Respuesta :
Answer:
The minimum speed that the car can have without slipping is [tex]14.05\frac{m}{s}[/tex]
Explanation:
First it’s important to remark two things 1) If the car doesn’t round the curve with enough speed it will slip down the curve, so to avoid that the static friction between the car and the street should be upwards the incline (see Figure 1) 2) Because of the car is moving on a curve it’s describing a circular motion and in this kind of motion there is always a radial acceleration pointing towards the center of the circumference (see figure 2). Knowing that, we sketch our free-body diagram in Figure 1 and use Newton’s second law on the y-axis:
[tex]\sum F_{y}=a_{y}=0[/tex] (1)
(the acceleration on the y-axis is zero because if the car doesn't slip it is not moving in the y direction)
[tex]N_{y}-W+f_{sy}=0[/tex] (2)
With Ny = N*cos(28°) the y component of the normal force, W=mg the weight of the car and fsy=fs*sin(28°) the y component of static friction, the magnitude of the static friction (fs) is μs*N = 0.34*N, so fsy=0.34*sin(28°). Using that on equation (2)
[tex]N\cos(28\text{\textdegree})-mg+0.34N\sin(28\text{\textdegree})=0[/tex] (3)
Solving for N:
[tex]N=\frac{mg}{\cos(28\text{\textdegree})+0.34N\sin(28\text{\textdegree})}[/tex] (4)
Similar using Newton’s second law on the x-axis:
[tex]N_{x}-f_{sx}=ma_{x}[/tex] (5)
[tex]N_{x}-f_{sx}=ma_{rad}[/tex] (6)
[tex]N\sin(28\text{\textdegree})-0.34*N\cos(28\text{\textdegree})=ma_{rad}[/tex] (7)
[tex]N(\sin(28\text{\textdegree})-0.34\cos(28\text{\textdegree}))==ma_{rad}[/tex] (8)
Using the fact that [tex]a_{rad}=\frac{v^{2}}{R}=\frac{v^{2}}{124m}[/tex] and using (4) in (8):
[tex]\frac{\cancel{m}g}{\cos(28\text{\textdegree})+0.34N\sin(28\text{\textdegree})}(\sin(28\text{\textdegree})-0.34\cos(28\text{\textdegree}))=\cancel{m}\frac{v^{2}}{124m}[/tex] (9)
Solving for v with [tex]g=9.81\frac{m}{s^{2}}[/tex]:
[tex]v=\sqrt{\frac{(\sin(28\text{\textdegree})-0.34\cos(28\text{\textdegree}))*9.81*124}{\cos(28\text{\textdegree})+0.34N\sin(28\text{\textdegree})}}\simeq14.05\frac{m}{s}[/tex] (10)
That is the minimum speed that the car can have without slipping.
Radius of curvature of car is the radius of circular path. The minimum speed that the car can have without slipping is 14.05 meter per second.
What is radius of curvature?
Radius of curvature is the radius of circle, which is made by the body travelling in the circular path.
Given information-
The radius of curvature of the road is 124 m
The banking angle is 28 degrees.
The coefficient of static friction is 0.34.
The net force acting on the body in the y axis is,
[tex]\sum F=N\cos(28^o)-mg+0.34N\sin (28^o)=0\\N=\dfrac{mg}{\cos(28)+0.34\sin(28)}[/tex]
Let the above equation is equation 1.
The net force acting on the body in the x axis by the newtons seconds law of motion is,
[tex]\sum F=N\sin(28^o)-0.34N\cos (28^o)=ma_{rad}[/tex]
Let the above equation is equation 2.
Now the value of [tex]a_{rad}[/tex] is,
[tex]a_{rad}=\dfrac{v^2}{124}[/tex]
Put the value of [tex]a_{rad}[/tex] and [tex]N[/tex] from equation 1 in the equation 2. Thus,
[tex]\dfrac{m\times9.8}{\cos(28^o)-0.34N\sin(28^o)}\times\sin(28^o)-0.34N\cos (28^o)=m\dfrac{v^2}{124}\\v \cong14.05 \rm m/s\\[/tex]
Thus the minimum speed that the car can have without slipping is 14.05 meter per second.
Learn more about the radius of curvature here;
https://brainly.com/question/24116470
