Question

A track consists of two circular parts ABC and CDE of equal radius 100 m and joined smoothly as shown in figure (7-E1). Each part subtends a right angle at its centre.A cycle weighing 100 kg together with the rider travels at a constant speed of 18 km/h on the track. (a) Find the normal contact force by the road on the cycle when it is at B and at D. (b) Find the force of friction exerted by the track on the tyres when the cycle is at B, C and D. (c) Find the normal force between the road and the cycle just before and just after the cycle crosses C. (d) What should be the minimum friction coefficient between the road and the tyre, which will ensure that the cyclist can move with constant speed ? Take $g=10m/s^2.$

Answer 1

Radius of the curves = 100 m

Weight = 100 kg

Velocity = 18 km/hr = 5 m/sec

(a) At B, $mg-\frac{m{v}^2}{r}=N$

$\Rightarrow N=(100\times 10)-(100\times \frac{25}{100})$

$=1000-25=975N$

At D, $N=mg+\frac{m{v}^2}{R}$

$=1000+25=1025N$

(b) At B & D, the cycle has no tendency to slide.

So at B & D, frictional force is zero.

$A{t}^{\prime }{C}^{\prime },mgsin\theta =f$

$\Rightarrow 1000\times \left(\frac{1}{\sqrt{2}}\right)=707N$

${}^{\prime }{C}^{\prime }mgcog\theta -N=\frac{m{v}^2}{r}$

$\Rightarrow N=mgcos\theta -\frac{m{v}^2}{r}$

$=707-25=682N$

$\left(ii\right)N-mgcos\theta =\frac{m{v}^2}{r}$

$\Rightarrow N=\frac{m{v}^2}{r}+mgcos\theta$

$=25+707=732N$

(d) To find out the minimum desired coefficient of friction, we have to consider a point just before c (where N is minimum).

Now $\mu N=mgsin\theta$

$\Rightarrow \mu \times 682=707$

So $\mu =1,037$