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 0.900 kg is welded to one end of the bar, and a uniform sphere with radius 6.00 cm and mass 0.580 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.

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

Answer 1

the center of gravity of the system is displaced towards the heavier sphere by a distance of x = 0.05686 m measured from the center of the bar.

Step-by-step explanation:

The center of mass of a body is given by

`x_(cm) = (1)/(M) \sum x_im_1`

where M is the total mass of the body

The center of gravity is the point where the weight of a body is applied, in the case of small and homogeneous bodies, the two points coincide, this is the value of the acceleration of gravity in the size of the body does not change

In our case we have three small bodies,

the center of mass and gravity of the bar is at its geometric center, if we place the origin at one end

x_{bar} = L / 2

x_{bar} = 0.400 / 2

x_{bar} = 0.200 m

the center of gravity of a sphere coincides with its geometric center

for the sphere 1 x_{cm1} = 0.0800 m

for the sphere 2 x_{cm2} = 0.0600 m

since they indicate that all objects are on the same horizontal line, we look for the center of gravity of the combined body

M = m₁ + m₂ + m₃

M = 0.300 + 0.900 + 0.580

M = 1.78 kg

`x_(cm) = (1)/(M) \ ( m_1x_1 + m_2x_2 + m_2x_3 )`

taken the zero at the ends of the bar

note that the distance to the heaviest sphere is negative because it is to the left of the reference system and the distance to the furthest sphere is the radius plus the length of the bar

x_{cm} =
`(1)/(1.78)` (0.9000 (-0.08) + 0.300 0.20 + 0.580 (0.40 + 0.06) )

x_{cm} =
`(1)/(1.78)` (-0.072 + 0.06 + 0.2668)

x_{cm} = 0.143 m

this distance for the center of gravity is from the origin of our reference system, if we change the origin to the geometric center of the bar

x_{cm} ’=
`(L)/(2) - x_(cm)`

x_{cm} ’= 0.20 - 0.143

x_{cm} ’= 0.05686 m

In other words, the center of gravity of the system is displaced towards the heavier sphere by a distance of x = 0.05686 m measured from the center of the bar.