Chapter 11.1, Problem 4PP

To determine

Translational and rotational kinetic energy.

Answer to Problem 4PP

Translational and rotational both kinetic energies are 225 J.

Explanation of Solution

Given:

Mass of wheel, $m=2\text{ kg}$

Linear speed ofwheel, $v=15\text{ m/s}$

Formula used:

  $K{E}_{trans}=\frac{1}{2}m{v}^2$

Here, $K{E}_{trans}$ is translational kinetic energy, m is mass and vis linear speed.

  $K{E}_{rot}=\frac{1}{2}I{\omega }^2$

Here, $K{E}_{rot}$ is rotational kinetic energy, I is moment of inertia and $\omega$ is angular velocity.

Calculation:

  $\begin{array}{l}K{E}_{trans}=\frac{1}{2}m{v}^2\\ =\frac{1}{2}\times 2\times {15}^2\\ \text{=225 J}\end{array}$

Moment of inertia of wheel is equal to,

  $I=m{r}^2$ …..(2)

And angular velocity is equal to,

  $v=r\omega \mathrm{....}\left(3\right)$

Now, putting the value of equation 2 in equation 1 and using equation 3:

  $\begin{array}{l}K{E}_{rot}=\frac{1}{2}\left(m{r}^2\right){\omega }^2\\ =\frac{1}{2}m{\left(r\omega \right)}^2\\ =\frac{1}{2}m{v}^2\\ =\frac{1}{2}\times 2\times {15}^2\\ =225\text{ J}\end{array}$

Conclusion:

Therefore, translational kinetic energy and rotational kinetic energy both are 225 J