Question

A structure consists of a 2.0 kg ball of radius 6.0 cm (mass uniformly distributed over the volume) and

a 500 g ball of radius 4.0 cm (mass uniformly distributed over the volume), connected by a 800 g, 40.0
cm long rod of nonuniform mass distribution. The rod's linear mass density (mass per length) is a =
cx, where x is measured from the left end the rod (connected to the 2.00 kg ball) and c is a constant (of
value 37.5 in the SI units).
(a) What are the units of c?
(b) What is the position of the center of mass of the rod with respect to the left end the rod?
(c) What is the position of the center of mass of the structure with respect to the center of the 2.00
kg ball?

Answer 1

The units for c are kg/m², and the center of mass of the rod and structure can be determined by integration and summation of each part's mass and distance from a reference point.

Step-by-step explanation:

Calculating the Center of Mass

To find the units of the constant c in the linear mass density equation a = cx, we need to understand that linear mass density has units of mass per length; in SI units, that's kilograms per meter (kg/m). Since x is a length with units of meters (m), the units for c must result in kg/m when multiplied by m. Hence, the units for c are kg/m².

To find the position of the center of mass of the rod, we use the integrated linear mass density over the length of the rod:
x₀ = (1/M)∫ (x ⋅ a(x)) dx, where M is the total mass of the rod and a(x) = cx. Using the given mass and length for the rod, along with the constant c, we solve the integral to find the center of mass.

Finally, the position of the center of mass of the entire structure in relation to the center of the 2.00 kg ball can be found by considering the mass and position of each component (the two balls and the rod) and using the formula:
x₀ = (1/Mₓ)Σ(mₗ ⋅ xₗ), where Mₓ is the total mass of the structure, and mₗ and xₗ are the mass and position of each part of the structure.