The temperature of a thin uniform circular disc, of one metre diameter is increased by 10$℃$. The percentage increase in moment of inertia of the disc about an axis passing through its centre and perpendicular to the circular face (linear coefficient of expansion=$11 imes 10^{- 6}$ $℃^{- 1}$)

Question

The temperature of a thin uniform circular disc, of one metre diameter is increased by 10$℃$. The percentage increase in moment of inertia of the disc about an axis passing through its centre and perpendicular to the circular face (linear coefficient of expansion=$11 	imes 10^{- 6}$ $℃^{- 1}$)

Dr. Abubacker · Accepted Answer

(c) Increase in area of disc
$\mathrm{\Delta}A = A(2\alpha)\mathrm{\Delta}t$
= $\pi(0.5)^{2}(2 	imes 11 	imes 10^{- 6}) 	imes 10$
= $0.000055\pi$
New area of the disc,
$A^{'} = A + \mathrm{\Delta}A$
$	herefore\ A^{'}\  = \pi(0.5)^{2} + 0.000055\pi$
Or $\pi{r'}^{2} = \ 0.250055\pi m^{2}$
Or ${r'}^{2} = \ 0.500055m$
Increase in moment of inertia,
$\frac{I^{'} - I}{I} = \ \frac{{(0.500055)}^{2} - (0.5)^{2}}{(0.5)^{2}}$
= 0.00022=0.022%

Explanation

Reviewed by Dr. Abubacker

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