state parallel axis theorem
Answers
Answer:
The parallel axis theorem, also known as Huygens–Steiner theorem, or just as Steiner's theorem,[1] named after Christiaan Huygens and Jakob Steiner, can be used to determine the mass moment of inertia or the second moment of area of a rigid body about any axis, given the body's moment of inertia about a parallel axis through the object's center of gravity and the perpendicular distance between the axes.
Explanation:
Suppose a body of mass m is made to rotate about an axis z passing through the body's centre of gravity. The body has a moment of inertia Icm with respect to this axis. The parallel axis theorem states that if the body is made to rotate instead about a new axis z′ which is parallel to the first axis and displaced from it by a distance d, then the moment of inertia I with respect to the new axis is related to Icm by
{\displaystyle I=I_{\mathrm {cm} }+md^{2}.} I=I_{\mathrm {cm} }+md^{2}.
Explicitly, d is the perpendicular distance between the axes z and z′.
The parallel axis theorem can be applied with the stretch rule and perpendicular axis theorem to find moments of inertia for a variety of shapes.
Parallel axes rule for area moment of inertia
Derivation
Edit
We may assume, without loss of generality, that in a Cartesian coordinate system the perpendicular distance between the axes lies along the x-axis and that the center of mass lies at the origin. The moment of inertia relative to the z-axis is
{\displaystyle I_{\mathrm {cm} }=\int (x^{2}+y^{2})\,dm.} I_{\mathrm {cm} }=\int (x^{2}+y^{2})\,dm.
The moment of inertia relative to the axis z′, which is a perpendicular distance d along the x-axis from the centre of mass, is
{\displaystyle I=\int \left[(x+d)^{2}+y^{2}\right]\,dm} {\displaystyle I=\int \left[(x+d)^{2}+y^{2}\right]\,dm}
Expanding the brackets yields
{\displaystyle I=\int (x^{2}+y^{2})\,dm+d^{2}\int dm+2d\int x\,dm.} {\displaystyle I=\int (x^{2}+y^{2})\,dm+d^{2}\int dm+2d\int x\,dm.}
The first term is Icm and the second term becomes md2. The integral in the final term is a multiple of the x-coordinate of the center of mass – which is zero since the center of mass lies at the origin. So, the equation becomes:
{\displaystyle I=I_{\mathrm {cm} }+md^{2}.} I=I_{\mathrm {cm} }+md^{2}.
Tensor generalization
Edit
The parallel axis theorem can be generalized to calculations involving the inertia tensor. Let Iij denote the inertia tensor of a body as calculated at the centre of mass. Then the inertia tensor Jij as calculated relative to a new point is
{\displaystyle J_{ij}=I_{ij}+m\left(|\mathbf {R} |^{2}\delta _{ij}-R_{i}R_{j}\right),} J_{ij}=I_{ij}+m\left(|\mathbf {R} |^{2}\delta _{ij}-R_{i}R_{j}\right),
where {\displaystyle \mathbf {R} =R_{1}\mathbf {\hat {x}} +R_{2}\mathbf {\hat {y}} +R_{3}\mathbf {\hat {z}} \!} \mathbf {R} =R_{1}\mathbf {\hat {x}} +R_{2}\mathbf {\hat {y}} +R_{3}\mathbf {\hat {z}} \! is the displacement vector from the centre of mass to the new point, and δij is the Kronecker delta.
Answer:
going to this theorem the moment of inertia of a body about an axis in its plane and parallel to the axis passing the centre of mass of the body is equal to the moment of inertia through the centre of mass of body plus the mass of the body and square of the distance between two axis
Proof:- let A and B with Axes about which we have to calculate moment inertia
let P and Q with Axis which is passing through the centre of the mass of body
let us consider a particle of mass M at a distance x from a p q access
now the moment of inertia is PQ=mx2
some information refers to the attachment..
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