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A circular lamina of radius a and centre O has a mass per unit area of kx2 , where x is the distance from O and k is a constant. If the mass of the lamina is; in terms of M and a the moment of inertia of the lamina about an axis through O and perpendicular to the lamina is I=x3Ma2. Find x.

Answer
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Hint: In order to find the moment of inertia of the given shape, we need the mass of the lamina to find the moment of inertia, so we will consider mass of a small element and in order to find the mass of the lamina we will integrate the mass for the element. Further we will use the formula of moment of inertia and finally compare with the given value.

Formula used- M=0adm,I=0a(dm.x2)

Complete Step-by-Step solution:
Let's consider the small element of lamina at x distance from the centre of thickness dx as shown below.
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The area of the element is
dA=2πxdx
The mass per unit area is given as
dmdA=kx2
The mass of lamina is,
M=0admM=0akx2.dAM=0akx2.(2πxdx)M=0a2πkx3.dxM=2πk[x44]0aM=πka42
The moment of inertia of lamina about an axis through O and perpendicular to the lamina is,
I=0a(dm.x2)I=0a(kx2dA.x2)I=0a(kx4dA)I=0a(kx4(2πxdx))I=2πk0ax5.dx
Solve further,
I=2πk[x66]0aI=πka63
As we know that:
M=πka42k=2Mπa4
Let us substitute the value of k in the equation of moment of inertia.
I=πka63I=π(2Mπa4)a63I=23Ma2
Thus, the required moment of inertia is 23Ma2
Hence, the value of x is 2.

Note- The moment of inertia of a solid body, otherwise known as the mass moment of inertia, angular momentum, or rotational inertia, is a quantity that specifies the torque necessary for a desired angular acceleration along a rotational axis; analogous to how gravity decides the force required for a desired acceleration. Students must remember the formula for moment of inertia for some common shapes and the general formula to find the moment of inertia.
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