Answer:
[tex]\frac{K_T}{K_R}=\frac{600J}{188.63J}=3.18[/tex]
Explanation:
To find the rotational kinetic energy you first calculate the angular acceleration by using the following formula:
[tex]\tau=I\alpha\\\\FR=I\alpha\\\\\alpha=\frac{FR}{I}[/tex]
F: force applied
R: radius of the wheel
I: moment of inertia
[tex]\alpha=\frac{(120N)(0.80m)}{26.88kgm^2}=3.57\frac{rad}{s^2}[/tex]
With this value you calculate the angular velocity:
[tex]\omega^2=\omega_o^2+2\alpha \theta\\[/tex]
you calculate how many radians the wheel run in 5.0m
[tex]\theta=\frac{2\pi (0.8m)}{5.0m}=\approx1.00rad[/tex]
[tex]\omega=\sqrt{2(3.57\frac{rad}{s^2})(1.00rad)}=2.67\frac{rad}{s}[/tex]
Next, you use the formula for the rotational kinetic energy:
[tex]K_R=\frac{1}{2}I\omega^2=(26.88)(2.67)^2 J = 188.63J[/tex]
For the transnational kinetic energy you use the following equation:
[tex]W=\Delta K_T[/tex] (net work equals the change in the kinetic energy).
By replacing the you obtain:
[tex]\Delta K_T=Fd=(120)(5.0)J=600J[/tex]
Finally, the ratio between translational rotational kinetic energy is:
[tex]\frac{K_T}{K_R}=\frac{600J}{188.63J}=3.18[/tex]
hence, translational kinetic energy is three times the rotational kinetic energy.
