Question

find the radius of gyration of a ring about an axis passing through the tangent in aplane

Answer 1

Answer:

MI about a diameter : I=

2

1

MR

2

by perpendicular axis theorem

MI about an axis passing through tangent to the ring: I=

2

1

MR

2

+MR

2

=

2

3

MR

2

by parallel axis theorem

For maximum radius of gyration ring should have maximum moment of inertia about a axis here, option [D] has maximum moment of inertial which is 2MR

2

Answer 2

Answer:

Moment of inertia at the centre and perpendicular to the plane of the ring.

So, about a point on the rim of the ring and the axis ⊥ to the plane of the ring, the moment of inertia

= mR2 + mR2 = 2mR2 (parallel axis theorem)

mK2 = 2mR2 (K = radius of the gyration)

K = √2R2 = √2 R.

Step-by-step explanation:

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