NEET-XI-Physics
07: System of particles and Rotational Motion
- #10(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 MR2/5, where M is the mass of the sphere and R is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to 1 be 1/4 MR2, find the moment of inertia about an axis normal to the disc passing through a point on its edge.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to 1 be 1/4 MR2, find the moment of inertia about an axis normal to the disc 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 MR2/5, where M is the mass of the sphere and R is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to 1 be 1/4 MR2, find the moment of inertia about an axis normal to the disc passing through a point on its edge.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to 1 be 1/4 MR2, find the moment of inertia about an axis normal to the disc passing through a point on its edge.Ans : (a) Moment of inertia of sphere about any diameter = 2/5 MR
2
Applying theorem of parallel axes,Moment of inertia of sphere about a tangent to the sphere = 2/5 MR2 +M(R)2 =7/5 MR2
(b) We are given, moment of inertia of the disc about any of its diameters = 1/4 MR2
(-i) Using theorem of perpendicular axes, moment of inertia of the disc about an axis passing through its centre and normal to the disc = 2 x 1/4 MR2 = 1/2 MR2.
(ii) Using theorem axes, moment of inertia of the disc passing through a point on its edge and normal to the dies = 1/2 MR2+ MR2 = 3/2 MR2.
(b) We are given, moment of inertia of the disc about any of its diameters = 1/4 MR2
(-i) Using theorem of perpendicular axes, moment of inertia of the disc about an axis passing through its centre and normal to the disc = 2 x 1/4 MR2 = 1/2 MR2.
(ii) Using theorem axes, moment of inertia of the disc passing through a point on its edge and normal to the dies = 1/2 MR2+ MR2 = 3/2 MR2.
(a) Moment of inertia of sphere about any diameter = 2/5 MR
2
Applying theorem of parallel axes,Moment of inertia of sphere about a tangent to the sphere = 2/5 MR2 +M(R)2 =7/5 MR2
(b) We are given, moment of inertia of the disc about any of its diameters = 1/4 MR2
(-i) Using theorem of perpendicular axes, moment of inertia of the disc about an axis passing through its centre and normal to the disc = 2 x 1/4 MR2 = 1/2 MR2.
(ii) Using theorem axes, moment of inertia of the disc passing through a point on its edge and normal to the dies = 1/2 MR2+ MR2 = 3/2 MR2.
(b) We are given, moment of inertia of the disc about any of its diameters = 1/4 MR2
(-i) Using theorem of perpendicular axes, moment of inertia of the disc about an axis passing through its centre and normal to the disc = 2 x 1/4 MR2 = 1/2 MR2.
(ii) Using theorem axes, moment of inertia of the disc passing through a point on its edge and normal to the dies = 1/2 MR2+ MR2 = 3/2 MR2.
- #10-aFind 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 MR2/5, where M is the mass of the sphere and R is the radius of the sphere.
(b) Given the moment of inertia of a disc of mass M and radius R about any of its diameters to 1 be 1/4 MR2, find the moment of inertia about an axis normal to the disc passing through a point on its edge.Ans : Moment of inertia of sphere about any diameter = 2/5 MR
2
Applying theorem of parallel axes,Moment of inertia of sphere about a tangent to the sphere = 2/5 MR2 +M(R)2 =7/5 MR2
(b) We are given, moment of inertia of the disc about any of its diameters = 1/4 MR2
(-i) Using theorem of perpendicular axes, moment of inertia of the disc about an axis passing through its centre and normal to the disc = 2 x 1/4 MR2 = 1/2 MR2.
(ii) Using theorem axes, moment of inertia of the disc passing through a point on its edge and normal to the dies = 1/2 MR2+ MR2 = 3/2 MR2.
- #10-bGiven the moment of inertia of a disc of mass M and radius R about any of its diameters to 1 be 1/4 MR2, find the moment of inertia about an axis normal to the disc passing through a point on its edge.Ans : We are given, moment of inertia of the disc about any of its diameters = 1/4 MR2
(-i) Using theorem of perpendicular axes, moment of inertia of the disc about an axis passing through its centre and normal to the disc = 2 x 1/4 MR2 = 1/2 MR2.
(ii) Using theorem axes, moment of inertia of the disc passing through a point on its edge and normal to the dies = 1/2 MR2+ MR2 = 3/2 MR2.