Question

a uniform disc of radius R lies in xy plane with with centre at origin its moment of inertia about the x axis = 2R and Y = 0 is equal to the moment of inertia about the axis Y = 2D and Z = 0 what is the d equal to?

Answer 1

moment of inertia application

Answer 2

Answer:

D = ±(√17/4)R

Explanation:

We know that moment of interia of a disc around the axis perpendicular to it's plane and passing through center is 1/2MR² and moment of interia around  it's diameter as axis if 1/4MR²

Now,

Look at the first picture, There are two planes x = 2R and y = 0 which gives a line.

This line AB is parallel to its axis(The axis which is perpendicular on the disc and passing through origin)  at the distance of 2R.

Thus, on applying parallel axes theorem for line AB,

$I_{AB}=I_{cm}+M(2R)^2\\\;\\I_{AB}=\frac{1}{2}MR^2+4MR^2$

Again,

Look at the second picture, There are two planes y = 2D and z = 0 which gives a line.

This line CD is parallel to the axes, the axes which passes through diameter of disc and at a distance of 2D.

Thus, on applying parallel axes theorem for the line CD,

$I_{CD}=I_{diameter}+M(2D)^2\\\;\\I_{CD}=\frac{1}{4}MR^2+4MD^2$

Now, According to question,

$I_{AB}=I_{CD}\\\;\\\frac{1}{2}MR^2+4MR^2=\frac{1}{4}MR^2+4MD^2\\\;\\\frac{1}{2}R^2+4R^2=\frac{1}{4}R^2+4D^2\\\;\\\frac{1}{2}R^2+4R^2-\frac{1}{4}R^2=4D^2\\\;\\\frac{1}{4}R^2+4R^2=4D^2\\\;\\\frac{17}{4}R^2=4D^2\\\;\\D^2=\frac{17}{16}R^2\\\;\\D=\pm\frac{\sqrt{17}}{4}R$