A solid cylindrical disk whose mass is 1.4 kg and radius is 8.5 cm, rolls across a horizontal table at a speed of 15 cm/...

Given:

Mass of the disk is $1.4kg$.

Radius of the disk is $r=8.5cm=8.5\times {10}^{-2}m$.

Rolls across a horizontal table at speed ${\stackrel{\rightharpoonup }{v}}_{com}=15\raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$s=15\times {10}^{-2}{\displaystyle \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}$}\right.$.

Calculation:

The instantaneous velocity at the top of the rolling cylindrical disk is given as:

$\stackrel{\rightharpoonup }{v}=2{\stackrel{\rightharpoonup }{v}}_{com}$

Here, ${\stackrel{\rightharpoonup }{v}}_{com}$ is velocity of center of mass given as $15\times {10}^{-2}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$. So,

$$\begin{gathered} \stackrel{\rightharpoonup }{v}=2\left(15\times {10}^{-2}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right) \\ \stackrel{\rightharpoonup }{v}=0.30\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.=30\raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Thus, the instantaneous velocity of the top of the rolling cylinder is $30\raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

The angular speed of the rolling disk is given as:

$\omega =\frac{v_{com}}{r}$

Substitute  for $8.5cm$ for $r$ and $15\raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ for ${\stackrel{\rightharpoonup }{v}}_{com}$.

$$\begin{gathered} \omega =\frac{15{\displaystyle \raisebox{1ex}{$cm$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}{8.5cm} \\ \omega =1.7647\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Thus, the angular speed of the rolling disk is $1.7647\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

The kinetic energy for the rolling disk is given as:

$$\begin{gathered} KE=\frac{1}{2}{I}_{com}{\omega }^2+\frac{1}{2}m{r}^2{\omega }^2 \\ KE=\frac{1}{2}{I}_{com}{\omega }^2+\frac{1}{2}m{v}_{com}^2 \left(v_{com}=\omega r\right) \end{gathered}$$

Here, $I_{com}$ is moment of inertia about center of mass given as:

$$\begin{gathered} I_{com}=\frac{1}{2}m{r}^2 \\ {I}_{com}=\frac{1}{2}\left(1.4kg\right){\left(8.5\times {10}^{-2}m\right)}^2 \\ {I}_{com}=5.0575\times {10}^{-3}kg m^2 \end{gathered}$$

Now, kinetic energy:

$$\begin{gathered} KE=\frac{1}{2}\left(5.0575\times {10}^{-3}kg m^2\right){\left(1.7647\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^2+\frac{1}{2}\left(1.4kg\right){\left(15\times {10}^{-2}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^2 \\ KE=\left(7.8749\times {10}^{-3}+15.75\times {10}^{-3}\right)J \\ KE=23.62\times {10}^{-3}J \end{gathered}$$

Thus, the kinetic energy of the rolling disk is $23.62\times {10}^{-3}J$.