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!

Torque and Moment of Inertia

  1. Apr 27, 2005 #1
    What would be the acceleration of a yo-yo if these are my datas:

    m= 10.456g
    r(small) = 0.7mm
    R= 1.9mm

    Now on I've had accelerations varying from 0.006m/s² to 930m/s² -yeah right...

    Worked on that prrety all of the time today and all I got left is a few so that I'm really in a hurry right now.
     
  2. jcsd
  3. Apr 27, 2005 #2
    You haven't enough data, as far as I can tell.
     
  4. Apr 28, 2005 #3

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Think of the torque of the centre of mass of the yo-yo about the point of tangential contact with the string:

    [tex]\tau = mgr_{small} = I\alpha[/tex]

    where I is the moment of inertia of the yo-yo about that small radius. Use the parallel axis theorem to work that out.

    Use the relationship between [itex]\alpha[/itex] and acceleration to find the acceleration of the centre of mass.

    Are you sure these dimensions are mm and not cm? That is one awfully small yo-yo!

    AM
     
  5. Apr 28, 2005 #4

    xanthym

    User Avatar
    Science Advisor

    SOLUTION HINTS:
    {Yoyo Small Radius} = r = (0.7 mm) = (7e(-4) meters)
    {Yoyo Large Radius} = R = (1.9 mm) = (1.9e(-3) meters)
    {Yoyo Mass} = (10.456 g) = (1.0456e(-2) kg)
    {Yoyo Moment of Inertia about CM Axis} = I = (1/2)*m*(R^2)
    {Force Acting Thru CM} = F = m*a
    {Gravitational Force on Yoyo} = m*g
    {String Tension on Yoyo} = T
    {Torque from String Tension about CM Axis} = τ = r*T

    For Linear Motion of CM:
    F = m*a =
    = T - mg
    ::: ⇒ a = (T/m) - g ::: <---- Eq #1

    For Rotational Motion about CM axis:
    τ = I*α = (1/2)*m*(R^2)*{-a/r} =
    = r*T ::: <---- Torque from tension "T" acting at radius "r"
    ::: ⇒ T = -(1/2)*m*a*(R/r)^2 ::: <---- Eq #2

    Placing Eq #2 into Eq #1:
    a = -(1/2)*a*(R/r)^2 - g
    ::: ⇒ a*{1 + (1/2)*(R/r)^2} = -g

    Solve for "a" in terms of "R" and "r" (and "g"), values for which are given in the problem statement.


    ~~
     
    Last edited: Apr 28, 2005
  6. Apr 28, 2005 #5
    something is wrong with your calculation
    I think T is confused with τ and R^2*-a/r is derived into a*(R/r)^2
    otherwise the answer makes sense either way (r² or r)
    lastly, you're corect sir Mason, I did translate my cm in mm AND add the mm sign therefore doing the process twice.
     
  7. Apr 28, 2005 #6

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Ok. So here is how I see it (which is equivalent to xanthym's approach):

    [tex]\tau = mgr = I\alpha = I(a/r) = (\frac{1}{2}mR^2 + mr^2)\frac{a}{r}[/tex]

    [tex]a = \frac{gr^2}{(\frac{1}{2}R^2 + r^2)} = \frac{2g}{(\frac{R^2}{r^2} + 2)}[/tex]

    I get a = 2.1 m/sec^2

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

Have something to add?



Similar Discussions: Torque and Moment of Inertia
Loading...