Question

A 1500 kg car drives around a flat, 50 m diameter track, starting from rest. The drive wheels supply a small but steady 525 N force in a forward direction. The coefficient of static friction between the tires and the road is 0.90. How many revolutions of the track have been made when the car slides out?

Answer 1

Answer:

No. of revolutions before the car slides off is 2

Solution:

As per the question:

Mass of the car, m = 1500 kg

Diameter of the track, d = 50 m

Force, F = 525 N

Coefficient of static friction, $\mu_{s} = 0.90$

Now,

The acceleration in the tangential direction is given by:

$F = ma_{T}$

$525 = 1500\times a_{T}$

$a_{T} = 0.35\ m/s^{2}$

Here, the centripetal force is given by the friction force:

$\frac{mv^{2}}{R} = \mu mg$

Thus

$v = \sqrt{\mu gR} = \sqrt{0.90\times 9.8\times 25} = 14.849\ m/s$

Time taken by the car is given by:

v = v' + at

v' = initial velocity = 0

Thus

$t = \frac{v}{a} = \frac{14.849}{0.35} = 42.426\ s$

Total Distance covered, d is given by:

$v^{2} = v'^{2} + 2ad$

$14.849^{2} = 0^{2} + 2\times 0.35d$

d = 341.99 m

Distance covered in 1 revolution is the circumference of the circle, d' = $2\pi R$

Now, the no. of revolutions is given by:

$n = \frac{d}{d'} = \frac{314.99}{2\pi \times 25} = 2$