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Torque and accleration

  1. Feb 6, 2010 #1
    could some on help me understand why the following is true?
    for some of them i wrote exactly what i don't understand, for others
    i really have no idea

    Two uniform rods are connected to a table by pivots at one end. Rod B is longer than rod A. Both are released simultaneously from an initial angle θ. Neglect air friction. NOTATION: CM = center of mass; α = angular acceleration; |ay|= size of downward acceleration.

    True: αA and αB are dependent on θ.
    just don't understand why.

    False: |ax| of the CM initially equals 0 for both rods.
    this would be false because of the centr. force right?

    True: |ay| is initially equal for the CM of A and B.
    that is because it is g right?
    True: Just before landing, the CM of B has a greater speed than the CM of A.
    why so? i found an explanation on the internet but there they comapered mgh to 0.5Iw^2
    and from there proved b's speed is larger therfore the answer.
    but why can i use such an equation? why do i not take into consideration the kinetic energy
    of center of mass and how do i define the height here? according to the center of mass?

    False: Rods A and B hit the table at the same time.

    False: αA and αB are the same initially.

    True: αA and αB both increase with time.

  2. jcsd
  3. Feb 6, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    How do you find the angular acceleration? What's Newton's 2nd law for rotation?

    Find the initial acceleration of the CM by first finding the initial angular acceleration.

    The initial acceleration is not g. (It's not in free fall.)

    If you measure I from the axis of rotation, you are automatically including the KE of the center of mass. The change in gravitational PE can be measured by the change in height of the center of mass.

    Figure out α as a function of θ.
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