Question

A machine part consists of a thin, uniform 4.00kg bar that is 1.50m long, hinged perpendicular to a similar vertical bar of mass 3.00kg and length 1.80m . The longer bar has a small but dense 2.00kg ball at one end.

By what distance will the center of mass of this part move horizontally and vertically if the vertical bar is pivoted counterclockwise through 90 degrees to make the entire part horizontal? Find the magnitude of horizontal displacement. Find the magnitude of vertical displacement.

Answer 1

The free body diagram for the given problem is shown in the following figure.

(xi-yi) (0.75 m, 0 m) (2(0 m, 0.9 m) (xsys) O (0 m, 1.80 m)

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Center of gravity coordinates for horizontal bar is, (1)150m0 (0.75 m,0 m) Center of gravity coordinates for vertical bar is )-00130 - (0m,0.90 m) Ceneter of gravity for denser ball is, (5%)-(0,180 m) Mass of the horizontal bar is, m, 4.00 kg Mass of the vertical bar is, m, -3.00 kg Mass of the denser ball is, m - 2.00 kg The x-component of the center of mass coordinates for the system is, mi+m2 +m3 (4.00 kg) (0.75)(3.00 kg) (0 m)+(2.00kg) (0m) (4.00 kg)+(3.00 kg)+(2.00kg) 0.333 m The y-component of the center of mass coordinates for the system is, (4.00 kg) (0.75m)+(3.00 kg) (0.90 m m)+(2.00 kg) (1.80m) (4.00kg)+(3.00kg)+(2.00 kg) 0.700 m

(r2y2) (xİyi ) (xivi) x1,y1 (0.9 m, 0 m) (1.80 m, 0 m) (0.75 m, 0 m)

Now the vertical bar is rotated 90° in counter clockwise direction. Hence, the new center of gravity corodinates are as shown in the above figure Hence, the new x-compoenent of the center of mass of the system is, (400 kg) (0.75 m)+(3.00 kg) (0.9 m(2.00kg) (1.80 m) 4.00kg) +(3.00 kg)(2.00kg) -1.033 m The new y-component of the center of mass coor dinates for the system is, (4.00 kg) (0 m)+(3.00 kg) (0 m m)(2.00 kg) (1m) (4.00 kg)+(3.00 kg)+(2.00kg) Threfore, the di stance of the center of mass corodinates in horizontal direction is cm 1.033 m-0.333 m 0.7m right in the vertical directi on: 0-0.7 m 0.7 m upward