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

“Theorem of Parallel 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 Parallel Axis”. The session will be useful for all the aspirants of MHT-CET 2022. The session will be in Marathi and English.

Theorem of Parallel Axis:-

Hello everyone, we know that in rotational motion the body has tendency to oppose any change in its state of motion. This property is called M.I. It is represented by mass & distribution of mass from axis of rotation given as,

As per the symmetry of object, magnitude of moment of inertia changes. Hence to find the moment of inertia for a body at different axes we need to apply theorem like parallel axis and perpendicular axis.

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

Statement: The M.I. of a body about any axis is equal to the sum of its M.I. about a parallel axis passing through its canter of mass & the product of the mass of the body & square of distance between the two axes.

i.e.                     I_o =I_c + Mh²

Proof: Consider a body of mass M is rotating about an axis passing through point ‘O’. Let I_o be the corresponding M.I. of body an axis passing through point ‘O’. Let ‘C’ be the Centre of mass of the body at distance h from O. Let I_c be the M.I. of the body about an axis passing through ‘C’.

Consider a small element of mass dm at point ‘P’ join P & OC. Draw PD perpendicular to OD.

M.I of element dm about an axis passing through C = CP²dm

So, M.I of body about an axis passing through  C

I_c= ∫CP²dm………………..(1)

∴ M.I of element about an axis passing through O = OP²dm

∴ M.I of body about an axis passing through point O

I_o= ∫ CP²dm    ………………..(2)

Now from figure, in ΔODP using Pythagoras theorem we get,

OP²= OD² + PD²

= (OC + CD)² + PD²

= OC² + 2OC.CD + CD² + PD²

In ΔCPD , CD² + PD² = CP²

∴ OP² = OC² + 2 OC.CD + CP²

∴ Equation (2) becomes

I₀ = ∫ (OC² + 2 OC.CD + CP²) dm

∴ I₀ =∫ OC²dm + OC.CD dm + C²dm

From fig OC =R      &

∴ ∫ OC²dm = I_c

While ∫ CDdm = 0 (C is center of mass) &∫ dm = M = mass of body

∴ I_o = Mh² + O + I_c

∴ I_o =  I_c + Mh²

Hence proved.

Let’s learn some important applications of parallel axe theorem….!

1. M.I. of uniform rod about a transverse axis passing through one end.

Transverse axis passing through one end is parallel to transverse axis passing through Centre. Let I₀ be M.I. of rod about parallel axis passing through Centre C.

According to principle of parallel axes.

I_o = I_c + MR²

Where R be the distance between two parallel axes.

2. M.I. of disc bout a tangent perpendicular to plane of disc

A tangent perpendicular to the plane of the disc is parallel to the axis passing through the centre and perpendicular to the plane .Let I_c be the M.I of disc about axis passing through centre & perpendicular to its plane

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