A track consists of two circular pars ABC andCDE of equal rdius 100 m and joined smoothly as shown in figure.Each part sutends a right ngle at its centre. A cycle weighing 100 kg together with rider travels at a constant speed of 18 km/h on the track. A. Findteh nromal contct force by tehroad on the cycle whenit is at B and at D. b.Findteh 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 teh road and teh 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 teh cyclist can move with constant speed? Take `g=10 m/s^2`

Explanation

`v = 18 kmh^(-2) = (1800)/(60 xx 60) = 5ms^(-1)`
a. At point B verious forces acting on the cyclist are
i Weight , `mg` vertically downwards
ii Normal force , `N_(B)` by the road upward
Since under the action of these two forces the cyclist moves in a circular path of radius r from the dynarrics of circular motion .
`(mv^(2))/(r ) = mg - N_(B)`
` rArr N_(B) = m(g - (v^(2))/(r )) = 100 ( 10- (5^(2))/(100)) = 975 N`
At point D various force acting on the cyclist are
i. Weight , `mg` vertically downwards
ii Normal force , `N_(D)` by the road upward
`N_(D) - mg = (mv^(2))/(r )`
` rArr N_(D) = mg (g +(v^(2))/(r )) = 100 (10+ (5^(2))/(100)) = 1025 N`
b. At point B and D. the tracks are almost horizontal therefore , there is no component of g along the track at those point that will changs the speed of the cyclist since the cyclist moves with a constant speed , the friction force the at these two points is neccessarily zero.
At point C however of `mg` along the track that tends to accelerate the motion of the cyclist ` = mg sin appha`
But since his speed remainn constant , a force equal and opposite to `mg sin alpha` must be acting on it and this force is the force of friction `mu N_(c)` where `mu` is the friction coefficient between the road and the type and `N_(c)` is the normal reaction at point `C`

Hence friction force at C is
`mu N_(c) = mg sin alpha = 100 xx 10 xx sin 45^(@) = 707.12 N`
c. Just before the cyclist crosses point C varius force acting on him are
i. His weight, `mg`, vertically downwards, and
ii. Normal force `N_(b)` of the track
Hence ,from the dynamics of circular motion,we have
`mg cos alpha - N_(b) = (mv^(2))/(r )`
` N_(b) = m( g cos alpha - (v^(2))/(r )) = 682.11 N`
Just after the cyclist crosses the point C varius forces acting on him are
i. Weight, `mg`, vertically downwards, and
ii. Normal force `N_(d)` of the track
Hence ,from the dynamics of circular motion ,
`N_(d) mg cos alpha = (mv^(2))/(r )`
` rArr N_(d) = m( g cos alpha + (v^(2))/(r )) = 7.32.11 N`
d. The tendecy of the cyclist to skid is maximum just before the point C , the point is so because the carve is steepest at this point , Force along the track at this point `= mg sin alpha`
And friction force `= mu N_(b)` where `N_(a)` is the normal force of of the track just before the point C
The cyclist can move with constant speed if the two forces are equal and opposite
`mu N_(b) = mg sin alpha rArr mu = 1.037`