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Chapter 1: Q30E (page 296)

Small blocks, each with mass m, are clamped at the ends and at the center of a rod of length L and negligible mass. Compute the moment of inertia of the system about an axis perpendicular to the rod and passing through

(a) the center of the rod and

(b) a point one-fourth of the length from one end.

Short Answer

Expert verified

The center of the rod is mL22.

thus, point one-fourth of the length from one end is 11mL216.

Step by step solution

01

Step:-1(a)

The blocks on the edge of the rod have a fixed time but the center is actually on the axis so it does not. And the formula is:

/=2mL22

02

Step:-2(a)

go toL2because blocks that distance from the rotating axis. And at the last moment of inertia to the axis located in the center of the rod.

l=mL22.

Hence , the center of the rod is mL22.

03

Step:-3 explanation 

One m mass is L / 4 and the other 3L / 4 rod has negligible weight, so one has two m points.

04

Step:-4 calculation 

I=2mL42+m3L42

I=11mL216

point one-fourth of the length from one end is 11mL216.

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