Question

The parallel axis theorem relates Icm, the moment of inertia of an object about an axis passing through its center of mass, to Ip, the moment of inertia of the same object about a parallel axis passing through point p. The mathematical statement of the theorem is Ip=Icm+Md2, where d is the perpendicular distance from the center of mass to the axis that passes through point p, and M is the mass of the object. Part A Suppose a uniform slender rod has length L and mass m. The moment of inertia of the rod about about an axis that is perpendicular to the rod and that passes through its center of mass is given by Icm=112mL2. Find Iend, the moment of inertia of the rod with respect to a parallel axis through one end of the rod. Express Iend in terms of m and L. Use fractions rather than decimal numbers in your answer. Part B Now consider a cube of mass m with edges of length a. The moment of inertia Icm of the cube about an axis through its center of mass and perpendicular to one of its faces is given by Icm=16ma2. (Figure 1) Find Iedge, the moment of inertia about an axis p through one of the edges of the cube Express Iedge in terms of m and a. Use fractions rather than decimal numbers in your answer.

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

part a.

$I = \frac{1}{3}ML^2$

part b.

I = $\frac{2}{3}Ma^2$

Explanation:

Part a.

For the center of mass the moment of inertia is:

$I_{cm}=\frac{1}{12}ML^2$

Using the parallel axis theorem, we get:

$I = I_{cm}+Md^2$

I = $\frac{1}{12}ML^2+Md^2$

Additionally,

d = L/2

so:

$I =\frac{1}{12}ML^2+M\frac{L^2}{4}$

$I = \frac{1}{3}ML^2$

part b.

For the center of mass the moment of inertia is:

$I_{cm}=\frac{1}{6}Ma^2$

Using the parallel axis theorem, we get:

$I = I_{cm}+Md^2$

I = $\frac{1}{6}Ma^2+Md^2$

Adittionally, in this case:

d = $\sqrt{(\frac{a}{2})^2+(\frac{a}{2})^2}=\sqrt{\frac{a^2}{2}}=\frac{a}{\sqrt{2}}$

so:

$I =\frac{1}{6}Ma^2+M(\frac{a}{\sqrt{2}})^2$

$I =\frac{1}{6}Ma^2+M\frac{a^2}{2}$

I = $\frac{2}{3}Ma^2$