Question 96: Centre of Mass and Rotational Motion

Question

A solid sphere of mass 5 kg and radius 10 cm is kept in contact with another solid sphere of mass 10 kg and radius 20 cm . The moment of inertia of this pair of spheres about the tangent passing through the point of contact is $\_\_\_\_$ $\mathrm{kg} \cdot \mathrm{m}^2$.

Accepted Answer

We have two solid spheres touching each other. The axis of rotation is the common tangent passing exactly through their point of contact.
Using the Parallel Axis Theorem.
$$ \mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Md}^2 $$
Where $I_{c m}$ is the moment of inertia about the centre of mass, $M$ is the mass of object whose centre of mass is at distance d from the axis of rotation.
For a uniform solid sphere the moment of inertia about its diametric centre axis is,
$$ \mathrm{I}_{\mathrm{cm}}=\frac{2}{5} \mathrm{Mr}^2 $$
The distance $d$ from the center to the tangent is the radius $r$.
$$ \mathrm{I}_{	ext {tangent }}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Mr}^2 $$
$$\Rightarrow $$ $$ \mathrm{I}_{	ext {tangent }}=\frac{2}{5} \mathrm{Mr}^2+\mathrm{Mr}^2=\frac{7}{5} \mathrm{Mr}^2 $$
For sphere $\mathrm{s}_1$
Mass $\mathrm{m}_1=5 \mathrm{~kg}$
Radius $\mathrm{r}_1=10 \mathrm{~cm}=0.1 \mathrm{~m}$
So, the moment of inertia of sphere $\mathrm{s}_1, \mathrm{I}_1=\frac{7}{5} \mathrm{~m}_1 \mathrm{r}_1^2$
$$ \mathrm{I}_1=\frac{7}{5} 	imes 5 	imes(0.1)^2 $$
$$\Rightarrow $$ $$ \mathrm{I}_1=7 	imes 0.01=0.07 \mathrm{~kg} \cdot \mathrm{~m}^2 $$
For Sphere $2 \mathrm{~s}_2$ :
Mass $\mathrm{m}_2=10 \mathrm{~kg}$
Radius $\mathrm{r}_2=20 \mathrm{~cm}=0.2 \mathrm{~m}$
So, the moment of inertia of sphere $\mathrm{s}_2, \mathrm{I}_2=\frac{7}{5} \mathrm{~m}_2 \mathrm{r}_2^2$
$$ I_2=\frac{7}{5} 	imes 10 	imes(0.2)^2 $$
$$\Rightarrow $$ $I_2=7 	imes 2 	imes 0.04$
$$\Rightarrow $$ $$ \mathrm{I}_2=\frac{1}{4} 	imes 0.04=0.56 \mathrm{~kg} \cdot \mathrm{~m}^2 $$
So, the total moment of inertia of two sphere system is,
$$ \mathrm{I}_{	ext {total }}=\mathrm{I}_1+\mathrm{I}_2 $$
$$ \begin{aligned} & \Rightarrow \mathrm{I}_{	ext {total }}=0.07+0.56 \ & \Rightarrow \mathrm{I}_{	ext {total }}=0.63 \mathrm{~kg} \cdot \mathrm{~m}^2 \end{aligned} $$
Therefore, the moment of inertia of given system of spheres about their common tangent is $0.63 \mathrm{~kg} \cdot \mathrm{~m}^2$.
Hence, the correct option is (D).

Answer

0.18

Answer

0.72

Answer

0.36

Answer

0.63

Centre of Mass and Rotational Motion

A solid sphere of mass 5 kg and radius 10 cm is kept in contact with another solid sphere of mass 10 kg and radius 20 cm . The moment of inertia of this pair of spheres about the tangent passing through the point of contact is \_\_\_\_ \mathrm{kg} \cdot \mathrm{m}^2.

Choose the correct answer: