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# physics !?

A uniform steel rod has mass 0.300 kg and length 40.0 cm and is horizontal. A uniform sphere with radius 8.00 cm and mass 0.300 kg is welded to one end of the bar, and a uniform sphere with radius 6.00 cm and mass 0.380 kg is welded to the other end of the bar. The centers of the rod and of each sphere all lie along a horizontal line.

How far is the center of gravity of the combined object from the center of the rod?

Express your answer with the appropriate units. Enter negative value if the center of gravity is toward the 0.300 kg sphere and positive value if the center of gravity is toward the 0.380 kg sphere.

### 2 Answers

- NCSLv 71 month agoFavorite Answer
divide the sum of the moments about the midpoint (easiest) and divide by the total mass. The bar contributes no moment about its center.

X_cm = (0.380kg*(20.0+6.00)cm - 0.300kg*(20.0+8.00)cm) / (0.300+0.300+0.380)kg

X_cm = 1.51 cm

Hope this helps!

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- az_lenderLv 71 month ago
Place the origin of your coordinate system at the center of the rod, and the lighter sphere at the "negative end" --

x-bar =

(0*0.30 - 24*0.30 + 23*0.38)/(0.98)

= 1.57.

The center of gravity is 1.57 cm to the right of the origin, that is, it is displaced towards the 0.380-kg sphere.

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