Respuesta :
Answer:
053.
Explanation:
Given that the radius of curvature of the path, R = 75 m.
Speed of the car on that path , [tex]v_1 = 35 m/s[/tex]
The centripetal force,[tex]F_c[/tex] acting on the body having mass, m, when it moves with the velocity v on curved path having radiusR
Ris[tex]F_c = mv^2/R[/tex]
Gravitational force, [tex]F_g[/tex] = mg.
Let tha angle of superelevation is [tex]\theta.[/tex]
As the car does not skid even with zero friction, so
[tex]mg\sin\theta = (mv_1^2)/2 \cos\theta[/tex]
[tex]\Rightarrow \tan\theta = v_1^2/2g=\cdots(i)[/tex]
On sunny day, let the minimun static friction coefficient between the wheels and the pavement is [tex]\mu.[/tex]
As [tex]v_2[/tex] = 118 m/s is greater than v_so the car tends to skid in upper direction and the frictional
force,f, will acts is downward direction.
As there is no skidding, so
[tex]f+ mg\sin\theta= (m(v_2)^2)/R\cos\theta[/tex]
[tex]\Rightarrow f=(m(v_2)^2)/R\cos\theta - mg\sin\theta[/tex]
where [tex]f= \mu N.[/tex]
[tex]So, \mu = \frac {(m(v_2)^2)/R\cos\theta - mg\sin\theta}{N} \cdots(ii)[/tex]
Where N is the normal reaction can be determined by balancing the force in perpendicular direction of the plane.
[tex]N= mg\cos\theta+\frac {m(v_2)^2}{R}\sin\theta[/tex]
From equation (ii)
[tex]\mu = \frac {(m(v_2)^2)/R\cos\theta - mg\sin\theta}{mg\cos\theta+\frac {m(v_2)^2}{R}\sin\theta}[/tex]
[tex]=\frac{-g\tan\theta+v^2/R}{v_2^2\tan\theta+g}[/tex]
[tex]= \frac {(m(v_2)^2)/R\cos\theta - mg\sin\theta}{mg\cos\theta+\frac {m(v_2)^2}{R}\sin\theta}[/tex]
[tex]=\frac{-g(v_1^2/2g)+v^2/R}{v_2^2(v_1^2/2g)+g}[/tex]
[tex]= \frac{v_2^2/R-v_1^2/R}{g+v_2^2/R\times v_1^2/Rg}[/tex]
[tex]=\frac{118^2/75-35^2/75}{9.81+118^2/75\times 35^2/(75\times 9.81)}[/tex]
=0.53
Hence, the minimum coefficient of friction is 0.53.
