Solveeit Logo
Login

Question

Question: A lamina is made by removing a small disc of 2R diameter from a bigger disc of uniform mass density ...

A lamina is made by removing a small disc of 2R diameter from a bigger disc of uniform mass density and radius 2R, as shown in figure. The moment of inertia of this lamina about the axis passing through O and P is Io{I_o} andIp{I_p}, respectively. Both these axes are perpendicular to the plane of the lamina. The ratio IpIo\dfrac{{{I_p}}}{{{I_o}}} to the nearest integer is:

A) 1337\dfrac{{13}}{{37}}
B) 3713\dfrac{{37}}{{13}}
C) 7331\dfrac{{73}}{{31}}
D) 813\dfrac{8}{{13}}

Explanation

Solution

First we found the moment of inertia about O and also found out the moment of inertia about P by subtracting the moment of inertia of cavity from the total part.

Complete step by step solution:

When we rotate this about O, then make a circle of radius 2R. Then first we find the moment of inertia of circle of radius 2Rand subtract moment of inertia of circle of radius R.
I0=12(4m)(2R2)−[12mR2+mR2]{I_0} = \dfrac{1}{2}\left( {4m} \right)\left( {2{R^2}} \right) - \left[ {\dfrac{1}{2}m{R^2} + m{R^2}} \right]
⇒8mR2−32mR2\Rightarrow 8m{R^2} - \dfrac{3}{2}m{R^2}
⇒132mR2\Rightarrow \dfrac{{13}}{2}m{R^2}

⇒IP=[12(4m)(2R)2+4m(2R)2]−[12mR2+m(5R)2]\Rightarrow {I_P} = \left[ {\dfrac{1}{2}\left( {4m} \right){{\left( {2R} \right)}^2} + 4m{{\left( {2R} \right)}^2}} \right] - \left[ {\dfrac{1}{2}m{R^2} + m{{\left( {\sqrt 5 R} \right)}^2}} \right]
⇒[12×4m×4R2+4m×4R2]−[12mR2+5mR2]\Rightarrow \left[ {\dfrac{1}{2} \times 4m \times 4{R^2} + 4m \times 4{R^2}} \right] - \left[ {\dfrac{1}{2}m{R^2} + 5m{R^2}} \right]
⇒[8mR2+16mR2]−[112mR2]\Rightarrow \left[ {8m{R^2} + 16m{R^2}} \right] - \left[ {\dfrac{{11}}{2}m{R^2}} \right]
⇒24mR2−112mR2\Rightarrow 24m{R^2} - \dfrac{{11}}{2}m{R^2}
⇒mR2[24−112]\Rightarrow m{R^2}\left[ {24 - \dfrac{{11}}{2}} \right]
⇒mR2[48−112]\Rightarrow m{R^2}\left[ {\dfrac{{48 - 11}}{2}} \right]
⇒mR2[372]\Rightarrow m{R^2}\left[ {\dfrac{{37}}{2}} \right]
⇒372mR2\Rightarrow \dfrac{{37}}{2}m{R^2}
Then the ratio of moment of inertia around P, to moment of inertia about O.
⇒IpI0=372132=3713\Rightarrow \dfrac{{{I_p}}}{{{I_0}}} = \dfrac{{\dfrac{{37}}{2}}}{{\dfrac{{13}}{2}}} = \dfrac{{37}}{{13}}
⇒IpI0=3713\Rightarrow \boxed{\dfrac{{{I_p}}}{{{I_0}}} = \dfrac{{37}}{{13}}}

Note: When we find out the moment inertia around P, then we take a small sphere of radius R, but we should take the sphere radius of 5R\sqrt 5 R.