Question 12.7: A car of mass M is driven by a motor that performs the torqu...
A car of mass M is driven by a motor that performs the torque 2D on the wheel axis. The radius of the wheels is R, and their moment of inertia is Θ= mR² (m is the reduced mass of the wheels).
(a) Determine the friction force f which acts on each wheel and causes the acceleration of the car. The street is assumed to be planar.
(b) Calculate the acceleration of the car if the torque 2D = 10³ J, M = 2 · 10³ kg, R = 0.5 m and m = 12.5 kg.
(a) Fig. 12.16 shows one of the wheels and the force f acting on it. Since the linear acceleration of the wheel center is the same as that of the center of gravity of the car a_{s} , one has
Ma_{s} = 2f−F. (12.48)
The factor 2 accounts for the fact that a car in general is driven by two wheels. F is a possible external force which impedes the motion (air resistance), and a_{s} is the acceleration of the car. For the torque relative to the axis, one obtains
4Θ\dot{ω} = 2(D −fR), (12.49)
where Θ is the moment of inertia of each of the four wheels, D is the accelerating torque, and −fR is the torque performed by the friction force on each wheel. The moment of inertia is Θ= mR². If the car does not glide, one has
\dot{ω} R = a_{s} , (12.50)
and with (12.49) and (12.48), it immediately follows that
\dot{ω} R = a_{s} = \frac{1}{2} \left( \frac{DR −fR^{2}}{mR^{2}}\right)=\frac{2f −F}{M}. (12.51)
Equation (12.51) finally yields for the friction force
f = \frac{1}{2} \frac{(2D/R)M +4mF}{M +4m} (12.52)
and by neglecting the backdriving force F (F = 0), we have
f = \frac{D/R}{1+(4m/M)}. (12.53)
(b) By replacing f in (12.49) by (12.53) and solving for a_{s} (F = 0), one finds the acceleration of the car
a_{s} = \frac{2D/R}{M +4m}= \frac{10^{3}/0.5}{2 · 10^{3} + 4 · 12.4}= \frac{10^{3}}{1025}≈ 1 \frac{m}{s^{2}}.With the numerical values from (b), the friction force f is given by
f ≈ \frac{D}{R}= 1000 N.