Part A Part B The parallel axis theorem relates Icm Icm, the moment of inertia o

Question

Part A

Part B

The parallel axis theorem relates Icm Icm, the moment of inertia of an object about an axis passing through its center of mass, to Ip 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 Ip = Icm + M d^2 , where d d is the perpendicular distance from the center of mass to the axis that passes through point p, and MM is the mass of the object. Part A Suppose a uniform slender rod has length L L and mass m 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=1/12mL^2 Icm = 1/12m L^2 . Find Iend Iend , the moment of inertia of the rod with respect to a parallel axis through one end of the rod. Express Iend Iend in terms of m m and L L. Use fractions rather than decimal numbers in your answer. Part B Now consider a cube of mass m m with edges of length a a. The moment of inertia Icm Icm of the cube about an axis through its center of mass and perpendicular to one of its faces is given by Icm=1/6ma^2 Icm = 1/ 6 m a^2 . (Figure 1) Find Iedge Iedge , the moment of inertia about an axis p through one of the edges of the cube Express Iedge Iedge in terms of m m and a a. Use fractions rather than decimal numbers in your answer.

Explanation / Answer

a) 1/12 M L^2 + M (L/2)^2 = I
Moment about C.M + M d^2   where d is distance from C.M.
I = M L^2 / 3
b) d = [(a/2)^2 + (a/2)^2]^1/2 = 2^1/2 a / 2
where d is distance of edge from center of cube
I = M (1/6 + 1/2) a^2 = 2 M a^2 / 3

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