Question

Q1.A uniform disc of radius R lies in xy plane with its centre at origin. It's moment of inertia about z axis is equal to it's moment of inertia about line y=x+c. The value of c is

Answer 1

Step-by-step explanation:

Answer:- c = ±R/√2

Algorithm for solution:- First of all we find our the interia of disc with respect to line y = m + c. To find this value we will apply Parallel Axes Theorem. Also, to evaluate this we draw a parallel line passing through origin and having same slope that i.e. y = x. We need find the distance between the lines y = x + c and y = x.

Further we would satisfy the given condition in the question.

Solution:-

We know that interia of the a uniform solid disc with respect to its diameter as axis is MR²/4 where R is radius of the disc and interia with respect to z axis is MR²/2

Now,

Let Distance between the lines y = x + c and y = x be p,

then

$\begin{gathered}p=\frac{1-1+c}{\sqrt{2}}-\frac{1-1+0}{\sqrt{2}}\\\;\\p=\frac{c}{\sqrt{2}}\end{gathered}$

Interia of disc around the line y = x +c will be,

$\begin{gathered}I_{line}=\frac{1}{4}MR^2+Mp^2\\\;\\I_{line}=\frac{1}{4}MR^2+M(\frac{c}{\sqrt{2}})^2\end{gathered}$

Now, according to given condition,

$\begin{gathered}I_z=I_{line}\\\;\\\frac{1}{2}MR^2 = \frac{1}{4}MR^2+M(\frac{c}{\sqrt{2}})^2\\\;\\\frac{1}{2}R^2=\frac{1}{4}R^2+\frac{c^2}{2}\\\;\\\frac{1}{4}R^2=\frac{c^2}{2}\end{gathered}$

$c=\pm\frac{R}{\sqrt{2}}$

Note:- 1)To find the value of p, we have subtracted distance of line y = x with respect to origin from the distance of line y = x + c with respect to origin.