Two Parallel Circular Wheels, Each Of Mass ‘M’ And Radius ‘R’ And Filled With Weightless Spokes Are Rigidly Joined Together By A Weightless Rod Of Length 2R Passing Through The Centre Of The Wheels Perpendicular To Their Planes. Calculate The Moment Of Inertia Of The Wheels About An Axis Passing Perpendicular Through The Centre Of The Rod.

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Ans.

Fig. 1

Here, two wheels each of mass M and radius R are rigidly connected by a weightless rod of length 2R.
The moment of inertia of each wheel about their diameter is \( \frac{1}{2}MR^2 \).
[ To know the derivation of moment of inertia of a wheel about its own diameter, ( CLICK HERE ) ]

By applying the theorem of parallel axes, the moment of inertia of each wheel about the axis \( yy' \) passing through the centre o and perpendicular to the length of the rod \( O_1O_2 \) is given by,

\( \frac{1}{2}MR^2+MR^2\\=\frac{3}{2}MR^2 \)

So the moment of inertia of the both wheel about the axis \( yy' \) is

\( 2\times{\frac{3}{2}MR^2}=3MR^2 \)

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