Question

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 0.500 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 a negative value if the center of gravity is toward the 0.500 kg sphere and a positive value if the center of gravity is toward the 0.380 kg sphere.

Answer 1

Answer:

The difference in distance between the Center of gravity of the rod and that of the combined object is  $X_{C.G} = -3.4915 cm$  

Explanation:

A sketch of the free body diagram is shown in the first uploaded image

Looking at the diagram we can intuitively say that the position of the center of gravity of the steel would be position x = 0cm

The length of the first mass $m_1$from the position of center of gravity of the rod is

          $L_1 = 28cm$

This obtained by adding the length of the rod from one edge to the center + the radius of $m_1$

The length of the first mass $m_2$from the position of center of gravity of the rod is

          $L_2 = 26cm$

This obtained by adding the length of the rod from one edge to the center + the radius of $m_2$

Now to obtain the difference in distance between the center of gravity of the rod and that of the combined object

       $X_{C.G} = \frac{(m_1 *L_1)(m_2*L_2) (m * 0)}{m_1 +m_2 + m}$

Where m is the mass of the rod

                $= \frac{(0.500 * -28.0)+(0.300 * 0 )+(0.38*26.0)}{0.500 +0.300 +0.380}$

                $= \frac{-14 + 9.88}{1.180} =-3.4915cm$