Question

From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through O is:

• A. 4 MR2
• B. 40/9 MR2
• C. 10 MR2
• D. 37/9 MR2

Answer 1

Answer: (A)

4 MR2

Answer 2

M.I. of complete disc about 'O' point

ITotal=½ (9M) R2 ...(i)

Radius of removed disc $= \frac { R } { 3 }$

Mass of removed disc $= \frac { 9 M } { 9 } = M$

[As M = p R2 t MR2]

M.I. of removed disc about its own axis

$= \frac { 1 } { 2 } M \left( \frac { R } { 3 } \right) ^ { 2 } = \frac { M R ^ { 2 } } { 18 }$

Moment of inertia of removed disc about 'O'

Iremoved disc= $= I _ { c m } + m x ^ { 2 }$ $= \frac { M R ^ { 2 } } { 18 } + M \left( \frac { 2 R } { 3 } \right) ^ { 2 }$ $= \frac { M R ^ { 2 } } { 2 }$

M.I. of complete disc can also be written as

ITotal= IRemoved disc+ IRemaining disc

ITotal= $\frac { \mathrm { MR } ^ { 2 } } { 2 }$ + IRemaining disc ...(ii)

Equating (i) and (ii) we get

$\frac { \mathrm { MR } ^ { 2 } } { 2 }$ + IRemaining disc= $\frac { 9 \mathrm { MR } ^ { 2 } } { 2 }$

IRemaining disc= $\frac { 9 \mathrm { MR } ^ { 2 } } { 2 }$ – $\frac { \mathrm { MR } ^ { 2 } } { 2 }$ = $\frac { 8 \mathrm { MR } ^ { 2 } } { 2 }$ = 4 MR²