Login/Sign Up
Home » (a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be , where is the mass of the sphere and is the radius of the sphere.(b) Given the moment of inertia of a disc of mass M and radius about any of its diameters to be , find its moment of inertia about an axis normal to the disc and passing through a point on its edge.
(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2 \mathrm{MR}^{2} / 5, where M is the mass of the sphere and \mathbf{R} is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius \mathbf{R} about any of its diameters to be \mathrm{MR}^{2} / 4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.
CBSE | Class 11 | NCERT | Physics | System of Particles and Rotational Motion
(a) Find the moment of inertia of a sphere about a tangent to the sphere, given the moment of inertia of the sphere about any of its diameters to be 2 \mathrm{MR}^{2} / 5, where M is the mass of the sphere and \mathbf{R} is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius \mathbf{R} about any of its diameters to be \mathrm{MR}^{2} / 4, find its moment of inertia about an axis normal to the disc and passing through a point on its edge.

The moment of inertia of a sphere about its diameter is =2 \mathrm{MR}^{2} / 5 and is also shown in the figure,

As the the theorem of parallel axes says, M.I of a sphere about a tangent to the sphere =2 \mathrm{MR}^{2} / 5 +M R^{2}=\left(7 M R^{2}\right) / 5

(b) Moment of inertia of a disc about its diameter is given as =\left(\mathrm{MR}^{2}\right) 1 / 4

(i)The theorem of the perpendicular axis states that the moment of inertia of a planar body about an axis passing through its center and perpendicular to the disc so we have =2 \times(1 / 4) \mathrm{MR}^{2}=\mathrm{MR}^{2} / 2. It is shown in the figure below,

(ii) Using the theorem of parallel axes:

Moment of inertia about an axis normal to the disc and going through a point on its circumference

=M R^{2} / 2+M R^{2}

=\left(3 \mathrm{MR}^{2}\right) / 2

i

More Material