Question

A track consists of two circular parts ABC and CDE of equal radius 100 m and joined smoothly as shown in figure. 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/hr on the track. 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 = 10 m/s²

• A.
• B.
• C. 1.037
• D. 0.732

Answer 1

The correct option is C

1.037

So if the cycle should not accelerate or to maintain constant speed its net force in tangential direction has to be 0
$\Rightarrow mgsin\theta =fr$
To ensure this maximum friction is needed ${\mu }_{s}N=mgsin\theta$
Also here mg cos$\theta -N=\frac{m{v}^2}{R}$
$\Rightarrow N=mgcos\theta -\frac{m{v}^2}{R}$
To find out minimum desired coefficient of friction, we have to consider a point just before 'C' (where N is minimum)
$A{t}^{\prime }{C}^{\prime },mgsin\theta =f\Rightarrow f=1000\times \frac{1}{\sqrt{2}}=707N$
Before 'C', $mgcos\theta -N=\frac{m{v}^2}{R}\Rightarrow N=mgcos\theta -\frac{m{v}^2}{R}=683N$
$\therefore Asf=mgsin\theta$
$\Rightarrow \mu N=mgsin\theta$
$\mu \left(683\right)=707$
$\Rightarrow \mu =1.037$