - 0.2001
    = 1.2049 N

     
  9. Feb 6, 2017 #8

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Looks good to me. One thing though: torque dimension is N m, not N
     
  10. Feb 6, 2017 #9
    Excellent, and thanks again for all your help!
     
  11. Feb 6, 2017 #10

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You're welcome. Good luck with your studies !
     
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: Linear acceleration, Angular acceleration, tension.
Loading...