Question

A barbell spins around a pivot at its center (Figure 9.16). The barbell consists of two small balls, each with mass$800g$at the ends of a very low mass rod whose length is$35cm$. The barbell spins with angular speed$40rad/s$.Calculate$K_{rot}$.

Short answer

$\begin{array}{rcl}\frac{E_g}{t}& =& 1.785·{10}^{17}\frac{\text{J}}{\text{s}}\\ \frac{N}{tA}& =& 3.84·{10}^{21}{\text{s}}^{-1}{\text{m}}^{-2}\\ & & \end{array}$

Step-by-step solution

Given Data

The mass of each ball is$m=0.8\text{kg}=800\text{g}$

The angular speed of each ball is${\omega }_0=40\text{rad}/\text{s}$

The length of the rod is$35\text{cm}=0.35\text{m}$.

Concept of the rotational kinetic energy

The rotational kinetic energy is given by,

$\mathit{K}{\mathit{E}}_{\mathbf{\text{rot}}}\mathbf{=}\frac{{\mathbf{L}}_{\mathbf{\text{rot}}}^{\mathbf{2}}}{\mathbf{2}\mathbf{I}}$

Where${\mathit{L}}_{\mathbf{r}\mathbf{o}\mathbf{t}}$is magnitude of the rotational angular momentum andis $\mathit{I}\mathbf{}$moment of inertia.

Determine the work done

The intensity is defined as a power per unit area so we can calculate the energy of the green photons that falls on Earth in one second by multiplying the intensity by the cross section area of the Earth. We multiply it by the cross section area because when we look at the sphere we only see a circle.

$\begin{array}{rcl}\frac{E_g}{t}& =& I·A\\ & =& 1400\frac{\text{J}}{{\text{sm}}^2}·\pi {\left(6370·{10}^3\text{m}\right)}^2\\ & =& 1.785·{10}^{17}\frac{\text{J}}{\text{s}}\\ & & \end{array}$

The frequency of the green photons is :

$f=550\text{THz}$

The energy of the single photon is:

$\begin{array}{rcl}E& =& f·h=550·{10}^{12}\text{Hz}·6.626·{10}^{-34}{\text{Js}}_{\text{s}}\\ & =& 3.643·{10}^{-19}\text{J}\\ & & \end{array}$

The number of photons that fall on an area of one square meter in one second is:

$I=\frac{NE}{tA}/:E\frac{N}{tA}=\frac{I}{E}=\frac{1400\text{J}/{\text{sm}}^2}{3.643·{10}^{-19}\text{J}}=3.84·{10}^{21}{\text{s}}^{-1}{\text{m}}^{-2}$

Thereore,

$$\begin{gathered} \frac{E_g}{t}=1.785·{10}^{17}\frac{\text{J}}{\text{s}} \\ \frac{N}{tA}=3.84·{10}^{21}{\text{s}}^{-1}{\text{m}}^{-2} \end{gathered}$$