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Chapter 5: Problem 567

Let I be the moment of inertia of a uniform square plate about an axis \(\mathrm{AB}\) that passes through its centre and is parallel to two of its sides \(\mathrm{CD}\) is a line in the plane of the plate that passes through the centre of the plate and makes an angle of \(\theta\) with \(\mathrm{AB}\). The moment of inertia of the plate about the axis \(\mathrm{CD}\) is then equal to.... \(\\{\mathrm{A}\\} \mathrm{I}\) \(\\{B\\} I \sin ^{2} \theta\) \(\\{C\\} I \cos ^{2} \theta\) \(\\{\mathrm{D}\\} I \cos ^{2}(\theta / 2)\)

Short Answer

Expert verified
The moment of inertia of the plate about the axis CD is equal to I (option A).

Step by step solution

01

Identify the given information

In this problem, we have a uniform square plate with moment of inertia I about an axis AB passing through its center. We are also given another axis CD that passes through the center of the plate and makes an angle θ with AB. Our task is to find the moment of inertia of the plate about the axis CD.
02

Set up axes and find initial moments of inertia

Let's set our coordinate axes by placing the origin at the center of the square plate. Assume the plate has sides of length "a" and mass "m", and the axis AB is parallel to y-axis while CD makes an angle θ with the y-axis. The moment of inertia for the plate about the y-axis is given as \(I_y = \frac{1}{12}ma^2\). Since AB is parallel to the y-axis, we can say that I = \(I_y = \frac{1}{12}ma^2\). Now, we need to find the moment of inertia about the x-axis passing through the center of the plate. The moment of inertia for the plate about the x-axis is given as \(I_x = \frac{1}{12}ma^2\).
03

Apply the Perpendicular Axis Theorem

The Perpendicular Axis Theorem states that the moment of inertia about an axis perpendicular to the plane of an object is equal to the sum of the moments of inertia about two mutually perpendicular axes lying in the plane of the object and passing through the same point. In our case, this point is the center of the square plate. Applying this theorem, we have: \(I_{CD} = I_x\cos^2\theta + I_y\sin^2\theta\)
04

Substitute the values and simplify

We already know the values of \(I_x\) and \(I_y\) from the previous step. Now, we will substitute their values in the equation and simplify it. \(I_{CD} = \left(\frac{1}{12}ma^2\right)\cos^2\theta + \left(\frac{1}{12}ma^2\right)\sin^2\theta\) Factor out \(\frac{1}{12}ma^2\): \(I_{CD} = \frac{1}{12}ma^2(\cos^2\theta + \sin^2\theta)\) We know that \(\cos^2\theta + \sin^2\theta = 1\). So, \(I_{CD} = \frac{1}{12}ma^2\) Since we are given that the moment of inertia I about the axis AB is equal to \(\frac{1}{12}ma^2\), our final answer becomes: \(I_{CD} = I\) So, the correct option is: \(\\{\mathrm{A}\\} \mathrm{I}\)

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Most popular questions from this chapter

A straight rod of length \(L\) has one of its ends at the origin and the other end at \(\mathrm{x}=\mathrm{L}\) If the mass per unit length of rod is given by Ax where \(A\) is constant where is its centre of mass. \(\\{\mathrm{A}\\} \mathrm{L} / 3\) \(\\{\mathrm{B}\\} \mathrm{L} / 2\) \(\\{\mathrm{C}\\} 2 \mathrm{~L} / 3\) \(\\{\mathrm{D}\\} 3 \mathrm{~L} / 4\)

One solid sphere \(\mathrm{A}\) and another hollow sphere \(\mathrm{B}\) are of the same mass and same outer radii. The moment of inertia about their diameters are respectively \(\mathrm{I}_{\mathrm{A}}\) and \(\mathrm{I}_{\mathrm{B}}\) such that... \(\\{\mathrm{A}\\} \mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{B}}\) \(\\{\mathrm{B}\\} \mathrm{I}_{\mathrm{A}}>\mathrm{I}_{\mathrm{B}}\) \(\\{\mathrm{C}\\} \mathrm{I}_{\mathrm{A}}<\mathrm{I}_{\mathrm{B}}\) $\\{\mathrm{D}\\}\left(\mathrm{I}_{\mathrm{A}} / \mathrm{I}_{\mathrm{B}}\right)=(\mathrm{d} \mathrm{A} / \mathrm{dB})$ (radio of their densities)

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