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Rolling Objects on an Incline

  1. Nov 17, 2006 #1
    Please check my setups and kindly tell me the correct method of finding the answers if possible.

    1) The radius of a 3.0 kg wheel is 6.0 cm. The wheel is released from rest a point A on a 30 degree incline. The wheel rolls without slipping and moves 2.4 m to point B in 1.20 s. The moment of inertia of the wheel is closest to:

    a.0.0057 kg*m^2
    b.0.0048 kg*m^2
    c.c. 0.0051 kg*m^2
    d.0.0054 kg*m^2
    e.0.0060 kg*m^2

    At first, I thought it would be easy by just using I = 0.5*m*r^2 = 0.0054 kg*m/s^2.

    Then, I thought of the conservation of energy.

    mgh = 0.5*mv^2 + 0.5*I*w^2, where w = v/r

    h = L*sin(30) = 1.2 m, where L = 2.4 m

    Is v = 2.4 m/ 1.2 s = 2m/s? w = (2 m/s)/(.06 m) = 33.33 rad/s
    I have all values so isolate I.
    I = (2*29.28 J)/ (w^2) = (2*29.28 J)/ (33.33 rad/s)^2 = 0.0521 kg*m^2

    This doesn’t match any of the choices.
    What did I do wrong?


    2) A hoop is released from rest at the top of a plane inclined at 24 degrees above horizontal. How long does it take the hoop to roll 24.0 m down the plane?

    a. 3.27 s
    b. 4.91 s
    c. 4.48 s
    d. 7.70 s
    e. 3.13 s


    I used conservation of energy for this part. I = M*r^2 for a hoop.

    If L, length of incline, is 24 m then hoop must travel, height = L*sin (24) = 9.76168 m.

    mgh = 0.5*mv^2 + 0.5*I*w^2, where w = v/r

    mgh = 0.5*mv^2 + 0.5(m*r^2)*(v/r)^2

    mgh = 0.5m*v^2 + 0.5*m*v^2

    mgh = m*v^2

    v = sqrt(g*h) ???

    If I do that, v = sqrt(9.8 m/s^2*9.76168 m) = 9.78998 m/s

    24 m/9.78998 m/s = 2.4514 s
    What did I do incorrectly??

    Thank you.
     
  2. jcsd
  3. Nov 17, 2006 #2
    You might need to check your moment of inertia. What does the wheel look like?
     
  4. Nov 17, 2006 #3
    Well, from the side view in the diagram it is a circle (like O for cylinders) that is rolling down an incline.
     
  5. Nov 17, 2006 #4

    Doc Al

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    Staff: Mentor

    prob #1

    Looks like a match to me. But that assumes the wheel is a perfectly symmetric disk.

    That's the average speed, not the final speed at the bottom. (It starts from rest.)
     
  6. Nov 17, 2006 #5

    Doc Al

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    Staff: Mentor

    prob #2

    You made the same error as in prob #1, only in reverse. (The v you calculated is the final speed, not the average speed.)
     
  7. Nov 17, 2006 #6
    For #1, I need to find the final speed, but, in #2, I need to find the average speed??:confused:
     
  8. Nov 17, 2006 #7

    Doc Al

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    Staff: Mentor

    Yes. In #1, you are finding the energy at the bottom--thus final speed is needed. In #2, you are using t = d/v, which is only true for constant speed or average speed.
     
  9. Nov 17, 2006 #8
    How do I find the final speed for #1 and the average speed for #2?


    For the final speed, do I have to use a kinematics equation?

    For #2, the initial speed is 0 m/s and the final is 9.78998 m/s.
    v^2 = 2*a*x
    a = (9.78998)^2/(2*24 m) = 1.9967 m/s^2

    v_f = a*t
    t = (9.78998 m/s)/(1.9967 m/s^2) = 4.903 s for #2 ?
     
    Last edited: Nov 17, 2006
  10. Nov 17, 2006 #9

    Doc Al

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    Staff: Mentor

    That's fine, but there's an easier way to find the average speed from the final speed. Since the acceleration is constant, V(ave) = (Vi + Vf)/2.
     
  11. Nov 17, 2006 #10
    For #1, I found:

    the final velocity is v_f = 2*2m/s = 4 m/s if v_i = 0.

    Now I have to change my kinetic engergy value.

    0.5*(3 kg)*(4 m/s)^2 = 24 J

    PE = 3Kg*9.8*1.3 m = 35.28 J

    11.28 J = (I*w^2)/2

    I= 2*11.28 J/(w^2)

    w = 4 m/s/0.06 m = 66.67 rad/s

    I = (2*11.28)/(66.67^2) = 0.005075 kg*m^2
     
  12. Nov 17, 2006 #11

    Doc Al

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    Staff: Mentor

    Looks OK to me.
     
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