Question

If an object has a moment of inertia $\mathbf{19}\mathbf{}\mathit{k}\mathit{g}\mathbf{·}{\mathit{m}}^{\mathbf{2}}$ and rotates with an angular speed of $\mathbf{70}\mathbf{}\mathit{r}\mathit{a}\mathit{d}\mathbf{/}\mathit{s}$, what is its rotational kinetic energy?

Short answer

The rotational kinetic energy is$4.6\times {10}^{4}J$

Step-by-step solution

Identification of the given data

• The moment of inertia is $I=19kg·m^2$
• An angular speed of $\omega =70rad/s$

Concept of rotational kinetic energy

The rotational kinetic energy means the kinetic energy due to the rotation of a System the system.

Determination of the rotational kinetic energy

The rotational kinetic energy,

${\mathit{K}}_{\mathbf{r}\mathbf{o}\mathbf{t}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{I}{\mathit{\omega }}^{\mathbf{2}}$

Substitute $\mathit{I}$and $\omega$in the above equation, we get:

$$\begin{gathered} K_{rot}=\frac{\left(19kg·m^2\right)}{2}\times {\left(70rad/s\right)}^2 \\ =\left(\frac{19\times 70\times 70}{2}\right)·\left(1kg·m^2\times {\left(1rad/s\right)}^2\times \frac{1J}{1kg·m^2·ra{d}^2·s^{-2}}\right) \\ =4.6\times {10}^4·\left(1J\right) \\ =4.6\times {10}^{4}J \end{gathered}$$

Hence, the rotational kinetic energy is $4.6\times {10}^{4}J$