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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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21 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
The moment of inertia of a disc about the tangent parallel to its plane is I. The moment of inertia of the disc tangent and perpendicular to its plane 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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