Answer:
[tex]\frac{KE_{Rotational}}{KE_{Total}} = 0.018[/tex]
Explanation:
To develop this exercise we proceed to use the kinetic energy equations,
In the end we replace
[tex]KE_{Total}=KE_{Translational}+KE_{Rotational}[/tex]
[tex]KE_{Total}=\frac{1}{2}m_{car}+4*\frac{1}{2}*I*(\frac{v}{r})^2[/tex]
Here
[tex]I=\frac{1}{2}m_{wheels}*r^2[/tex] meaning the 4 wheels,
So replacing
[tex]KE_{Rotational}=4\frac{1}{2}*(\frac{1}{2}m_{wheels}*r^2)*(\frac{v}{r})^2=m*v^2[/tex]
So,
[tex]\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}*v^2}{\frac{1}{2}m_{car}*v^2+m_{wheels}*v^2}[/tex]
[tex]\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}}{\frac{1}{2}m_{car}+m_{wheels}}[/tex]
[tex]\frac{KE_{Rotational}}{KE_{Total}} = \frac{10}{545+10}[/tex]
[tex]\frac{KE_{Rotational}}{KE_{Total}} = 0.018[/tex]
