JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Rotational Mechanics

A lamina is made by removing a small disc of diameter

2 R

from a bigger disc of uniform mass density and radius

2 R

, as shown in the figure. The moment of inertia of this lamina about axes passing through

O

and

P

I_{O}

and

I_{P}

, respectively. Both these axes are perpendicular to the plane of the lamina. The ratio

\frac{I _{P}}{I _{O}}

to the nearest integer is

Answer: 3

Solution

Let the mass of the big disc be

4 m

, then the removed disc has mass

m

. For the big disc:

I_{O, bi g} = \frac{4 m ( 2 R ) ^{2}}{2} = 8 m R^{2}

I_{P, bi g} = 8 m R^{2} + 4 m (2 R)^{2} = 24 m R^{2}

. For the removed disc:

I_{O, r e m o v e d} = \frac{m R ^{2}}{2} + m (R)^{2} = \frac{3 m R ^{2}}{2}

I_{P, r e m o v e d} = \frac{m R ^{2}}{2} + m (5 R^{2}) = \frac{11 m R ^{2}}{2}

. Thus

I_{O} = 8 m R^{2} - \frac{3 m R ^{2}}{2} = \frac{13 m R ^{2}}{2}

and

I_{P} = 24 m R^{2} - \frac{11 m R ^{2}}{2} = \frac{37 m R ^{2}}{2}

. The ratio is

\frac{I _{P}}{I _{O}} = \frac{37}{13} \approx 2.85 \approx 3

JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Integer answer