First slide

Moment of inertia

Question

Given a uniform disc of mass M and radius R. A small disc of radius R/2 is cut from this disc in such a way that the distance between the centers of the two discs is R/2. Find the moment of inertia of the remaining disc about a diameter of the original disc perpendicular to the line connecting the centers of the two discs.

Moderate

A

$3 M R^{2} / 32$
B

$5 M R^{2} / 16$
C

$11 M R^{2} / 64$
D

$N o n e o f t h e s e$

Solution

Mass of cut disc: $m_{1} = M / 4$
Moment of inertia of original disc about axis 1:
$I = \frac{1}{4} M R^{2}$
Moment of Inertia of small disc about axis 1:
$\begin{array}{l} I^{'} = \frac{1}{4} m_{1} {(\frac{R}{2})}^{2} + m_{1} {(\frac{R}{2})}^{2} \\ or I^{'} = \frac{5}{16} m_{1} R^{2} = \frac{5}{64} M R^{2} \\ Required moment of inertia: \\ I - I^{'} = \frac{1}{4} M R^{2} - \frac{5}{64} M R^{2} = \frac{11}{64} M R^{2} \end{array}$

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