Chapter 18.4, Problem 1PP

Determine the kinetic energy of the 100-kg object.

To determine

The kinetic energy of the object:

Answer to Problem 1PP

The kinetic energy of the disk is $900\text{J}$.

Explanation of Solution

Given:

The mass of disk is $100\text{kg}$.

The angular velocity of the disk is $\omega =3\text{rad/s}$.

The radius of the disk is $2\text{m}$.

Draw the free body diagram of the rod as shown in Figure (1a).

Refer Figure (1a).

Write the formula for mass moment of inertia $\left(I\right)$ of the disk about it is center.

$I=\frac{1}{2}m{r}^2$

Here, $m$ is the mass and $r$ is the radius of the disk.

Write the formula for kinetic energy $\left(T\right)$

(Rotation about fixed axis).

$T=\frac{1}{2}I{\omega }^2$

Substitute $\frac{1}{2}m{r}^2$ for $I$.

$T=\frac{1}{2}\left(\frac{1}{2}m{r}^2\right)\omega$ (I)

Here, $\omega$ is the angular velocity.

Conclusion:

Calculate the kinetic energy of the disk.

Substitute $100\text{kg}$ for $m$, $2\text{}\text{m}$ for $r$ and $3\text{rad/s}$ for $\omega$ in Equation (I).

$\begin{array}{c}T=\frac{1}{2}\left[\frac{1}{2}\left(100\text{kg}\right){\left(2\text{m}\right)}^2\right]{\left(3\text{rad/s}\right)}^2\\ =\frac{1}{2}\left(200\right)\left(9\right)\\ =900\text{J}\end{array}$

Thus, the kinetic energy of the disk is $900\text{J}$.

To determine

The kinetic energy of the object:

Answer to Problem 1PP

The kinetic energy of the disk is $800\text{J}$.

Explanation of Solution

Given:

The mass of rod is $100\text{kg}$.

The angular velocity of the rod is $\omega =2\text{rad/s}$.

The length of the rod is $6\text{m}$.

Draw the free body diagram of the rod as shown in Figure (1b).

Refer Figure (1b).

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

$I_O=I_G+m{l}_{OG}{}^2$                     (I)

Here, $I_G$ is the moment of inertial about the centroid $G$ and $I_G=\frac{1}{12}m{l}^2$.

Substitute $\frac{1}{12}m{l}^2$ for $I_G$ in Equation (I).

$I_O=\frac{1}{12}m{l}^2+m{l}_{OG}{}^2$

Here, $I_O$ is the mass moment of inertia about point $O$, $m$ is the mass, $l$ is the total length of the rod and $l_{OG}$, distance between the point $O\text{and}G$.

Write the formula for kinetic energy $\left(T\right)$.

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

Substitute $\frac{1}{12}m{l}^2+m{l}_{OG}{}^2$ for $I_O$.

$T=\frac{1}{2}\left(\frac{1}{12}m{l}^2+m{l}_{OG}{}^2\right){\omega }^2$ (I)

Here, $\omega$ is the angular velocity.

Conclusion:

Refer Figure (1b).

Calculate the kinetic energy of the rod.

Substitute $100\text{kg}$ for $m$, $6\text{}\text{m}$ for $l$, $1\text{m}$ for $l_{OG}$ and $2\text{rad/s}$ for $\omega$ in Equation (I).

$\begin{array}{c}T=\frac{1}{2}\left[\frac{1}{12}\left(100\text{kg}\right){\left(6\text{m}\right)}^2+\left(100\text{kg}\right){\left(1\text{m}\right)}^2\right]{\left(2\text{rad/s}\right)}^2\\ =\frac{1}{2}\left(400\right)\left(4\right)\\ =800\text{J}\end{array}$

Thus, the kinetic energy of the rod is $800\text{J}$.

To determine

The kinetic energy of the object:

Answer to Problem 1PP

The kinetic energy of the disk is $1200\text{J}$.

Explanation of Solution

Given:

The mass of disk is $100\text{kg}$.

The angular velocity of the disk is $\omega =3\text{rad/s}$.

The radius of the disk is $2\text{m}$.

Draw the free body diagram of the disk as shown in Figure (1c).

Refer Figure (1c).

Write the formula for mass moment of inertia $\left(I\right)$ of the disk about it is center.

$I=\frac{1}{2}m{r}^2$

Here, $m$ is the mass and $r$ is the radius of the disk.

Write the formula for kinetic energy $\left(T\right)$.

$T=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

Here, $v=r\omega$.

Substitute $r\omega$ for $v$ and $\frac{1}{2}m{r}^2$ for $I$.

$T=\frac{1}{2}m{\left(r\omega \right)}^2+\frac{1}{2}\left(\frac{1}{2}m{r}^2\right){\omega }^2$ (I)

Here, $m$ is the mass, $r$ is radius, $I$ is the mass moment of inertia and $\omega$ is the angular velocity.

Conclusion:

Refer Figure (1c).

Calculate the kinetic energy of the disk.

Substitute $100\text{kg}$ for $m$, $2\text{}\text{m}$ for $r$ and $3\text{rad/s}$ for $\omega$ in Equation (I).

$\begin{array}{c}T=\left[\frac{1}{2}\left(100\text{kg}\right){\left(2\text{m}\times 2\text{rad/s}\right)}^2\right]+\frac{1}{2}\left[\frac{1}{2}\left(100\text{kg}\right){\left(2\text{m}\right)}^2\right]{\left(2\text{rad/s}\right)}^2\\ =800+400\\ =1200\text{J}\end{array}$

Thus, the kinetic energy of the disk is $1200\text{J}$.

To determine

The kinetic energy of the object:

Answer to Problem 1PP

The kinetic energy of the rod is $600\text{J}$.

Explanation of Solution

Given:

The mass of rod is $100\text{kg}$.

The angular velocity of the rod is $\omega =2\text{rad/s}$.

The length of the rod is $3\text{m}$.

Draw the free body diagram of the rod as shown in Figure (1d).

Refer Figure (1d).

Write the formula for mass moment of inertia $\left(I_O\right)$ of the rod having the rotation axis at one end.

$I_O=\frac{1}{3}m{l}^2$

Here, $m$ is the mass and $l$ is the length of the rod about mass center O.

Write the formula for kinetic energy $\left(T\right)$

(Rotation about fixed axis).

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

Substitute $\frac{1}{3}m{l}^2$ for $I_O$.

$T=\frac{1}{2}\left(\frac{1}{3}m{l}^2\right){\omega }^2$ (I)

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

Conclusion:

Refer Figure (1d).

Calculate the kinetic energy of the rod.

Substitute $100\text{kg}$ for $m$, $3\text{}\text{m}$ for $l$ and $2\text{rad/s}$ for $\omega$ in Equation (I).

$\begin{array}{c}T=\frac{1}{2}\left[\frac{1}{3}\left(100\text{kg}\right){\left(3\text{m}\right)}^2\right]{\left(2\text{rad/s}\right)}^2\\ =\frac{1}{2}\left(300\right)\left(4\right)\\ =600\text{J}\end{array}$

Thus, the kinetic energy of the rod is $600\text{J}$.

To determine

The kinetic energy of the object:

Answer to Problem 1PP

The kinetic energy of the disk is $4800\text{J}$.

Explanation of Solution

Given:

The mass of disk is $100\text{kg}$.

The angular velocity of the disk is $\omega =4\text{rad/s}$.

The radius of the disk is $2\text{m}$.

Draw the free body diagram of the disk as shown in Figure (1e).

Refer Figure (1e).

Write the formula for mass moment of inertia $\left(I\right)$ of the disk about it is center.

$I=\frac{1}{2}m{r}^2$

Here, $m$ is the mass and $r$ is the radius of the disk.

Write the formula for kinetic energy $\left(T\right)$.

$T=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

Here, $v=r\omega$.

Substitute $r\omega$ for $v$ and $\frac{1}{2}m{r}^2$ for $I$.

$T=\frac{1}{2}m{\left(r\omega \right)}^2+\frac{1}{2}\left(\frac{1}{2}m{r}^2\right){\omega }^2$ (I)

Here, $m$ is the mass, $r$ is radius, $I$ is the mass moment of inertia and $\omega$ is the angular velocity.

Conclusion:

Refer Figure (1e).

Calculate the kinetic energy of the disk.

Substitute $100\text{kg}$ for $m$, $2\text{}\text{m}$ for $r$ and $4\text{rad/s}$ for $\omega$ in Equation (I).

$\begin{array}{c}T=\left[\frac{1}{2}\left(100\text{kg}\right){\left(2\text{m}\times 4\text{rad/s}\right)}^2\right]+\frac{1}{2}\left[\frac{1}{2}\left(100\text{kg}\right){\left(2\text{m}\right)}^2\right]{\left(4\text{rad/s}\right)}^2\\ =3200+1600\\ =4800\text{J}\end{array}$

Thus, the kinetic energy of the disk is $4800\text{J}$.

To determine

The kinetic energy of the object:

Answer to Problem 1PP

The kinetic energy of the disk is $3200\text{J}$.

Explanation of Solution

Given:

The mass of rod is $100\text{kg}$.

The angular velocity is $\omega =4\text{rad/s}$.

Draw the free body diagram of the object as shown in Figure (1f).

Refer Figure (1f).

Here, the ends of the rod are connected to two rods of same length. Hence the rod travels in circular motion.

Consider as the mass travels in a radius of $r=2\text{m}$.

Write the formula for kinetic energy $\left(T\right)$

(Rotation about fixed axis).

$T=\frac{1}{2}m{v}^2$

Here, $v=r\omega$.

Substitute $r\omega$ for $v$.

$T=\frac{1}{2}m{\left(r\omega \right)}^2$ (I)

Here, $m$ is the mass, $r$ is radius and $\omega$ is the angular velocity.

Conclusion:

Refer Figure (1f).

Here $r=2\text{m}$.

Calculate the kinetic energy of the disk.

Substitute $100\text{kg}$ for $m$, $2\text{}\text{m}$ for $r$ and $4\text{rad/s}$ for $\omega$ in Equation (I).

$\begin{array}{c}T=\frac{1}{2}\left(100\text{kg}\right){\left[\left(4\text{rad/s}\right)\left(2\text{m}\right)\right]}^2\\ =50\left(64\right)\\ =3200\text{J}\end{array}$

Thus, the kinetic energy of the disk is $3200\text{J}$.