Advertisements
Advertisements
Question
Prove the theorem of perpendicular axes.
(Hint: Square of the distance of a point (x, y) in the x–y plane from an axis through the origin perpendicular to the plane is x2 + y2).
Advertisements
Solution 1
The theorem of perpendicular axes states that 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 the perpendicular axis and lying in the plane of the body.
A physical body with centre O and a point mass m,in the x–y plane at (x, y) is shown in the following figure.
Moment of inertia about x-axis, Ix = mx2
Moment of inertia about y-axis, Iy = my2
Moment of inertia about z-axis, Iz = `m(sqrt(x^2 + y^2))^2`
Ix + Iy = mx2 + my2
= m(x2 + y2)
`= m(sqrt(x^2 + y^2))`
`I_x + I_y = I_z`
Hence the theorem is proved
Solution 2
The theorem of perpendicular axes: According to this theorem, the moment of inertia of a plane lamina (i.e., a two dimensional body of any shape/size) about any axis OZ perpendicular to the plane of the lamina is equal to sum of the moments of inertia of the lamina about any two mutually perpendicular axes OX and OY in the plane of lamina, meeting at a point where the given axis OZ passes through the lamina. Suppose at the point ‘R’ m{ particle is situated moment of inertia about Z axis of lamina
= moment of inertia of body about r-axis
= moment of inertia of the body about y-axis.
RELATED QUESTIONS
State an expression for the moment of intertia of a solid uniform disc, rotating about an axis passing through its centre, perpendicular to its plane. Hence derive an expression for the moment of inertia and radius of gyration:
i. about a tangent in the plane of the disc, and
ii. about a tangent perpendicular to the plane of the disc.
Prove the theorem of parallel axes about moment of inertia
State Brewster's law.
A string of length ℓ fixed at one end carries a mass m at the other. The string makes 2/π revolutions/sec around the vertical axis through the fixed end. The tension in the string is ______.
State and explain the theorem of parallel axes.
When a 12000 joule of work is done on a flywheel, its frequency of rotation increases from 10 Hz to 20 Hz. The moment of inertia of flywheel about its axis of rotation is ______. (π2 = 10)
A wheel of moment of inertia 2 kg m2 is rotating at a speed of 25 rad/s. Due to friction on the axis, it comes to rest in 10 minutes. Total work done by friction is ______.
A solid sphere of mass 'M' and radius 'R' is rotating about its diameter. A disc of same mass and radius is also rotating about an axis passing through its centre and perpendicular to the plane but angular speed is twice that of the sphere. The ratio of kinetic energy of disc to that of sphere is ______.
From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?
A uniform cylinder of mass M and radius R is to be pulled over a step of height a (a < R) by applying a force F at its centre 'O' perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). The minimum value of F required is:
A metal ring has a moment of inertia 2 kg·m2 about a transverse axis through its centre. It is melted and recast into a thin uniform disc of the same radius. What will be the disc's moment of inertia about its diameter?
Prove the theorem of perpendicular axes about the moment of inertia.
For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is:
A thin rod of length ‘4L’ and mass ‘4m’ is bent at the points as shown in the figure. The moment of inertia of the rod about an axis passing through point ‘O’ and perpendicular to plane of the paper is:
