Question 385: Centre of Mass and Rotational Motion

Question

A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is $\sqrt{n} \mathrm{~cm}$. The value of $n$ is

Accepted Answer

To find the radius of gyration, we first need to determine the moment of inertia (I) of the solid sphere about the given axis using the Parallel Axis Theorem.
Radius of sphere $=R=10 \mathrm{~cm}$
Distance of the axis from the centre $=\mathrm{d}=15 \mathrm{~cm}$
Let the mass of the sphere be M .
The moment of inertia of a solid sphere about an axis passing through its centre is :
$$ \mathrm{I}_{\mathrm{cm}}=\frac{2}{5} \mathrm{MR}^2 $$
According to the Parallel Axis Theorem, the moment of inertia about an axis at a distance d is :
$$ \mathrm{I}=\mathrm{I}_{\mathrm{cm}}+\mathrm{Md}^2 $$
$$\Rightarrow $$ $$ \mathrm{I}=\frac{2}{5} \mathrm{MR}^2+\mathrm{Md}^2 \ldots (i)$$
The radius of gyration k is defined by the relation :
$$ \mathrm{I}=\mathrm{Mk}^2 \ldots (ii)$$
$$\Rightarrow $$ $\mathrm{Mk}^2=\frac{2}{5} \mathrm{MR}^2+\mathrm{Md}^2$
$\Rightarrow \mathrm{k}^2=\frac{2}{5} \mathrm{R}^2+\mathrm{d}^2$
Substituting the values, $\mathrm{R}=10 \mathrm{~cm}$ and $\mathrm{d}=15 \mathrm{~cm}$ :
$$ \mathrm{k}^2=\frac{2}{5}(10)^2+(15)^2 $$
$$\Rightarrow $$ $\mathrm{k}^2=\frac{2}{5}(100)+225$
$$\Rightarrow $$ $\mathrm{k}^2=2(20)+225$
$$ \mathrm{k}^2=40+225=265 $$
The question states that the radius of gyration $\mathrm{k}=\sqrt{\mathrm{n}} \mathrm{cm}$.
Therefore, $\mathrm{k}^2=\mathrm{n}=265 \Rightarrow \mathrm{n}=265$
Therefore, the value of n is 265.
Hence, the correct answer is 265 .

Centre of Mass and Rotational Motion

A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is \sqrt{n} \mathrm{~cm}. The value of n is

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