A 4-wheel drive car of mass M = 2100 kg accelerates from rest to 100 km/h in 3:4
ID: 1370604 • Letter: A
Question
A 4-wheel drive car of mass M = 2100 kg accelerates from rest to 100 km/h
in 3:40 seconds, and we will assume that the acceleration is constant. In the
following, ignore resistance from the air and the rolling friction of the wheels
against the surface. The wheels are rolling without sliding, and they have
a diameter of D = 50:0 cm. Each wheel has a mass of m = 18:0 kg, and
has a moment of inertia around their centre of mass corresponding to a disc,
I = 1=2mR2. The mass of the wheels is included in the total 2100 kg. The
wheels are taken to be identical and carry the same amount of weight. The
gravitational constant is g = 9:80 m/s2.
a) What is the required torque that the engine has to provide for each wheel,
to have this acceleration? Provide both the algebraic expression, and the
numerical result.
b) What is the required coefficient of static friction between road and wheel,
to avoid spinning? Provide both the algebraic expression, and the numerical
result.
c) What is the total kinetic energy of the car as it reaches 100 km/h?
Explanation / Answer
Answer:
a) a=(vf-vi)/(tf-ti)
= (1000*1000)/(3600*3.40)
=8.17 m/s^2
F=ma=2100 kg ×8.17 m/s^2 = 17157 N
F=mr
=F/mR= 17157 N / 0.25 m *2100 kg
= 32.68 rad/s^2
= I
=1/2mR^2*
=0.5×2100 kg ×(0.25 m)^2 * 32.68 rad/s^2
= 2144.63 N*m
/4=536.16 N*m
b) Coefficient of static friction between road and wheel
=a/g
=8.17 m/s^2 / 9.8 m/s^2
= 0.834
c) K=1/2Mv^2
K= 0.5×2100×(1000*1000/3600)^2
K = 29166.7 J