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Need help with torque and angular inertial problems

  1. Nov 11, 2005 #1
    Figure 10-43 shows particles 1 and 2, each of mass m, attached to the ends of a rigid massless rod of length L1 + L2, with L1 = 19 cm and L2 = 79 cm. The rod is held horizontally on the fulcrum and then released.
    [​IMG]
    a.What is the magnitude of the initial acceleration of particle 1?
    b.What is the magnitude of the initial acceleration of particle 2?

    I have no idea where to begin with this problem when the only thing they give me is length. Should mass be neglected in the equation for angular inertia?
    In Figure 10-53, two blocks, of mass m1 = 310 g and m2 = 630 g, are connected by a massless cord that is wrapped around a uniform disk of mass M = 500 g and radius R = 12.0 cm. The disk can rotate without friction about a fixed horizontal axis through its center; the cord cannot slip on the disk. The system is released from rest.
    [​IMG]
    Find the magnitude of the acceleration of the blocks.
    a= 2.635 m/s^2 (I got the right answer, but I don't specifically understand the steps to getting it)
    b. Find the tension of T1
    c. Find the tensin of T2

    For the last problem in finding the tension, why does the following not work?
    -Fg + T1 = ma
    T1 = Fg + ma
    T1 = (.310 kg * 9.8m/s^2) + (.310 kg * 2.635 m/s^2)
     
  2. jcsd
  3. Nov 11, 2005 #2

    Doc Al

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    Staff: Mentor

    How can you ignore the mass when talking about inertia?!
    Apply Newton's 2nd law for rotational motion: [itex]\tau = I \alpha[/itex]

    Show how you solved for the acceleration.
    Assuming you have the correct acceleration (I haven't checked) then that should give you the correct tension. Show how you solved for the acceleration.
     
    Last edited: Nov 11, 2005
  4. Nov 11, 2005 #3
    Thank you for the reply. You are right though, I did get the correct answer for the tension. I have no idea what I did wrong before. I must had punched in the wrong values or something.
    How did I get the acceleration? Professly, I would like to understand this better. I originally got help with this problem, but the person never really explained his steps.
    F = ma
    F = (m2 - m1)
    F = (M1 + M2)a + Ia (angular) / R I'm unsure how Ia/R comes into play
    (m1 + m2 + 0.5M)a = (m2 - m1)g
    a = (g(m2 -m1)) / (m1 + m2 + .5M)
    a = 2.635 m/s^2


    I know you need mass for t = Ia , but the fact that there is no value for m is what makes me confused as to what to do. I am surmizing that 9.8 m/s^2 factors into that problem somehow?
     
  5. Nov 11, 2005 #4

    Doc Al

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    Staff: Mentor

    That's one way to solve it, but I don't recommend it. Instead I'd ask you to analyze each object separately, applying Newton's 2nd law to each. You'll end up with 3 equations (one for each mass and one for the disk) that you can solve for a, T1, & T2. Give it a shot and see how you do.

    You'll need to use [itex]a = \alpha R[/itex], to relate the linear and angular accelerations. (9.8 m/s^2 is the acceleration due to gravity!) You are given the masses of each object. (What's the rotational inertia of a disk?)
     
  6. Nov 11, 2005 #5
    Oh I'm sorry by my last statement of "I know you need mass for t = Ia , but the fact that there is no value for m is what makes me confused as to what to do. I am surmizing that 9.8 m/s^2 factors into that problem somehow?"
    I was referring to this problem.
    Figure 10-43 shows particles 1 and 2, each of mass m, attached to the ends of a rigid massless rod of length L1 + L2, with L1 = 19 cm and L2 = 79 cm. The rod is held horizontally on the fulcrum and then released.
    [​IMG]
    a.What is the magnitude of the initial acceleration of particle 1?
    b.What is the magnitude of the initial acceleration of particle 2?
     
  7. Nov 11, 2005 #6

    Doc Al

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    Staff: Mentor

    You are given all the mass there is. (The rod has no mass; all the mass is at the ends of the rod.) What's the rotational inertia of a mass at the end of massless rod? (Hint: All the mass is concentrated at a single point. How far is the mass from the axis of rotation?)

    And 9.8 m/s^2 is still the acceleration due to gravity! Of course you'll need it.
     
  8. Nov 11, 2005 #7
    I still don't have a clue as to what I should do. Okay providing I'm on the right track. I should start out with...
    T=Ia
    T = mr^2a
    a = T / mr^2


    The problem is I don't know what else to do from here. Since the rod is massless does this mean that m is ignored or something? There's no force mentioned in the problem so that the net torque can be solved. The only thing I have is the radius.
     
  9. Nov 12, 2005 #8

    Doc Al

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    Staff: Mentor

    You are on the right track. I'll rewrite that as follows so as not to confuse angular acceleration with linear acceleration:
    [tex]\tau = I \alpha[/tex]

    One mass is a distance [itex]L_1[/itex] from the axis; the other, [itex]L_2[/itex]. Find the total rotational inertia of both masses.
    The mass of the rod is assumed to be negligible; but it does serve to hold the two masses (m) in place.
    If there's mass, then there's gravity.

    Once you find the angular acceleration, you can find the linear acceleration using:
    [tex]a = \alpha R[/tex]
     
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