A uniform disc of radius R lies in the x-y pl | vedclass

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(A) The moment of inertia of a uniform disc of mass $M$ and radius $R$ about the $z$-axis (passing through the center and perpendicular to the plane)

Vedclass · Accepted Answer

$R/\sqrt{2}$

Vedclass · Answer

(A) The moment of inertia of a uniform disc of mass $M$ and radius $R$ about the $z$-axis (passing through the center and perpendicular to the plane) is $I_z = \frac{1}{2}MR^2$.The line $y = x + c$ can be rewritten as $x - y + c = 0$. The perpendicular distance $d$ of this line from the origin $(0,0)$ is $d = \frac{|0 - 0 + c|}{\sqrt{1^2 + (-1)^2}} = \frac{|c|}{\sqrt{2}}$.According to the parallel axis theorem,the moment of inertia about any line in the $x-y$ plane parallel to a diameter is $I = I_{cm} + Md^2$,where $I_{cm}$ is the moment of inertia about a diameter,which is $\frac{1}{4}MR^2$.Thus,$I_{line} = \frac{1}{4}MR^2 + M\left(\frac{c}{\sqrt{2}}\right)^2 = \frac{1}{4}MR^2 + \frac{Mc^2}{2}$.Given that $I_{line} = I_z$,we have $\frac{1}{4}MR^2 + \frac{Mc^2}{2} = \frac{1}{2}MR^2$.Subtracting $\frac{1}{4}MR^2$ from both sides,we get $\frac{Mc^2}{2} = \frac{1}{4}MR^2$.Simplifying,$c^2 = \frac{R^2}{2}$,which gives $c = \pm \frac{R}{\sqrt{2}}$.

$A$ uniform disc of radius $R$ lies in the $x-y$ plane with its centre at the origin. Its moment of inertia about the $z$-axis is equal to its moment of inertia about the line $y = x + c$. The value of $c$ is

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