Question

3-Calculate the M.I.of thin uniform ring about an axis tangent to ring in aplane of ring .if its M.I.about an axis passing through the centre & perpendicular to its plane is 4kg-m2 *

1 point

3kg-m2

6kg-m2

9kg-m2

12kg-m2

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Answer 1

Answer:

M.I. of ring about it's center and perpendicular to it's axis is MR^2

therefore

MR^2 = 4

now using perpendicular axis theorem

the M.I. of the ring about it's diameter in it's plane would be MR^2 / 2

now using parallel axis theorem

about it's tangent in it's plane

M.I. would be (MR^ 2 / 2) + MR^ 2

now put value

M.I. = 6

Answer 2

Given: M.I.about an axis passing through the centre and perpendicular to its plane is 4kg-m^2

To find: M.I.of thin uniform ring about an axis tangent to ring in the plane of ring

Explanation: The moment of inertia(M.I) about an axis passing through the centre and perpendicular to its plane is given by the formula $m {r}^{2}$ where m is mass of the ring and r is the radius of the ring.

But is given that M.I. through centre and perpendicular to ring is 4 kg m^2.

Therefore,

$m {r}^{2} = 4 \: kg \: {m}^{2}$

Now, using perpendicular axis theorem which states that M.I. about an axis perpendicular to plane is twice the M.I. in the plane.

Therefore,

M.I. passing through centre in plane of ring

= M.I. passing through centre and perpendicular to ring/2

=$\frac{m {r}^{2} }{2}$

Now using parallel axis theorem which states that M.I. along any axis is the sum of M.I. about centre and product of mass and square of radius between centre and axis.

M.I. passing through tangent in plane of ring

=$\frac{m {r}^{2} }{2} + m {x}^{2}$

where x is the distance between centre and tangent.

Distance between centre and tangent is equal to the radius of the ring(r).

Using formula:

=$\frac{m {r}^{2} }{2} + m {r}^{2}$

=$\frac{3}{2} m {r}^{2}$

Using $m {r}^{2} = 4$

=$\frac{3}{2} m {r}^{2} =\frac{3}{2} \times 4 =6$

Therefore, M.I about an axis tangent to ring in the plane of the ring is 6 kg m^2.