Respuesta :
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, [tex]\mu_{s} = 0.90[/tex]
Now,
The acceleration in the tangential direction is given by:
[tex]F = ma_{T}[/tex]
[tex]525 = 1500\times a_{T}[/tex]
[tex]a_{T} = 0.35\ m/s^{2}[/tex]
Here, the centripetal force is given by the friction force:
[tex]\frac{mv^{2}}{R} = \mu mg[/tex]
Thus
[tex]v = \sqrt{\mu gR} = \sqrt{0.90\times 9.8\times 25} = 14.849\ m/s[/tex]
Time taken by the car is given by:
v = v' + at
v' = initial velocity = 0
Thus
[tex]t = \frac{v}{a} = \frac{14.849}{0.35} = 42.426\ s[/tex]
Total Distance covered, d is given by:
[tex]v^{2} = v'^{2} + 2ad[/tex]
[tex]14.849^{2} = 0^{2} + 2\times 0.35d[/tex]
d = 341.99 m
Distance covered in 1 revolution is the circumference of the circle, d' = [tex]2\pi R[/tex]
Now, the no. of revolutions is given by:
[tex]n = \frac{d}{d'} = \frac{314.99}{2\pi \times 25} = 2[/tex]
