Question

A uniform cylinder of radius $\mathbf{10}\mathbf{cm}$and mass $\mathbf{20}\mathbf{}\mathbf{kg}$ is mounted so as to rotate freely about a horizontal axis that is parallel to and $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{cm}$ from the central longitudinal axis of the cylinder.

(a) What is the rotational inertia of the cylinder about the axis of rotation?

(b) If the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates, what is the angular speed of the cylinder as it passes through its lowest position?

Short answer

1. The rotational inertia of the cylinder about the axis of the cylinder is $0.15{\text{\hspace{0.17em}kg.m}}^2$.
2. The angular speed of the cylinder as it passes through its lowest position is$11\text{rad}/\text{s}$

Step-by-step solution

Given

The Radius of the cylinder is,$\mathrm{R}=10\text{\hspace{0.17em}cm}=0.10\text{\hspace{0.17em}m}$.

The mass of the cylinder is,$\mathrm{m}=20\text{\hspace{0.17em}kg}$.

The central longitudinal axis of the cylinder is, $\mathrm{d}=5.0\text{\hspace{0.17em}cm}=0.05\text{\hspace{0.17em}m}$.

Understanding the concept

Find therotational inertia of the cylinder about the axis of the cylinder using the parallel axis theorem. Then, using the law of conservation of energy, findtheangular speed of the cylinder as it passes through its lowest position.

Formula:

$\mathbf{I}\mathbf{=}{\mathbf{I}}_{\mathbf{com}}\mathbf{+}{\mathbf{md}}^{\mathbf{2}}\mathbf{}\mathbf{}$

${\mathbf{E}}_{\mathbf{i}}\mathbf{=}{\mathbf{E}}_{\mathbf{f}}$

(a) Calculate the rotational inertia of the cylinder about the axis of rotation

According to the parallel axis theorem,

$\mathrm{I}={\mathrm{I}}_{\mathrm{com}}+{\mathrm{md}}^2$

In this case,

$\mathrm{I}=\frac{1}{2}{\mathrm{mR}}^2+{\mathrm{md}}^2$

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{I}=\frac{1}{2}\left(20\text{\hspace{0.17em}kg}\right){\left(0.1\text{\hspace{0.17em}m}\right)}^2+\left(20\text{\hspace{0.17em}kg}\right){\left(0.05\text{\hspace{0.17em}m}\right)}^2\\ \mathrm{I}=0.15{\text{\hspace{0.17em}kg.m}}^{\text{2}}\end{array}$

Therefore, rotational inertia of the cylinder about the axis of the cylinder is,$0.15{\text{\hspace{0.17em}kg.m}}^{\text{2}}$ .

Step 4(b): Calculate the angular speed of the cylinder as it passes through its lowest position if the cylinder is released from rest with its central longitudinal axis at the same height as the axis about which the cylinder rotates.

The cylinder is released from rest, let’s say, fromtheheight h.

So, its initial K.E is zero.

According to the conservation of energy,

${\mathrm{E}}_{\mathrm{i}}={\mathrm{E}}_{\mathrm{f}}$

$\mathrm{K}.{\mathrm{E}}_{\mathrm{i}}+\mathrm{P}.{\mathrm{E}}_{\mathrm{i}}=\mathrm{K}.{\mathrm{E}}_{\mathrm{f}}+\mathrm{P}.{\mathrm{E}}_{\mathrm{f}}$

In this case,

$\begin{array}{c}0+\mathrm{mgh}=\frac{1}{2}{\mathrm{I\omega }}^2+0\\ {\mathrm{\omega }}^2=\frac{2\mathrm{mgh}}{\mathrm{I}}\end{array}$

Substitute all the value in the above equation.

$\begin{array}{c}\mathrm{\omega }=\sqrt{\frac{2\left(20\text{\hspace{0.17em}kg}\right)(9.8\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}})(0.050\text{\hspace{0.17em}m})}{(0.15{\text{\hspace{0.17em}kg.m}}^{\text{2}})}}\\ \mathrm{\omega }=11.4\text{\hspace{0.17em}rad}/\text{s}\\ \approx 11\text{\hspace{0.17em}rad}/\text{s}\end{array}$

Therefore, angular speed of the cylinder as it passes through its lowest position is$11\text{\hspace{0.17em}rad}/\text{s}$ .