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प्रश्न
Prove the theorem of parallel axes about moment of inertia
उत्तर
The moment of inertia of a body about any axis is equal to the sum of its moment of inertia about a parallel axis through its centre of gravity and the product of the mass of the body and the square of the distance between the two axes.
Proof :
Let us consider a body having its centre of gravity at G as shown in Fig.. The axis XX′ passes through the centre of gravity and is perpendicular to the plane of the body. The axis X1X1′ passes through the point O and is parallel to the axis XX′ . The distance between the two parallel axes is x.
Let the body be divided into large number of particles each of mass m . For a particle P at a distance r from O, its moment of inertia about the axis X1OX1′ is equal to m r 2.
The moment of inertia of the whole body about the axis X1X1′ is given by,
I = Σ mr2 ???(1)
From the point P, drop a perpendicular PA to the extended OG and join PG.
In the ∆OPA,
OP 2 = OA2 + AP 2
r2 = x2 + 2xh + h2 + AP2 ???(2)
But from ∆ GPA,
GP 2 = GA2 + AP 2
y 2 = h 2 + AP 2 ..(3)
Substituting equation (3) in (2),
r 2 = x 2 + 2xh + y 2 ..(4)
Substituting equation (4) in (1),
I0 = Σ m (x2 + 2xh + y2)
= Σmx2 + Σ2mxh + Σmy2
= Mx2 + My2 + 2xΣmh
Here My2 = IG is the moment of inertia of the body about the line passing through the centre of gravity. The sum of the turning moments of all the particles about the centre of gravity is zero, since the body is balanced about the centre of gravity G.
Σ (mg) (h) = 0 (or) Σ mh = 0 [since g is a constant]
equation (5) becomes, I0= Mx2 + IG
Thus the parallel axes theorem is proved.
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संबंधित प्रश्न
State the theorem of perpendicular axes about moment of inertia.
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.
State Brewster's law.
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).
Prove the theorem of parallel axes.
(Hint: If the centre of mass is chosen to be the origin `summ_ir_i = 0`
Find the radius of gyration of circular ring of radius r about a line perpendicular to the plane of the ring and passing through one of its particles.
Answer in brief:
State the conditions under which the theorems of parallel axes and perpendicular axes are applicable. State the respective mathematical expressions.
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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?
Two particles A and B having equal charges are placed at a distance d apart. A third charged particle placed on the perpendicular bisection of AB at distance x. The third particle experiences maximum force when ______.
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?
A uniform disc of mass 10 kg and radius 60 cm rotates about an axis perpendicular to its plane and passing through its centre at 1200 rpm. Calculate its rotation kinetic energy. [Take π2 = 10]
State and prove the theorem of the parallel axis about the moment of inertia.
Prove the theorem of perpendicular axes about the moment of inertia.
