# Car problem-rotational motion?

• Newton=boss
In summary, the total kinetic energy of a car weighing 1900 kg, with four tires each weighing 40 kg and having a diameter of 0.80 m, traveling at 60 km/h can be found by using the formula K=1/2 mv^2. To determine the fraction of kinetic energy in the tires and wheels, the rotational kinetic energy stored in the rotation of the wheels must also be taken into account, using the formula K=1/2 Iω^2. If the car is initially at rest and is pulled by a tow truck with a force of 2500 N, the acceleration of the car can be found using the formula F=ma, ignoring frictional losses. If the rotational inertia of

## Homework Statement

The 1900 kg mass of a car includes four tires, each of mass (including wheels) 40 kg and diameter 0.80 m. Assume each tire and wheel combination acts as a solid cylinder.
(a) Determine the total kinetic energy of the car when traveling 60 km/h.

(b) Determine the fraction of the kinetic energy in the tires and wheels.
%

(c) If the car is initially at rest and is then pulled by a tow truck with a force of 2500 N, what is the acceleration of the car? Ignore frictional losses.

(d) What percent error would you make in part (c) if you ignored the rotational inertia of the tires and wheels?

## Homework Equations

I would understand where to go in step two. For step 1 i was using K=1/2 mv^2
For step 3 i used F=ma. Apparantly I am totally wrong.

## The Attempt at a Solution

Help. The formulas I use are giving me incorrect results. about 200,000 J is what I got

You're forgetting the rotational kinetic energy stored in the rotation of the wheels. In addition to the $$\frac{mv^2}{2}$$ there should be a $$\frac{I\omega^2}{2}$$ term.

the rotational kinetic energy can be found from
I=1/2*40*4*.4^2?
and
60*1000/3600=v
omega=v/0.4?

check that. I was able to solve part a)..thankyou!
I do not even know how to start on the next 3 parts..if you could point me where to go I would be indebted.

wouldnt b simply be to divide the rotational Kinetic energy by the total kinetic energy? apparently not..