JEE MAIN - Physics (2023 - 25th January Evening Shift - No. 25)
If a solid sphere of mass 5 kg and a disc of mass 4 kg have the same radius. Then the ratio of moment of inertia of the disc about a tangent in its plane to the moment of inertia of the sphere about its tangent will be $$\frac{x}{7}$$. The value of $$x$$ is ___________.
Answer
5
Explanation
Solid Sphere :
_25th_January_Evening_Shift_en_25_1.png)
$$ \begin{aligned} & I_{\text {tangent }}=I_{\mathrm{cm}}+m R^{2} \\\\ & =\frac{2}{5} m R^{2}+m R^{2}=\frac{7}{5} m R^{2} \\\\ & =7 R^{2} \quad(m=5 \mathrm{~kg}) \end{aligned} $$
CIRCULAR DISC :
_25th_January_Evening_Shift_en_25_2.png)
$$ \begin{aligned} I_{\text {disc }} & =I_{\mathrm{cm}}+m R^{2} \\\\ & =\frac{m R^{2}}{4}+m R^{2} \\\\ & =\frac{5}{4} m R^{2} \\\\ & =5 R^{2} \end{aligned} $$
$\frac{l_{\text {disc }}}{l_{\text {tangent }}}=\frac{5}{7}$
Concept :
1. For Solid Sphere :
Position of the axis of rotation : About its diametric axis which passes through its centre of mass
_25th_January_Evening_Shift_en_25_3.png)
Moment of Inertia (I) = $${2 \over 5}M{R^2}$$
Radius of gyration (K) = $$\sqrt {{2 \over 5}} R$$
Position of the axis of rotation : About a tangent to the sphere
_25th_January_Evening_Shift_en_25_4.png)
Moment of Inertia (I) = $${7 \over 5}M{R^2}$$
Radius of gyration (K) = $$\sqrt {{7 \over 5}} R$$
2. For CIRCULAR DISC :
Position of the axis of rotation : About an axis perpendicular to the plane and passes through the centre
_25th_January_Evening_Shift_en_25_5.png)
Moment of Inertia (I) = $${1 \over 2}M{R^2}$$
Radius of gyration (K) = $${R \over {\sqrt 2 }}$$
Position of the axis of rotation : About the diametric axis
_25th_January_Evening_Shift_en_25_6.png)
Moment of Inertia (I) = $${1 \over 4}M{R^2}$$
Radius of gyration (K) = $${R \over { 2 }}$$
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