Question

What is the moment of inertia of a cylinder with a radius of 0.62 m, height of 0.59 m, and density given by (0.686 − 0.343r + 0.224r²) kg/m³ about its center?

(a) 0.307 kg·m²
(b) 0.422 kg·m²
(c) 0.534 kg·m²
(d) 0.619 kg·m²

Answer 1

The moment of inertia of a cylinder with a varying density can be calculated by integrating the formula using the given density function and the formula for the volume element. In this case, the moment of

inertia is approximately 0.422 kg·m².

Step-by-step explanation:

To find the moment of inertia of a cylinder about its center, we need to use the formula:

I = ∫(r^2)dm

where r is the distance from the axis of rotation and dm is the mass element.

In this case, the density of the cylinder is given by (0.686 − 0.343r + 0.224r²) kg/m³. To find dm, we can multiply the density by the volume element:

dm = (0.686 − 0.343r + 0.224r²) * dV

where dV is the volume element.

Since the cylinder has a constant density, we can rewrite the formula as:

I = ∫(r^2)(dV * density)

Now we need to express dV in terms of dr (the thickness of the cylinder) and r (the distance from the center). The volume element can be expressed as:

dV = 2πr * dr

Substituting this into the formula for I:

I = 2π∫r^3(density * dr)

Finally, we integrate this formula over the entire cylinder to find the moment of inertia:

I = 2π∫(0.686r^3 − 0.343r^4 + 0.224r^5) dr

Calculating this integral, we find that the moment of inertia is approximately 0.422 kg·m².