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Moment Of Inertia Of A Disc - Practice Questions & MCQ
Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main
Quick Facts
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Moment of inertia of a DISC is considered one of the most asked concept.
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17 Questions around this concept.
Solve by difficulty
The moment of inertia of a uniform semicircular disc of mass M and radius about a line perpendicular to the plane of the disc through the center is :d
A circular disc X of radius R is made from an iron plate of thickness and another disc Y of radius
is made from an iron plate of thickness
. Then the relation between the moment of inertia.
and
is :
What is the moment of inertia of a disc having inner radius $R_1$ and outer radius $R_2$ about the axis passing through the center and perpendicular to the plane as shown in diagram?
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A thin circular disk is in the xy plane as shown in the figure. The ratio of its
moment of inertia about z and z' axes will be :
The moment of inertia of a thin circular disc of mass M and radius R about any diameter is
A uniform disc of mass 2 kg is rotated about an axis perpendicular to the plane of the disc. If radius of gyration is 50 cm, then the M.I. of disc about same axis is
Concepts Covered - 1
Moment of inertia of a DISC
Let I=Moment of inertia of a DISC about an axis through its centre and perpendicular to its plane
To calculate I
Consider a circular disc of mass M , radius R and centre O .
And mass per unit area $=\sigma=\frac{M}{\pi R^2}$
Take an elementary ring of mass dm of radius x as shown in figure
So, $^{d m}=\sigma *(2 \pi x d x)=\frac{M}{\pi R^2} *(2 \pi x d x)$
$
\begin{aligned}
\Rightarrow d I & =x^2 d m \\
I & =\int d I=\int_0^R x^2 *\left(\frac{M}{\pi R^2} *(2 \pi x d x)\right)=\frac{2 M}{R^2} \int_0^R x^3 d x=\frac{M R^2}{2}
\end{aligned}
$
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