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Moment Of Inertia Of A Rod - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Moment of inertia of a Rod is considered one of the most asked concept.

  • 12 Questions around this concept.

Solve by difficulty

The moment of inertia of a uniform cylinder of length land radius R about its perpendicular bisector is I.  What is the ratio \frac{l}{R} such that the moment of inertia is minimum?

Four identical thin rods each of mass M and length l, form a square frame. Moment of inetia of this frame aboit an axis through the centre of the square and perpendicular to its plane is

There is a rod of length L and mass M; one line is passing through the center of the rod and another line is passing through the end of the rod; The radius of gyrations ratio of this rod for the given cases (concerning the line positions) center/end is 

A thin rod of length \mathrm{l} is suspended from a point from on its length which is at a distance \mathrm{x} from its centre, for what value of \mathrm{x} time period of oscillation will be minimum -
 

A thin rod of length \mathrm{l}  is suspended from a point from on its length which is at a distance \mathrm{x} from its centre, for what value of \mathrm{x} time period of oscillation will be minimum-
 

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A thin rod \mathrm{A B} (of mass \mathrm{m} and length \mathrm{l}) is sliding keeping in contact with a vertical wall and a smooth surface as shown. \mathrm{MI} of the rod about instantaneous axis of rotation at the instant shown is: -

A tube of length 1m is filled completely with an incompressible liquid with of mass 1kg and closed at both ends. The tube is rotated in a horizontal plane about one of its ends with a uniform angular velocity of 2rad/s. The force exerted by the liquid ath the other end is 

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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.

dm = \lambda.dx = \frac{M}{L}.dx

       \Rightarrow dI= x^2dm

Now integrate this dI  between the limits  x = -\frac{L}{2}\ to \ \frac{L}{2}

I=\int dI=\int x^2dm=\int_{\frac{-L}{2}}^{\frac{L}{2}} \frac{M}{L}x^2*dx= \frac{M}{L}\int_{\frac{-L}{2}}^{\frac{L}{2}}x^2dx=\frac{ML^2}{12}

 

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Moment of inertia of a Rod

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