Chapter 18, Problem 1FP

The 80-kg wheel has a radius of gyration about its mass center O of k_o = 400 mm. Determine its angular velocity after it has rotated 20 revolutions starting from rest.

To determine

The angular velocity of the wheel after it has rotated 20 revolutions starting from rest.

Answer to Problem 1FP

The angular velocity of the wheel after it has rotated 20 revolutions starting from rest is $24.3\text{rad/s}$.

Explanation of Solution

Given:

The mass of wheel is $80\text{kg}$.

The radius of gyration of wheel about its mass center $O$ is $k_O=400\text{mm}$.

The radius of the wheel is $0.6\text{m}$.

The force acting on the wheel is $P=50\text{N}$.

Write the formula for mass moment of inertia $\left(I_O\right)$.

$I_O=m{k}_O$ (I)

Here, $m$ is the mass and $k_O$ is the radius of gyration about mass center O.

Write the formula for kinetic energy $\left(T\right)$ (Rotation about fixed axis).

$T=\frac{1}{2}{I}_O{\omega }^2$ (II)

Here, $I_O$ is the mass moment of inertia and $\omega$ is the angular velocity.

Write the formula for work done by the couple moment. $\left(M\right)$.

$U_M=P\cdot r\cdot \theta$ (III)

Here, $P$ is the force or load, $r$ is the radius and $\theta$ is the angular displacement in $\text{radians}$.

Write the formula for Principle of Work and Energy.

$T_1+{\displaystyle \sum U_{1-2}=T_2}$ (IV)

Here, $T_1$ is initial kinetic energy, $T_2$ is final kinetic energy and ${{\displaystyle \sum U}}_{1-2}$ is total work done by all the forces.

Conclusion:

Calculate the mass moment of inertia $\left(I_O\right)$.

Substitute $80\text{kg}$ for $m$ and $400\text{mm}$ for $k_O$ in Equation (I).

$\begin{array}{c}{I}_O=80\text{kg}\times {\left(\text{400}\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}^2\\ =12.8\text{kg}\cdot {\text{m}}^2\end{array}$

Calculate the initial and final kinetic energy of the wheel.

At initial:

When the wheel starts from rest.

The initial kinetic energy becomes zero.

$T_1=0$

At final:

Substitute $12.8\text{kg}\cdot {\text{m}}^2$ for $I_O$ in Equation (II).

$\begin{array}{c}{T}_2=\frac{1}{2}\left(12.8\text{kg}\cdot {\text{m}}^2\right){\omega }^2\\ =6.4{\omega }^2\end{array}$

Calculate the angular displacement $\left(\theta \right)$.

$\begin{array}{c}\theta =\text{No}\text{.of revolutions}\times \text{2}\pi \\ \text{= 20}\times \text{2}\pi \\ \text{= 40}\pi \end{array}$

Calculate the work done by the couple moment $\left(M\right)$.

Substitute $50\text{N}$ for $P$, $0.6\text{m}$ for $r$ and $40\pi$ for $\theta$ in Equation (III).

$\begin{array}{c}{U}_M=50\text{N}\times \text{0}\text{.6}\text{}\text{m}\times \text{40}\pi \\ \text{=3769}\text{.9111}\text{J}\end{array}$

Here, the total work done is ${{\displaystyle \sum U}}_{1-2}=U_M$.

${{\displaystyle \sum U}}_{1-2}=3769.9111\text{J}$

Calculate the angular velocity of the wheel after it has rotated 20 revolutions starting from rest.

Substitute $0$ for $T_1$, $3769.9111\text{J}$ for ${\displaystyle \sum U_{1-2}}$ and $6.4{\omega }^2$ for $T_2$ in Equation (IV).

$\begin{array}{l}0+3769.9111\text{J}=6.4{\omega }^2\\ {\omega }^2=\frac{3769.9111}{6.4}\\ \omega =\sqrt{589.0486}\\ \omega =24.2703\text{rad/s}\end{array}$

Thus, the angular velocity of the wheel after it has rotated 20 revolutions starting from rest is $24.3\text{rad/s}$.