First slide

Moment of inertia

Question

Mass per unit area of a circular disc of radius a depends on the distance r from its centre as $σ (r) = A + B r$ . The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its center is:

Moderate

A

$2 π a^{4} (\frac{A}{4} + \frac{a B}{5})$
B

$2 π a^{4} (\frac{a A}{4} + \frac{B}{5})$
C

$π a^{4} (\frac{A}{4} + \frac{a B}{5})$
D

$2 π a^{4} (\frac{A}{4} + \frac{B}{5})$

Solution

$\begin{array}{l} G o a t d i s \tan c e o f r a n d t a k e a n e l e m e n t o f t h i c k n e s s d r . \\ A r e a o f e l e m e n t = d A = (2 π r) d r \\ G i v e n : S u r f a c e m a s s d e n s i t y = α = A + B r \end{array}$
$\begin{array}{l} M a s s o f e l e m e n t = d m = α d A \\ ∴ T o t a l m a s s = \int d m = \int (A + B r) 2 π r d r \\ M o m e n t o f I n e r t i a a b o u t c e n t e r = I = \int d m r^{2} = \int_{0}^{a} (A + B r) 2 π r^{3} d r = 2 π (A \frac{a^{4}}{4} + B \frac{a^{5}}{5}) = 2 π a^{4} (\frac{A}{4} + \frac{a B}{5}) \end{array}$

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