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

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Question

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

Vedclass · Accepted Answer

$R/\sqrt{2}$

Vedclass · Answer

(B) The moment of inertia of a uniform disc of mass $M$ and radius $R$ about the $z$-axis (passing through the centre and perpendicular to the plane) is $I_z = \frac{1}{2}MR^2$.The line is given by $x - y + c = 0$. The perpendicular distance $d$ from the origin $(0,0)$ to this line 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 axis parallel to a diameter is $I = I_{cm} + Md^2$,where $I_{cm}$ is the moment of inertia about the diameter,which is $I_{cm} = \frac{1}{4}MR^2$.Thus,the moment of inertia about the line $y = x + c$ is $I = \frac{1}{4}MR^2 + M\left(\frac{c}{\sqrt{2}}\right)^2 = \frac{1}{4}MR^2 + \frac{Mc^2}{2}$.Equating the two moments of inertia:$\frac{1}{2}MR^2 = \frac{1}{4}MR^2 + \frac{Mc^2}{2}$$\frac{1}{4}MR^2 = \frac{Mc^2}{2}$$c^2 = \frac{R^2}{2}$$c = \pm \frac{R}{\sqrt{2}}$.Since the options provide $R/\sqrt{2}$,the correct value is $R/\sqrt{2}$.

$A$ 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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