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

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* a_(T)`

`a_(T) = 0.35\ m/s^(2)`

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

`(mv^(2))/(R) = \mu mg`

Thus

`v = √(\mu gR) = √(0.90* 9.8* 25) = 14.849\ m/s`

Time taken by the car is given by:

v = v' + at

v' = initial velocity = 0

Thus

`t = (v)/(a) = (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* 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 = (d)/(d') = (314.99)/(2\pi * 25) = 2`