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Homework Help: Rotational Dynamics

  1. Nov 26, 2005 #1
    Hello there, ive got a problem here.
    A solid cylinder of radius 0.8 m and mass 4.0 kg is at rest. A 4.0 N m torque is applied to the cylinder about an axis through its centre.

    Calculate the angular velocity of the cylinder, in rad s-1 , after the torque has been applied for 2.0 s.

    how do i go about solving this problem. what should i be considering first.
    thanks in advance
  2. jcsd
  3. Nov 26, 2005 #2


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    Science Advisor

    Think of Newton's second law for rotational systems. The linear version is F=ma. What is the rotational equivilent?
  4. Nov 26, 2005 #3
    torque=(m.r^2).angular acceleration
    since torque=F.r=ma.r
    where a=r.angular acceleration.
  5. Nov 26, 2005 #4
    I'm really not sure what you just wrote there, though I have a feeling you're missing something. If I'm correct, you have the form of the equation as

    [tex]\sum \tau = (m r^2) \alpha[/tex]

    It also looks like you have your definition of torque mixed up. Torque is defined as such: [tex]\tau = R cross s[/tex]. (what exactly is the tex for cross product x as opposed to curly mathematical x?) Since the cross product of R and s is ||R|| ||s|| sin(angle between r and s), if R and s are at a 90 degree angle it reduces down to the [tex]\tau = F r[/tex] that you stated, but only if they meet at a 90 degree angle. (You can, of course, take components to find the perpundicular and parallel forces ;) )

    However, your statement that the sum of the torques is equal to mr^2*alpha is false in all but one case. The proper general form is

    [tex]\sum \tau = I \alpha[/tex]

    where I is the moment of inertia of the body.
    Last edited: Nov 26, 2005
  6. Nov 27, 2005 #5


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    Homework Helper

    Somehow, you got this far into your physics course and you STILL ignore the Sum:
    [tex]\Sigma \vec{F} = m \vec{a}[/tex] , not just F = ma .

    In rotation, rotational Inertia depends on shape, which can change! Learn:
    [tex]\Sigma \vec{\tau} = \frac{\Delta \vec{L}}{\Delta t} = sometimes = I \vec{\alpha} [/tex] (if shape is constant).
    each [tex]\vec{\tau} = \vec{r} \times \vec{F} [/tex] helps to cause the effect, on the right-hand-side.
    (NEVER ignore the Delta, which operates on the L and the t).
    In your case, [tex]L = I \omega[/tex] , where solid disk (cylinder) Inertia is
    (look up a table in your book. BOOKMARK it. Shapes differ!) I = ½ M R^2 .
    Last edited: Nov 27, 2005
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