Question

show how to calculate the :-

1. Moment of Inertia of a Uniform Rod about a Perpendicular Bisector.

2. Moment of Inertia of a Circular Ring about its Axis.​

Answer 1

Answer:

When the temperature of the rod is increased by ΔT, the increase in the moment of inertia of the rod about the same axis is (Here, α is the coefficient of linear expansion of the rod).

Moment of inertia of a uniform rod of mass M, length L about its perpendicular bisector is

$i = \frac{1}{12} m {l}^{2}$

Moment of Inertia of ring about an axis perpendicular to its plane is

$i = m {r}^{2}$

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept\;:-}}}$

Here the Concept of Moment of Inertia has been used. We see that we are given the points at which we have to find the Moment of Inertia of different solids . Firstly we will obtain the dimensions of the solids and then differentiate and integrate them to find our answer .

Let's do it !!

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★ Solution :-

1) Moment of Inertia of a Uniform Rod about a Perpendicular Bisector ::

• Let the mass of this uniform rod be 'M'

• Let the length of this rod be 'L'

• Origin of bisection be 'O'

From figure (attachment), we see that

• Point at which moment of Inertia to be calculated = AB

We see that this rod is uniform. So there is a mass element dm (shaded portion) between x and (x + dx) .

Also the rod is uniform thus its mass is equal at all points. So Linear Mass Density that is mass per unit length remains constant .

So, using differentiation of element dm, we get

$\\\;\sf{\pink{:\rightarrow\;\;\dfrac{M}{L}\;=\;\bf{\dfrac{dm}{dx}}}}$

Then on cross multiplication,

$\\\;\bf{:\rightarrow\;\;dm\;=\;\bf{\dfrac{M}{L}\:dx}}$

Now we need to find the Moment of Inertia of dm . This is given as,

✒ Moment of Inertia of dm = dl = dm • x²

✒ dl = dm x²

On integrating all these equations, we get,

$\\\;\displaystyle{\sf{:\Longrightarrow\;\;I\;=\;\bf{\int_{-L/2} ^{+L/2}\;dl}}}$

$\\\;\displaystyle{\sf{:\Longrightarrow\;\;I\;=\;\bf{\dfrac{M}{L}\;\times\;\int^{+L/2} _{-L/2}\;x^{2}\:dx}}}$

Here x = -L/2 shows the left end of the rod and then x changing from -L/2 to +L/2 shows the element covers the entire rod

$\\\;\displaystyle{\sf{:\Longrightarrow\;\;I\;=\;\bf{\dfrac{M}{L}\;\times\;\bigg[\dfrac{x^{3}}{3}\bigg]^{+L/2} _{-L/2}}}}$

$\\\;\displaystyle{\sf{:\Longrightarrow\;\;I\;=\;\bf{\dfrac{M}{3L}\;\times\;\bigg[\bigg(\dfrac{L}{2}\bigg)^{3}\;-\;\bigg(\dfrac{-L}{2}\bigg)^{3}\bigg]}}}$

$\\\;\displaystyle{\sf{:\Longrightarrow\;\;I\;=\;\bf{\dfrac{M}{3L}\;\times\;\bigg[\bigg(\dfrac{L}{2}\bigg)^{3}\;+\;\bigg(\dfrac{L}{2}\bigg)^{3}\bigg]}}}$

$\\\;\displaystyle{\sf{:\Longrightarrow\;\;I\;=\;\bf{\dfrac{M}{3L}\;\times\;\bigg[\dfrac{L^{3}}{8}\;+\;\dfrac{L^{3}}{8}\bigg]}}}$

$\\\;\displaystyle{\sf{:\Longrightarrow\;\;I\;=\;\bf{\dfrac{M}{3L}\;\times\;\bigg[\dfrac{2L^{3}}{8}\bigg]}}}$

$\\\;\displaystyle{\sf{:\Longrightarrow\;\;I\;=\;\bf{\dfrac{M}{3L}\;\times\;\dfrac{L^{3}}{4}}}}$

$\\\;\displaystyle{\sf{:\Longrightarrow\;\;I\;=\;\bf{R^{2}\;\times\;\dfrac{M}{2\pi}\;\times\;\bigg[\theta\bigg]^{2\pi} _{0}}}}$

• Here θ is in radians .

From this eventually, we get,

$\\\;\displaystyle{\bf{:\Longrightarrow\;\;I\;=\;\bf{\blue{\dfrac{1}{12}\:ML^{2}}}}}$

$\\\;\underline{\boxed{\tt{Moment\;\;of\;\;Inertia,\;I\;=\;\bf{\blue{\dfrac{1}{12}\:ML^{2}}}}}}$

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2.) Moment of Inertia of a Circular Ring about its Axis ::

• Let there be a line perpendicular to the plane of the ring through its centre O .

• Let the radius of the ring be 'R'

• Let the mass of the ring be 'M'

• All the elements are 'dm'

From the figure (attachment) we see that all the elements are at same distance from the axis of rotation, R .

Since, the circular ring is uniform thus its Linear Mass Density is constant.

✒ Circumference of ring = 2πr

Now differentiating Linear Mass Density, we get,

$\\\;\sf{\orange{:\rightarrow\;\;\dfrac{M}{2\pi R}\;=\;\bf{\dfrac{dm}{d\theta}}}}$

On cross multiplication,

$\\\;\bf{:\rightarrow\;\;dm\;=\;\bf{\dfrac{M}{2\pi R}\:d\theta}}$

Similar like above, Moment of Inertia is given as,

$\\\;\displaystyle{\sf{:\rightarrow\;\;I\;=\;\sf{\int\;R^{2}\:dm}}}$

On integrating this with Limits, we get,

$\\\;\displaystyle{\sf{:\rightarrow\;\;I\;=\;\sf{R^{2}\;\int^{2\pi} _{0}\;\bigg(\dfrac{M}{2\pi}\bigg)\:d\theta}}}$

From this we see, limits as θ to 2π includes whole mass of the ring.

$\\\;\displaystyle{\sf{:\Longrightarrow\;\;I\;=\;\bf{\int _{0} ^{2\pi R}\;\:\dfrac{MR}{2\pi}\;dx}}}$

$\\\;\displaystyle{\sf{:\Longrightarrow\;\;I\;=\;\bf{\dfrac{MR}{2\pi}\;\int _{0} ^{2\pi R}\;\:dx}}}$

$\\\;\displaystyle{\sf{:\Longrightarrow\;\;I\;=\;\bf{\dfrac{MR}{2\pi}\;\bigg[\theta\bigg]_{0} ^{2\pi R}}}}$

$\\\;\displaystyle{\sf{:\Longrightarrow\;\;I\;=\;\bf{\dfrac{MR}{2\pi}\;(2\pi R\;-\;0)}}}$

$\\\;\displaystyle{\bf{:\Longrightarrow\;\;I\;=\;\bf{\purple{MR^{2}}}}}$

$\\\;\underline{\boxed{\tt{Moment\;\;of\;\;Inertia,\;I\;=\;\bf{\purple{MR^{2}}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;I_{z}\;=\;I_{x}\;+\;I_{y}}$

$\\\;\sf{\leadsto\;\;Rotational\;K.E.\;=\;\dfrac{1}{2}\:I\omega^{2}}$

$\\\;\sf{\leadsto\;\;L\;=\;I\omega}$

$\\\;\sf{\leadsto\;\;\tau\;=\;I\alpha}$

$\\\;\sf{\leadsto\;\;\tau\;=\;F\:d}$