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Rotational motion question

  1. Oct 20, 2008 #1
    1. There is a figure showing two thin beams joined at right angles. The vertical beam is 15.0 kg and 1.00m long and the horizontal beam is 25.0 kg and 2.00m long.

    Part a) askes to find the center of gravity to the two joined beams, taking the origin at the corner where the beams join.

    Part b) calculating the gravitational torque.

    2. Relevant equations: Xcg= (x1m1 + x2m2 + ...)/ (m1 + m2 +...)

    Ycg= (y1m1 + y2m2 + ...)/(m1 + m2 + ...)

    torque net= t1 +t2



    Tnet =T1 + T2

    3. I have no idea where to start this problem. We do problems similar in class but with a mass at each end of a "massless" rod.

    Starting off the right angle is oriented so the 25 kg 2.00 m bar would run along the + x axis and the 15 kg 1.00m bar would run along the + y axis.

    I set x1 m1 t1 as the 25 kg 2m bar.

    To calculate the x coord for center of gravity I did Xcg= (25.0kg)(2.00m)+(0m)(15.0kg)/(25.0kg+15.0kg)

    = 1.25m
    I used zero as m2 because I did not think it would contribute to the x coord system.

    Ycg= (1.0kg)(15.0kg)+(0m)(25.0kg)/(25.0kg + 15.0kg) = .375m

    so for part A I got (1.25m+.375m)

    Part B)

    I set toque 1 for the 2m bar. T1= -x1m1g (neg becuase it want to rotate CW) so T1= -25kg(1.25m)(9.8m/s^2) = -306.3 Nm

    torque 2 was for 1.0m bar. T2= +x2m2g T2 (I figured it would want to undergo CCW rotation motion so I thought it would be +) T2= +15kg(1.25m)(9.8m/s^2) =183.75 Nm

    Toque net =T1+T2 -306.3 Nm + 183.8Nm = -122.5 Nm

    I would really like some help in the equations and setting up. I feel like 122.5 Nm is to large of a torque.
  2. jcsd
  3. Oct 20, 2008 #2
    Your center of mass is wrong. You can calculate the center of mass of a bunch of beams by treating each beam as if it were a point particle located at its own center of mass, which should be half-way along the beam. In other words, use the half way points of the beams rather than the end points of the beams.
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