Question 435: Centre of Mass and Rotational Motion

Question

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 ___________.

Accepted Answer

Solid Sphere :
$$
\begin{aligned}
& I_{	ext {tangent }}=I_{\mathrm{cm}}+m R^{2} \\
& =\frac{2}{5} m R^{2}+m R^{2}=\frac{7}{5} m R^{2} \\
& =7 R^{2} \quad(m=5 \mathrm{~kg}) 
\end{aligned}
$$
CIRCULAR DISC :
$$
\begin{aligned}
I_{	ext {disc }} & =I_{\mathrm{cm}}+m R^{2} \\
& =\frac{m R^{2}}{4}+m R^{2} \\
& =\frac{5}{4} m R^{2} \\
& =5 R^{2}
\end{aligned}
$$
$\frac{l_{	ext {disc }}}{l_{	ext {tangent }}}=\frac{5}{7}$
Concept :
1. For Solid Sphere :
Position of the axis of rotation :
About its diametric
 axis which passes
through its centre
of mass
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
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
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
Moment of Inertia (I) = $${1 \over 4}M{R^2}$$
Radius of gyration (K) = $${R \over { 2 }}$$

Centre of Mass and Rotational Motion

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 ___________.

Integer / Numeric Answer Type