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Angular Acceleration Caused by Axehead on a Rotating Grindstone (Disk)

  1. Jul 17, 2014 #1
    1. The problem statement, all variables and given/known data
    PROBLEM:
    You have a horizontal grindstone (a disk) that is mass m, has a radius r, and is turning at f in the positive direction. You then press a steel axe against the edge with a force of F in the radial direction.

    y5gp4ep3.s2p.png

    RANDOMIZED VARIABLES:
    m= 95 kg
    r= 0.33 m
    f= 92 rpm
    F= 25 N

    a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone in rad/s2
    ANSWER: α = -0.319
    See section 3 for explanation​

    b) What is the number of turns, N, that the stone will make before coming to rest?
    SUBMISSION HISTORY:
    XN = 0.003
    XN = 0.017

    2. Relevant equations

    τ=rFsin(θ)
    τ=Iα
    α=τNET/I
    ↓ Moment of Inertia of a uniform disk about its center of mass
    I=[itex]\frac{1}{2}[/itex]mR2
    Fk= μkF
    ωfi+αΔt
    θfiiΔt+[itex]\frac{1}{2}[/itex]αΔt2

    3. The attempt at a solution

    I solved part a) by doing the following
    Kinetic Friction Force​
    Fk= μkF
    Torque by Fk
    τ=rFsin(θ) = rFk = rμkF
    Magnitude of Angular Acceleration (will be negative)​
    α=τNET/I = rμkF/[itex]\frac{1}{2}[/itex]mr2
    = (0.33 m)(0.2)(25 N) / [itex]\frac{1}{2}[/itex](95 kg)(0.33 m)2
    = -0.319 rad / s2

    OKAY. Now onto the next part. Part b).
    First I took the initial velocity, which is in rotations per minute, and converted it into radians per second
    [STRIKE]rot[/STRIKE]/[STRIKE]min[/STRIKE] ·2∏r/[STRIKE]rot[/STRIKE] · 1 [STRIKE]min[/STRIKE]/60 s
    =∏/30 rad/s

    so if ωi = ∏/30 rad/s
    & ωf = 0 rad/s
    I can solve for time by using the formula ωfi+αΔt
    0 rad/s = ∏/30 rad/s + -0.319 rad/s2Δt
    -∏/30 rad/s = -0.319 rad/s2Δt
    Δt = 0.3283 s

    now using this time, I plug it and the rest of my variables into the formula θfiiΔt+[itex]\frac{1}{2}[/itex]αΔt2
    =(∏/30 rad/s)(0.3283 s) + [itex]\frac{1}{2}[/itex](-0.319 rad/s2)(0.3283 s)2
    =0.03438 rad + -0.0172 rad
    =0.01719 rad
    1rot/2∏rad · 0.01719 rad = 0.002736 rot

    I tried entering that, and it was incorrect. Then I tried entering the answer I got in radians and that was incorrect too.. So I guess I'm just confused.
     
  2. jcsd
  3. Jul 17, 2014 #2
    Oh wow okay

    I read back over my post and I already realized what I did... so I'm sorry.

    for the initial speed I didn't convert it to radians per second... I converted one rotation to radians per second.... smoothe.
    Lemme try that again
    f=92 [STRIKE]rot[/STRIKE]/[STRIKE]min[/STRIKE] ·2∏r/[STRIKE]rot[/STRIKE] · 1 [STRIKE]min[/STRIKE]/60 s
    =92∏/30 rad/s
    = 3.07∏ rad/s

    so if ωi = 3.07∏ rad/s
    & ωf = 0 rad/s
    I can solve for time by using the formula ωfi+αΔt
    0 rad/s = 3.07∏ rad/s + -0.319 rad/s2Δt
    -3.07∏ rad/s = -0.319 rad/s2Δt
    Δt = 30.2 s

    now using this time, I plug it and the rest of my variables into the formula θfiiΔt+[itex]\frac{1}{2}[/itex]αΔt2
    =(3.07∏ rad/s)(30.2 s) + [itex]\frac{1}{2}[/itex](-0.319 rad/s2)(30.2 s)2
    =291.27 rad + -145.47 rad
    =145.8 rad
    1rot/2∏rad · 145.8 rad = 23.205 rot

    okay yes this is correct. CRISIS AVERTED SORRY
     
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