Question

1. A car travels at constant speed around a horizontal circular corner of radius 5 m. n (a) Given that the car just starts to skid if its speed is 12 km/h, find the frictional force acting on the car. (b) Assuming the same frictional force is acting, calculate the car's smallest possible turning radius if the speed is 30 km/h. (c) Calculate the turning radius for the car travelling at 12 km/h in wet conditions where the frictional force is halved.

Answer 1

1(a):-Speed of car v=12 km/h=12x 5/18=3.33m/s

frictional force acting on the car at the will be equal to the centripetal force required to rotate the car at the circular corner i.e.

      frictional force;

$f_{s}=\frac{mv^{2}}{r}$

$\Rightarrow f_{s}=\frac{m\times (3.33)^{2}}{5}$

$\mathbf{\Rightarrow f_{s}=2.22m}$ newton

(where m is the mass of car)

---------------------------------------------------------------------------------------------------------------

(b) Now velocity of car v=30km/h=30 x 5/18m/s=8.33m/s

$f_{s}=\frac{mv^{2}}{r}$

since frictional force is same as in part (a) so-

$\mathbf{\Rightarrow 2.22m=\frac{m\times (8.33)^{2}}{r}}$

$\mathbf{\Rightarrow r=\frac{(8.33)^{2}}{2.22}}$

$\mathbf{\Rightarrow r=31.26meter}$

-----------------------------------------------------------------------------------------------------------

(C)Now speed v=12km/h=12x5/18m/s=3.33m/s

Now friction force is halved so new frictional force-

$f_{s}'=\frac{f_{s}}{2}$

$f_{s}'=\frac{2.22m}{2}=1.11m$

Now using the equation-

$f_{s}'=\frac{mv^{2}}{r}$

$\Rightarrow r=\frac{mv^{2}}{f_{s}'}$

$\Rightarrow r=\frac{m\times (3.33)^{2}}{1.11m}$

$\mathbf{\Rightarrow r=9.99meter\approx 10 meter}$

my2

m x (3.33)2

2.221m

my2

2.22 mx (8.33)

(8.33)- 2.33

222m 1.11m

mu2

my2

m x (3.33)2 1.11m

We were unable to transcribe this image