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# Small blocks, each with mass m, are clamped at the ends ## Problem 32E Chapter 9

University Physics | 13th Edition

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Problem 32E

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.

Step-by-Step Solution:

Solution 32E Step 1: Equation for the moment of inertia, I = MR 2 Where, M - Mass of the body R - Distance of the mass from the axis Step 2: a) in the first case, we can represent the system as in the diagram 2 Then, the moment of inertia of mass “m” at the end, I end= m(L/2) Iend= mL /4 One more mass is held at another end. Therefore, the total moment of inertia, I’ = 2 × mL /4 2 I’ = mL /2 Since, the axis is passing through the center, the moment of inertia due to central mass will be zero. 2 The moment of inertia of the system, I’ = mL /2

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##### ISBN: 9780321675460

The full step-by-step solution to problem: 32E from chapter: 9 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. Since the solution to 32E from 9 chapter was answered, more than 252 students have viewed the full step-by-step answer. The answer to “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.” is broken down into a number of easy to follow steps, and 59 words. University Physics was written by and is associated to the ISBN: 9780321675460. This textbook survival guide was created for the textbook: University Physics, edition: 13. This full solution covers the following key subjects: rod, center, mass, length, axis. This expansive textbook survival guide covers 26 chapters, and 2929 solutions.

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