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Moment Of Inertia Of A Rod - 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 Rod is considered one of the most asked concept.
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10 Questions around this concept.
Solve by difficulty
The moment of inertia of a uniform cylinder of length and radius
about its perpendicular bisector is
. What is the ratio
such that the moment of inertia is minimum?
Three identical rods, each of length l are joined to form a rigid equilateral triangle. Its radius of gyration about an axis passing through a corner and perpendicular to the plane of the triangle is
The linear mass density of a thin rod AB of length L varies from A to B as $\lambda(x)=\lambda_0\left(1+\frac{x}{L}\right)$, where x is the distance from A. If M is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is :
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The moment of inertia of a thin rod of mass M and length l about an axis passing through one of its and perpendicular to length is.
Moment of inertia of a thin rod of mass M and length l about an axis perpendicular to the rod and passing through one end is equal to
Concepts Covered - 1
Moment of inertia of a Rod
Let I=Moment of inertia of a ROD about an axis through its centre and perpendicular to it
To calculate I (Moment of inertia of rod)
Consider a uniform straight rod of length L, mass M and having centre C
mass per unit length of the rod$
=\lambda=\frac{M}{L}
$
Take a small element of mass dm with length dx at a distance x from the point C .
$
\begin{aligned}
d m & =\lambda \cdot d x=\frac{M}{L} \cdot d x \\
\Rightarrow d I & =x^2 d m
\end{aligned}
$
Now integrate this dl between the limits
$
\begin{aligned}
& \text { Now integrate this dl between the limits } x=-\frac{L}{2} \text { to } \frac{L}{2} \\
& I=\int d I=\int x^2 d m=\int_{\frac{-L}{2}}^{\frac{L}{2}} \frac{M}{L} x^2 * d x=\frac{M}{L} \int_{\frac{-L}{2}}^{\frac{L}{2}} x^2 d x=\frac{M L^2}{12}
\end{aligned}
$
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