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Torque/angular acceleration

  1. Jan 12, 2004 #1
    A piece of metal is pressed against the rim of a 1.6kg, 19-cm-diameter grinding wheel that is turning at 2400 rev/min. The metal has a coeffiecient of friction of .85 with respect to the wheel. When the motor is cut off, with how much force must you press to stop the wheel in 20 sec.?

    Not sure I did this right, here's what I did:
    2400 rev/min = [tex] 80 \pi[/tex] rad/sec. = [tex] \omega_0 [/tex].
    t=20, and [tex] \omega_f = \omega_0 + \alpha t [/tex]. After plugging everything in (w_f=0) I got [tex] \alpha = -40 \pi [/tex] rad/sec^2. Since [tex] a = r \alpha [/tex], I got a = [tex] -3.4 \pi [/tex] m/sec^2. Then somehow from there I jumped to an answer of [tex] 4.624 \pi [/tex] N. I don't think that's right.
     
  2. jcsd
  3. Jan 12, 2004 #2

    Doc Al

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    Staff: Mentor

    [itex]\alpha = -4\pi[/itex]. Now that you've calculated [itex]\alpha[/itex], how do you relate this to force? Make use of [itex]\tau = I\alpha[/itex] to find the torque and thus the frictional force. You will need to know the rotational inertia (I) of a disk; look it up.
     
  4. Jan 12, 2004 #3
    So once I find the linear force of the wheel do I just multiply that by the coefficient of friction to find the force that I push with?
     
  5. Jan 12, 2004 #4

    Doc Al

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    Staff: Mentor

    No. The force you press with is the normal force; the force producing the torque on the wheel is the friction. Ffriction=μN.
     
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