Question

A uniform thin ring of radius r and mass M suspended in vertical from a point in its circumference it's time period of oscillation

Answer 1

Answer:

Correct option is

B

2π2g3R

The time period of the disc is 2π(3r/2g)

We know that the time period of an object,

T=2π(1/mgL)

where,

I= moment of inertia from the suspended point

L= distance of its centre from suspended point =r

we know that, the moment of inertia of disc about its centre =mr2/2

using parallel axis theoram the moment of inertia from a point in its periphery,

I=mr2+mr2/2=3mr2/2

putting the values in the above equation we get,

2π(3mr2/2mgr)

=2π(3r/2g) 

therefore, the time period of the disc is 2π(3r/2g)

Hence,

option (B) is correct answer.

Answer 2

Answer:

B

2π

2g

3R

Explanation:

The time period of the disc is 2π

(3r/2g)

We know that the time period of an object,

T=2π

(1/mgL)

where,

I= moment of inertia from the suspended point

L= distance of its centre from suspended point =r

we know that, the moment of inertia of disc about its centre =mr

2

/2

using parallel axis theoram the moment of inertia from a point in its periphery,

I=mr

2

+mr

2

/2=3mr

2

/2

putting the values in the above equation we get,

2π

(3mr

2

/2mgr)

=2π

(3r/2g)

therefore, the time period of the disc is 2π

(3r/2g)