Question

A thin rod of length ‘4L’ and mass ‘4m’ is bent at the points as shown in the figure. The moment of inertia of the rod about an axis passing through point ‘O’ and perpendicular to plane of the paper is:

Options

• `(m L^2)/3`

• `(10 m L^2)/3`

• `(m L^2)/12`

• `(m L^2)/24`

Answer 1

`bb((10 m L^2)/3)`

Explanation:

After bending, the length of each segment is ‘L’ as shown in the figure. Consider only the system of two segments forming the V shape (viz., OAB).

For the axis passing through Ο and perpendicular to the plane of paper,

M.I. of rod OA = `(m L^2)/3`    ...(i)

Now, distance of the centre of rod AB from O is:

h = `sqrt (L^2/4 + L^2)`

= `sqrt ((5 L^2)/4)`

= `(sqrt 5 L)/2`

Using the theorem of parallel axes:

M.I. of AB = `(m L^2)/12 + mh^2`

= `(m L^2)/12 + (5 m L^2)/4`

= `(4 m L^2)/3`    ...(ii)

From equations (i) and (ii),

I_OAB = `(m L^2)/3 + (4 m L^2)/3`

= `(5 m L^2)/3`

Similarly,

I_OCD = `(5 m L^2)/3`

∴ I_net = `(10 m L^2)/3`