# MHT-CET Physics Crash Course “Theorem of Perpendicular Axis” Class 12

## “Theorem of Perpendicular Axis” MHT-CET Class 12 Physics Crash Course| Chapter Rotational Dynamics

In this session G.S. Khairnar discussed the theory and MCQ’s of “Theorem of Perpendicular Axis”. The session will be useful for all the aspirants of MHT-CET 2022. The session will be in Marathi and English.

### Theorem of Perpendicular Axis:-

Hello everyone, in previous articles we have learnt about parallel axes theorem and some important applications of it.

According to parallel axes theorem we have,

Io =Ic + Mh2

But in some cases, the axes of rotation may be perpendicular to each other, in such cases we can’t apply parallel axes theorem. So let’s study alternative for parallel axes theorem, i.e. perpendicular axes theorem.

Let’s study the theorem of perpendicular axis……!

Statement: : The M.I. of a plane lamina about an axis perpendicular to its plane us equal to the sum of its M.I. about two mutually perpendicular axes in the plane of the lamina & intersecting at the point where perpendicular axes cuts the lamina.

i.e. Iz = Ix + Iy

Proof: Consider a plane lamina in the horizontal XY plane. OX & OY are two mutually perpendicular axes in the plane of lamina. Axis OZ is perpendicular to the plane of the lamina. Let Ix, Iy& Iz be the M.I of the lamina about x, y & Z axes respectively. Consider a small element of mass dm is situated at point ‘P’. Join OP &Perpendicular from P to X & Y aces.

The element dm is at distance y from X axis.

∴ M.I. of element about X – axis = dm y2

∴ M.I. of lamina about X – axis

Ix =∫ dm y2     ………(1)

The element dm is at distance x from Y – axis

∴ M.I. of element about Y – axis =  dm x2

∴ M.I. of lamina about Y- axis

Iy = ∫ dm x2     ………(2)

The element dm is at distance OP from Z – axis Hence proved.

Let’s study the some applications of theorem of perpendicular axis……!

1. M.I. of ring about any diameter:

The M.I. of ring about an axis passing through Centre& perpendicular to the plane is MR2. This axis is perpendicular to diameter of the ring.

From fig  Iz=MR2, By perpendicular axes theorem, 2. M.I. of disc about any diameter:

Any axis passing through centre & in the plane of the disc is coinciding with diameter.

Consider two such axes X & Y in the plane of the disc. Let Io be the M.I. of disc about tangent Let Ix, Iy& Iz be the M.I. of disc about X, Y & Z axes respectively. Hence according to principle of perpendicular axes. ### ♦जिल्हा नुसार जाहिराती ♦

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ठाणे वर्धा वाशिम यवतमाल बीड

### ♦शिक्षणानुसार जाहिराती ♦

७ वी (7th) दहावी (SSC) बारावी (HSC) डिप्लोमा आय.टी.आय पदवी
पदव्युत्तर शिक्षण बी.एड एम.एड एल.एल.बी / एल.एल.एम बीएससी एमबीए
बीसीए एमसीए बी.कॉम एम.कॉम GNM/ANM एमएससी
बी.फार्म एम.फार्म बी.ई एम.ई BAMS/BHMS एम.बी.बी.एस / एम.डी
बी.टेक एम.टेक MS-CIT