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Torque Lab

  1. Nov 20, 2007 #1
    1. I'm doing a lab with a coffee can rolling down a ramp. I'm asked to 'determine average torque' acting on the can while its rolling down the ramp. Then I'm asked to compare this value with I*alpha (angular acceleration) (I'm guessing because torque = I*alpha). So I need to calculate torque two ways, but I have no idea how!?

    I know the angle of incline, the mass of the coffee can, the radius of the coffee can, the distance it was displaced in meters and radians, the angular and translation velocities (and accelerations) and its moment of inertia. Any ideas!?


    2. Relevant equations



    3. The attempt at a solution

    I said

    Fnet = Torque - Ff

    this is just so horribly wrong.
     
  2. jcsd
  3. Nov 21, 2007 #2

    andrevdh

    User Avatar
    Homework Helper

    To calculate the average torque use

    [tex]\tau _a = I_{COM} \alpha[/tex]

    for the angular acceleration one have that

    [tex]\alpha = \frac{a_{COM}}{R}[/tex]

    the linear acceleration of the COM of the can down the ramp can be evaluated from

    [tex]a_{COM} = \frac{g\sin(\theta)}{1 + \frac{I_{COM}}{MR^2}}[/tex]


    The experimental determination of the average torque can be again be determined from the same equation by determining [tex]a_{COM}[/tex] experimentally

    [tex]L = \frac{1}{2}a_{COM}t_L ^2[/tex]

    where [tex]L[/tex] is the distance down the ramp and [tex]t_L[/tex] the time it takes to cover this distance. From which [tex]a_{COM}[/tex] can be calculated.
     
    Last edited: Nov 21, 2007
  4. Nov 21, 2007 #3
    [tex]
    \theta = \omega_i + \frac{\alpha t^2}{2} = \frac{\alpha t^2}{2}
    [/tex]
    [tex]
    2\theta/t^2 = \alpha
    [/tex]
    [tex]
    \theta = S_{ramp}/R_{can}
    [/tex]
    [tex]\alpha = 2(S_{ramp}/R_{can}) / t^2[/tex]


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


    [tex]mgh - \frac{1}{2}mv^2 = \frac{1}{2}I \omega^2[/tex]
    [tex]d = v_it + \frac{1}{2}at^2 = \frac{1}{2}at^2[/tex]
    [tex]2d/t^2 = a[/tex]
    [tex]v_f = v_i + at = 2d/t[/tex]

    [tex]mgh - \frac{1}{2}m(2d/t)^2 = \frac{1}{2}I\omega^2[/tex]

    [tex]\omega_f = \omega_i + \alpha t = \alpha t[/tex]
    [tex]\omega_f = 2(S_{ramp}/R_{can}) / t[/tex]

    [tex]2(2(S_{ramp}/R_{can}) / t)^2(mgh - \frac{1}{2}m(2d/t)^2)= I[/tex]

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

    [tex]2(2(S_{ramp}/R_{can}) / t)^2(mgh - \frac{1}{2}m(2d/t)^2) * 2(S_{ramp}/R_{can}) / t^2 = \tau[/tex]

    Or something to that effect. This wasn't meant to learn from, it was meant to inspire. I'm sure I've made more than a few mistakes (point them out, I have an important physics exam in 2 days).
     
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