First slide

Moment of inertia

Question

Two circular discs A and B are of equal masses and thicknesses but made of metal with densities d_A and d_B (d_A > d_B).If their moments of inertia about an axis passing
through their centers and perpendicular to circular faces are I_A and I_B, then

Moderate

A

$I_{A} = I_{B}$
B

$I_{A} > I_{B}$
C

$I_{A} < I_{B}$
D

$I_{A} \geq I_{B}$

Solution

Let M be the mass of each disc. Let R_A and R_B, be the radii of discs A and B, respectively. Then $M = π R_{A}^{2} t d_{A} = π R_{B}^{2} t d_{B}$
$\begin{array}{l} As d_{A} > d_{B}, so R_{A}^{2} < R_{B}^{2} . \\ Now, \\ I_{A} = \frac{1}{2} M R_{A}^{2}, I_{B} = \frac{1}{2} M R_{B}^{2} \\ \Rightarrow \frac{I_{A}}{I_{B}} = \frac{R_{A}^{2}}{R_{B}^{2}} < 1, i.e., I_{A} < I_{B} \end{array}$

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