91Ó°ÊÓ

A rod of mass is $m=1.2kg$, then the angular speed is${\mathrm{Ó\neg }}_0=50\text{rad}/\text{s}$ and Length of rotation is .$0.7m$

Rotational kinetic energy $\left({\text{K}}_{\text{rot}}\right)$and replacing the moment of inertia$\left(I_{\text{rod}}\right)$ for the appropriate expression for a rod.

\({{\rm{K}}_{{\rm{rot}}}} = \frac{1}{2}{{\rm{I}}_{{\rm{rod}}}}{\omega ^2}\)

${\mathbf{\text{I}}}_{\mathbf{\text{rod}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{12}}{\mathbf{\text{ML}}}^{\mathbf{2}}$

(a)

Let us know the definition of rotational kinetic energy$\left(K_{rot}\right)$ and replacing the moment of inertia$\left(I_{rot}\right)$for the appropriate expression for a rod.

$\begin{array}{rcl}{\text{K}}_{\text{rot}}& =& \frac{1}{2}{\text{I}}_{\text{rod}}{\mathrm{Ó\neg }}^2\\ {\text{I}}_{\text{rod}}& =& \frac{1}{12}{\text{ML}}^2\end{array}$

Finally, we calculate $\left(K_{rot}\right)$ with the given values

$\begin{array}{rcl}& ⇒& {\text{K}}_{\text{rot}}=\frac{1}{2}\left(\frac{1}{12}{\text{ML}}^2\right){\mathrm{Ó\neg }}^2\\ & =& \frac{1}{24}{M}^2{\mathrm{Ó\neg }}^2\\ & =& \frac{1}{24}(1.2\text{kg})(0.7\text{m}{)}^2{\left(50\frac{\text{rad}}{\text{s}}\right)}^2\\ & =& 61.3\text{J}\\ & & \end{array}$

(b)

Substitute the formula with the values.

$K=\frac{1}{2}m{v}^2=\frac{1}{2}(1.2\text{kg})(8\text{m}/\text{s}{)}^2=38.4\text{J}$

Plug in the values to the second formula.

$K_{\text{tot}}=K_{\text{rot}}+K_{\text{trans}}=61.25\text{J}+38.4\text{J}=99.7\text{J}$

The total kinetic energy is, $99.7J$.