Rotation Ques 30
- A thin circular plate of mass $M$ and radius $R$ has its density varying as $\rho(r)=\rho _0 r$ with $\rho _0$ as constant and $r$ is the distance from its centre. The moment of inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is $I=a M R^{2}$. The value of the coefficient $a$ is
(a) $\frac{1}{2}$
(b) $\frac{3}{5}$
(c) $\frac{8}{5}$
(d) $\frac{3}{2}$
(2019 Main, 8 April I)
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Answer:
Correct Answer: 30.(*)
Solution:
Formula:
Moment of Inertia of Different Objects:](/important-formula/physics/rigid_body_dynamics)
- Consider an elementary ring of thickness $d x$ and radius $r$.
Moment of inertia of this ring about a perpendicular axis through the centre is
$$ d I _c=d m \cdot x^{2}=\rho _0 x(2 \pi x) d x \cdot x^{2}=2 \pi \rho _0 x^{4} d x $$
Moment of inertia of this elementary ring about a perpendicular axis at a point through the edge, (by parallel axes theorem)
$$ \begin{alignedat} d I & =d m x^{2}+d m y^{2} \ & =2 \pi \rho _0 x^{2} d x+2 \pi \rho _0 R^{2} x^{2} d x \end{aligned} $$
Moment of inertia of a complete disc is $\frac{1}{2}MR^2$
$$ \begin{alignedat} I & =\int _0^{R} d I=\int _0^{R} 2 \pi \rho _0 x^{2} d x+\int _0^{R} 2 \pi \rho _0 R^{2} x^{2} d x \\ & =\frac{2 \pi \rho _0 R^{5}}{5}+\frac{2 \pi \rho _0 R^{5}}{3}=\frac{16 \pi \rho _0 R^{5}}{15} \\ \therefore \quad a & =\frac{16}{15} \text { (No option matches) } \end{aligned} $$
BETA
