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Chapter 5: Problem 658

The moment of inertia of a thin rod of mass \(\mathrm{M}\) and length \(\mathrm{L}\) about an axis passing through the point at a distance $\mathrm{L} / 4$ from one of its ends and perpendicular to the rod is \(\\{\mathrm{A}\\}\left[\left(7 \mathrm{ML}^{2}\right) / 48\right]\) \\{B \\} [ \(\left[\mathrm{ML}^{2} / 12\right]\) \(\\{\mathrm{C}\\}\left[\left(\mathrm{ML}^{2} / 9\right]\right.\) \(\\{\mathrm{D}\\}\left[\left(\mathrm{ML}^{2} / 3\right]\right.\)

Short Answer

Expert verified
The moment of inertia of a thin rod of mass M and length L about an axis passing through the point at a distance L/4 from one of its ends and perpendicular to the rod is: \(\\{A\\}\left[\left(7 \mathrm{ML}^{2}\right) / 48\right]\).

Step by step solution

01

Recall the parallel axis theorem

The parallel axis theorem states that the moment of inertia I about any axis can be found using the moment of inertia Ic about a parallel axis passing through the center of mass and the mass M of the object: \[ I = I_{c} + Md^2\] where d is the distance between the two parallel axes.
02

Calculate the moment of inertia about the center of mass

First, we need to find the moment of inertia of the thin rod about a parallel axis passing through the center of mass. For a thin uniform rod rotating about an axis perpendicular to the rod at its midpoint, the moment of inertia is: \[I_c = \frac{1}{12}ML^2\] We will use this equation in the parallel axis theorem to find the moment of inertia about the given axis.
03

Calculate the distance between the axes

The problem states that the axis passes through a point at distance L/4 from one end. Since the axis is perpendicular to the rod, and the axis through the center of mass is also perpendicular, we can say that the two axes are parallel to each other. The distance d between the axes is the distance from the center of mass to the given axis, which is: \[d = \frac{L}{4}\]
04

Use the parallel axis theorem to find the moment of inertia

Now, we use the parallel axis theorem to find the moment of inertia I about the given axis: \[ I = I_c + Md^2 = \frac{1}{12} ML^2 + M\left(\frac{L}{4}\right)^2\] \[ I = \frac{1}{12} ML^2 + \frac{1}{16}ML^2\] \[ I = \left(\frac{1}{12}+\frac{1}{16}\right)ML^2 = \frac{7}{48}ML^2\]
05

Compare the result with the given options

We found that the moment of inertia of the thin rod about the given axis is (7/48)ML^2. Comparing this result with the given options, we can see that it matches option A. So the correct answer is: \(\\{A\\}\left[\left(7 \mathrm{ML}^{2}\right) / 48\right]\)

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