Question

Find the moment of inertia of disc of massM and radius R about the standard axis

Answer 1

Answer:

The moment of inertia of remaining disk = Moment of inertia of whole disc - Moment of inertia of removed small disc

moment of inertia of any disk about its own central axis = 1/2 mr2

Moment of inertia of total disk of mass 9M and radius R = (1/2) 9M R2

Iwhole = (1/2) 9M R2

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Let us calculate the mass of the removed disk and the location of its centre of mass

mass density = 9M / area of disk = 9M/ piR2

mass of small disk of radius R/3 = area of small disk x mass density = pi(R/3)2 x 9M/ piR2

We get the mass of the removed small disk = M

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Its centre of mass is located at R - R/3 = 2R/3 from the main axis

its moment of inertia can be calculated using parallel axis theorem

moment of inertia of removed disk = moment of inertia of small disk about its own axis + moment of inerita of its centre of mass about the main axis

Iremoved = (1/2) M (R/3)2 + M (2R/3)2

= MR2 /18 + 4MR2 /9

Iremoved = MR2 /2

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hence the moment of inertia of remaining disk = Iwhole - Iremoved

= (1/2) 9M R2 - MR2 /2

= 4MR2

Answer 2

The moment of inertia of a disc about its diameter = MR2 / 4

According to the theorem of perpendicular axis, the moment of inertia of a planar body (lamina) about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two perpendicular axes concurrent with perpendicular axis and lying in the plane of the body.

The M.I. of the disc about its centre = MR2 / 4 + MR2 / 4 = MR2 / 2

The situation is shown in the given figure.

Applying the theorem of parallel axes:

The moment of inertia about an axis normal to the disc and passing through a point on its edge

= MR2 / 2 + MR2 = 3MR2 / 2