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Step-by-step solution
Step 1: Moment of inertia of a square plate about its center
For a uniform square plate of side $a$ and mass $m$, the moment of inertia about an axis perpendicular to the plane and passing through its center is given by Icenter=16ma2.
Step 2: Identify the distance between the center and a corner
The distance from the center of the square to any corner is half the diagonal of the square, which is a2+a22=2 a2=a2.
Step 3: Apply the parallel axis theorem
The parallel axis theorem states that if $I_\text{center}$ is the moment of inertia of a body about an axis through its center of mass, then the moment of inertia about a parallel axis passing through a point a distance $d$ away is I=Icenter+md2. Here, d=a2.
Step 4: Substitute the values
So the moment of inertia of the square plate about an axis perpendicular to its plane and passing through one corner is Icorner=16ma2+ma22. Simplifying, Icorner=16ma2+m a22=16ma2+12ma2. Combine the terms: Icorner=16ma2+36ma2=46ma2=23ma2.
Final Answer
Hence, the moment of inertia of the uniform square plate about the given axis is 23ma2.
Step-by-step solution
Step 1: Moment of inertia of a square plate about its center
For a uniform square plate of side $a$ and mass $m$, the moment of inertia about an axis perpendicular to the plane and passing through its center is given by Icenter=16ma2. I_\text{center} = \frac{1}{6} m a^2.
Step 2: Identify the distance between the center and a corner
The distance from the center of the square to any corner is half the diagonal of the square, which is a2+a22=2 a2=a2. \frac{\sqrt{a^2 + a^2}}{2} = \frac{\sqrt{2}\,a}{2} = \frac{a}{\sqrt{2}}.
Step 3: Apply the parallel axis theorem
The parallel axis theorem states that if $I_\text{center}$ is the moment of inertia of a body about an axis through its center of mass, then the moment of inertia about a parallel axis passing through a point a distance $d$ away is I=Icenter+md2. I = I_\text{center} + m d^2. Here, d=a2. d = \frac{a}{\sqrt{2}}.
Step 4: Substitute the values
So the moment of inertia of the square plate about an axis perpendicular to its plane and passing through one corner is Icorner=16ma2+ma22. I_\text{corner} = \frac{1}{6} m a^2 + m \left(\frac{a}{\sqrt{2}}\right)^2. Simplifying, Icorner=16ma2+m a22=16ma2+12ma2. I_\text{corner} = \frac{1}{6} m a^2 + m \, \frac{a^2}{2} = \frac{1}{6} m a^2 + \frac{1}{2} m a^2. Combine the terms: Icorner=16ma2+36ma2=46ma2=23ma2. I_\text{corner} = \frac{1}{6} m a^2 + \frac{3}{6} m a^2 = \frac{4}{6} m a^2 = \frac{2}{3} m a^2.
Final Answer
Hence, the moment of inertia of the uniform square plate about the given axis is 23ma2. \frac{2}{3} m a^2.
