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Homework Help: Angular Speed of a bicycle

  1. Mar 21, 2008 #1
    [SOLVED] Angular Speed

    [tex]\tau[/tex] = r * F

    1. The problem statement, all variables and given/known data

    youve got our bicycle upside-down for repairs with its 66 cm diameter wheel spinning freely at 230 rpm. the mass of the wheel is 1.9 kg and is concentrated mostly at the rim. you hold a
    wrench against the tire for 3.1s with a normal force of 2.7 N. if the coefficient of friction between the wrench and the tire is 0.46 what is the FINAL ANGULAR SPEED OF THE WHEEL?

    2. Relevant equations
    diameter = 66 cm
    radius = .033m
    mass = 1.9 kg
    Norm Force = 2.7 N
    friction coef = .46
    [tex]\omega[/tex]0 = 230 rpm
    [tex]\omega[/tex]0 = 24.1 rad/s
    w = ?

    I = Mr^2

    [tex]\omega[/tex] = [tex]\omega[/tex]0 + [tex]\alpha[/tex] * t

    [tex]\tau[/tex] = r * F * sin(theta)

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

    v=[tex]\omega[/tex] * r

    v=[tex]\omega[/tex] * r

    3. The attempt at a solution

    [tex]\tau[/tex] = r * F

    where F is the Friction Force

    r*F = M * r^2 * [tex]\alpha[/tex]

    we can solve for [tex]\alpha[/tex] and it = 19.08

    [tex]\omega[/tex] = ( [tex]\omega[/tex]0 + [tex]\alpha[/tex] * t )

    [tex]\omega[/tex] = 85.48 rad/s
    that means it sped up...

    idk how to do this problem keep getting confused with the answers im getting

    any help?

    thanks in advance for your time and effort.
    Last edited: Mar 21, 2008
  2. jcsd
  3. Mar 22, 2008 #2
    initial angular speed= 230 rpm.

    torque = r * f * sin 90 degrees

    >> torque= .33 * .46 * 2.7 ( radius * frictional force)

    calculate torque.

    Now the big ques is:
    What is the moment of inertia of a wheel?
  4. Mar 22, 2008 #3

    Doc Al

    User Avatar

    Staff: Mentor

    You have a typo here that messed you up later. Radius = 0.33m.
    Perfectly correct, except that you used the wrong value for the radius.

    Note that the angular acceleration created by the friction is opposite to the initial angular velocity, so the wheel slows down.
  5. Mar 22, 2008 #4
    im having trouble interpreting my answer

    [tex]\omega[/tex]0 = 24.1 rad/s

    asuming that is right

    [tex]\omega[/tex] = [tex]\omega[/tex]0 + [tex]\alpha[/tex] * t

    then i put it in this eqn to find [tex]\omega[/tex] = 30.238 rad/s

    so next i would take [tex]\omega[/tex] - [tex]\omega[/tex]0 to find the change over 3.1 sec?

    thus producing an answer of 6.138 rad/s

    and then they want to know final angular speed so 24.1 - 6.138 = 17.962 rad/ s

    ^^^^that was the more logical approach ^^^^

    vvvv you can also say that vvvv

    this one just produces the same answer negative, so thats why it dosnt make sense

    [tex]\omega[/tex] = 24.1 rad/s

    [tex]\omega[/tex] = [tex]\omega[/tex]0 + [tex]\alpha[/tex] * t

    solve for [tex]\omega[/tex]0 and you get -17.962 rad/s, but then you gotta change the sign, which u can because it asks for speed
    Last edited: Mar 22, 2008
  6. Mar 22, 2008 #5

    Doc Al

    User Avatar

    Staff: Mentor

    That's the initial angular speed.

    As I already pointed out, the acceleration is negative. Done correctly, this will give you the final speed.

    The change is just alpha*time. What is your value for alpha?

    Seems like you're doing a bit of extra work.
    Not sure what you're doing here. Why would you solve for the initial speed? That's given.
  7. Mar 22, 2008 #6
    [tex] \omega [/tex] will not increase. It will decrease. You have to put a negative sign before [tex]\alpha[/tex] in that equation.
  8. Mar 22, 2008 #7
    that makes sense! and it produces the 17.962 rad/s instantly! thx for the help

    doc al, since i forgot to change accel to negative i showed how to produce the same answer, but with a lot more work, lol and i was just throwing out that idea for the 2nd part, i knew it didnt make sense but it was worth mentioning.
    Last edited: Mar 22, 2008
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