Physics Question- 2 Masses on Balance?

Two blocks each of mass 1.7 kg are suspended from the ends of a rigid weightless rod of length l1 = 176.0 cm and l2 = 13.0 cm. The rod is held in the horizontal position shown and then released. Calculate the initial acceleration of the mass attached to l1. Assume up is positive and down is negative.

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  • 5 years ago
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    I think your lenghts are backwards. From the picture l1 < l2, but in the problem it says l1 > l2. I will go by the picture.

    Sum of the moments about the pivot point = I*alpha

    Find the moment of inertia:

    Treat the blocks as point masses.

    I_point mass = m*r^2

    I_LHS = m*l1^2

    I_RHS = m*l2^2

    I_total = I_RHS + I_LHS = m*l1^2 + m*l2^2 = m*(l1^2 + l2^2)

    What are the moments? Moments that produce CCW spin will be treated as positive. The mass on the left hand side produces a moment of m*g*l1, which makes the system want to spin CCW (positive). The mass on the right hand side produces a moment of m*g*l2, hich makes the system want to spin CW (negative).

    Sum of the moments (CCW = +) = m*g*l1 - m*g*l2 = I_total*alpha

    m*g*l1 - m*g*l2 = I_total*alpha

    alpha = (m*g*l1 - m*g*l2) / I_total = m*g*(l1 - l2) / (m*(l1^2+l2^2)) = g (l1-l2) / (l1+l2)^2

    Plug in numbers, remember to convert cm to m:

    alpha = -5.13 rad/s^2

    The negative means that the system is going to rotate in the clockwise direction. This means that the LHS is rising.

    The tangential/agular acceleration relatonship

    a = alpha*r

    The radius in this case is l1

    a = alpha*l1

    Sub in alpha, but since we know it is going up, treat alpha as a positive number so our acceleration will be positive (up).

    a = 5.13 rad/s^2 * 0.13 m = 0.667 m/s^2 <--- ANS

    Hope I helped explain it.

    • RickB
      Lv 7
      5 years agoReport

      I am skeptical. Clearly the moments are caused by the tensions in the string, T1 and T2; that means the CCW moment is (T1)(l1) and the CW moment is (T2)(l2). But these tensions do NOT equal mg; otherwise the masses would not be accelerating! Not sure what the resolution is here.

  • 5 years ago

    The moment of inertia of this rod is 1.7*0.13²+1.7*1.76² = 5.30

    The net torque = 1.7g(1.76-0.13) = 27.16

    We know that Net Torque = I *α => α = T/I = 27.2/5.3 = 5.13 rad/s²

    a = r*α = 1.76*5.13 = 9.03m/s²

    Notice that if you use 0.13m instead of 1.76m a = 0.667m/s²

    which makes sense, the further from the point of rotation, the greater the tangential

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