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Found in: Page 292

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# (a) Show that the rotational inertia of a solid cylinder of mass M and radius R about its central axis is equal to the rotational inertia of a thin hoop of mass M and radius about its central axis. (b) Show that the rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given by ${\text{k}}{=}\sqrt{\frac{\text{I}}{\text{M}}}$The radius k of the equivalent hoop is called the radius of gyration of the given body.

1. We can show that rotational inertia of a solid cylinder of mass M and radius R about its central axis is equal to the rotational inertia of a thin hoop of mass M and radius about its central axis.
2. We can show that the rotational inertia I of any given body of mass M about any given axis is equal to rotational inertia of an equivalent hoop about that axis, if the hoop has same mass M and radius k given by$\text{k}=\sqrt{\frac{\text{I}}{\text{M}}}$
See the step by step solution

## Step 1: Given

$\text{k}=\sqrt{\frac{\text{I}}{\text{M}}}$

## Step 2: Understanding the concept

We can use the concept of inertia of the cylinder and the hoop. Also we use the concept of radius of gyration. For the given objects, the masses are the same, so we can find the relation between their radii. Also using the equation of radius of gyration, we find the k in terms of I and M.

Formulae:

$\text{I}={\text{MR}}^{2}$

$\text{I}={\text{Mk}}^{2}$

## Step 3: (a) To show that the rotational inertia of the solid cylinder about its central axis is equal to that of the thin hoop about its central axis

Rotational inertia of the solid cylinder equal to rotational inertia of thin hoop:

From the book, table 10-2, we get the equation of inertia of cylinder and hoop:

${\text{I}}_{c}=\frac{1}{2}{\text{MR}}^{2}$

and${\text{I}}_{h}={\text{Mr}}^{2}$

We can write $\text{r}={\text{R}}_{h}$

Both the bodies have the same mass, so the inertia will be the same, we get

$\frac{1}{2}{\text{MR}}^{2}={\text{MR}}_{h}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{{\text{R}}^{2}}{2}={\text{R}}_{h}^{2}\phantom{\rule{0ex}{0ex}}⇒{\text{R}}_{h}=\frac{\text{R}}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}$

## Step 4: (b) To show that the rotational inertia I of any given body of mass M about any given axis is equal to the rotational inertia of an equivalent hoop about that axis, if the hoop has the same mass M and a radius k given byk=IM

Rotational inertia of any given body is equal to rotational inertia of an equivalent hoop of mass M and radius k:

From the equation of radius of gyration, we can write,

$\text{I}={\text{Mk}}^{2}\phantom{\rule{0ex}{0ex}}⇒{\text{k}}^{2}=\frac{\text{I}}{\text{M}}\phantom{\rule{0ex}{0ex}}⇒\text{k}=\sqrt{\frac{\text{I}}{\text{M}}}$