9303753286127016 Consider a uniform square plate of side a and mass m. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corner is:

Consider a uniform square plate of side a and mass m. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corner is:

 We can easily find R using the Pythagoras theorem,

⇒R=√ a2−(a/2)=a√2

We know that for a square plate, the moment of inertia along a perpendicular axis passing through the centre of mass is,

⇒ Iperpendicular = ma2/6

So, using the parallel axis theorem, we get

⇒Iparallel = Iperpendicular + MR2

Here Iparallel is the moment of inertia along the parallel axis, Iperpendicular is the moment of inertia along the axis through the centre of mass, M is the mass of the object and R is the distance between the centre of mass and the parallel axis.

Substituting the value of R and Iperpendicular we get ⇒ Iparallel = Iperpendicular + MR2

∴Iparallel =ma2/6 + ma2/(√2)2 = 2/3ma2

Balkishan Agrawal

At the helm of GMS Learning is Principal Balkishan Agrawal, a dedicated and experienced educationist. Under his able guidance, our school has flourished academically and has achieved remarkable milestones in various fields. Principal Agrawal’s vision for the school is centered on providing a nurturing environment where every student can thrive, learn, and grow.

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