Answer :
Kinetic energy due to linear motion:
[tex]E= \frac{1}{2}mv^2= \frac{1}{2}3.7(3.2)^2=18.944J[/tex]
The angular velocity is found by
[tex]v=r \omega [/tex]
[tex]\omega = \frac{v}{r}= \frac{3.2}{0.48} =6.667rad/s[/tex]
Kinetic energy due to rotational motion:
[tex]E= \frac{1}{2}I \omega^2 = \frac{1}{2}0.340992(6.667)^2=7.5776J[/tex]
Then
[tex]E_{total}=18.944J+ 7.5776J=26.5216J[/tex]
[tex]E= \frac{1}{2}mv^2= \frac{1}{2}3.7(3.2)^2=18.944J[/tex]
The angular velocity is found by
[tex]v=r \omega [/tex]
[tex]\omega = \frac{v}{r}= \frac{3.2}{0.48} =6.667rad/s[/tex]
Kinetic energy due to rotational motion:
[tex]E= \frac{1}{2}I \omega^2 = \frac{1}{2}0.340992(6.667)^2=7.5776J[/tex]
Then
[tex]E_{total}=18.944J+ 7.5776J=26.5216J[/tex]