Answer :
Answer:
24.3 degrees
Explanation:
A car traveling in circular motion at linear speed v = 12.8 m/s around a circle of radius r = 37 m is subjected to a centripetal acceleration:
[tex]a_c = \frac{v^2}{r} = \frac{12.8^2}{37} = 4.43 m/s^2[/tex]
Let α be the banked angle, as α > 0, the outward centripetal acceleration vector is split into 2 components, 1 parallel and the other perpendicular to the road. The one that is parallel has a magnitude of 4.43cosα and is the one that would make the car slip.
Similarly, gravitational acceleration g is split into 2 component, one parallel and the other perpendicular to the road surface. The one that is parallel has a magnitude of gsinα and is the one that keeps the car from slipping outward.
So [tex] gsin\alpha = 4.43cos\alpha[/tex]
[tex]\frac{sin\alpha}{cos\alpha} = \frac{4.43}{g} = \frac{4.43}{9.81} = 0.451[/tex]
[tex]tan\alpha = 0.451[/tex]
[tex]\alpha = tan^{-1}0.451 = 0.424 rad = 0.424*180/\pi \approx 24.3^0[/tex]
Using Newton's second law and free-body diagram the angle with which the curve is to be banked is obtained as [tex]24.31^\circ[/tex].
Newton's Second Law of Motion
This problem can be analysed using a free-body diagram.
The acceleration in the horizontal direction (radial diretion) is the centripetal acceleration.
Applying Newton's second law of motion in the x-direction, we get;
[tex]\sum F = ma[/tex]
[tex]\implies N\,sin\, \theta =\frac{mv^2}{r}[/tex]
Now, applying Newton's second law of motion in the y-direction, we get;
[tex]\implies N\,cos \, \theta=mg[/tex]
Dividing both the equations, we get;
[tex]\frac{N\,sin\, \theta}{N\,cos\, \theta} =\frac{mv^2}{r\, mg}[/tex]
[tex]\implies tan\theta=\frac{v^2}{rg}=\frac{12.8^2}{37\times9.8} =0.4518[/tex]
[tex]\implies \theta=tan^{-1}(0.4518)=24.31^\circ[/tex]
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