An automobile has a total mass of 1300 kg. It accelerates from rest to 65 km/h in 14 s. Assume each wheel is a uniform 30 kg disk. Find, for the end of the 14 s interval, (a) the rotational kinetic energy of each wheel about its axle, (b) the total kinetic energy of each wheel, and (c) the total kinetic energy of the automobile

Question

irishboy123 · Answer

start like this, for one wheel....

$T_{rot} = \dfrac{1}{2} I \omega^2 =  \dfrac{1}{2} \left( \dfrac{1}{2}mr^2\right) \left( \dfrac{v}{r} \right)^2$

you don't need the wheel's radius

michele_laino · Answer

I think that the kinetic $KE$ energy of a single wheel can be written like below:
$$\huge \begin{gathered}
  KE = \frac{1}{2}I{\omega ^2} + \frac{1}{2}m{v^2} =  \hfill \
   \hfill \
   = \frac{1}{2}I{\left( {\frac{v}{R}} \right)^2} + \frac{1}{2}m{v^2} =  \hfill \
   \hfill \
   = \frac{1}{2}\left( {\frac{I}{{{R^2}}} + m} \right){v^2} \hfill \ 
\end{gathered} $$
since, with respect to an inertila system whichis at rest with respect to the road, such wheel has a kinetic energy due to its rotational motion, plus a kinetic energy due to its traslational motion. Here $R$ is the radius of the wheel

anonymous · Answer

Is my anwer correct? a. 2445.0352 J  b. 7335.1056 J c . 221683.1901 J?