<p>I need to derive an equation for max. speed of a car around a banked curve wi

Question

<p>I need to derive an equation for max. speed of a car around a banked curve with friction. I need a free body design. I think I have this part right. I need to split the vectors into components and solve for speed around the curve. Once I get this equation I am to test it by letting the coefficient of friction = 0. This gives me the un-banked equation I think. I then need to set the angle = 0 and test to see if I get one these results.</p>
<p>1).show when &#952; becomes 0....v=&#8730;&#956;sgr</p>
<p>2).show when &#956;s becomes 0...v=&#8730;grtan&#952;</p>
<p>&#160;</p>
<p>Thanks</p>

Explanation / Answer

Vmax = (R * g * tan ?)^0.5 Without friction: V max = [r * g * (sin ? + µ * cos ?) ÷ (cos ? – µ * sin ?)]^0.5 V min = [r * g * (sin ? – µ * cos ?) ÷ (cos ? + µ * sin ?)]^0.5 Force centripetal force = mass * (velocity^2 ÷ radius) The centripetal force is a horizontal force directed toward the center of the curve. You need to determine the sum of the forces that are directed toward the center of the curve. The road exerts a force on the tires of the car. This force is perpendicular to the inclined surface, and is labeled N. For the car to remain at the same vertical position, the sum of the vertical forces must equal 0 N. Without friction: The horizontal component of this force = N * sin ? Eq. #1 m * (v^2 ÷ r) = N * sin ? With Friction: The friction force on an inclined plane = µ * N The horizontal component of friction force = µ * N * cos ? Centripetal force = Sum of these 2 horizontal forces m * (v^2 ÷ r) = N * sin ? + µ * N * cos ? The vertical component of N = N * cos ? The vertical component of friction force = -µ * N * sin ? Weight is a negative vertical force = -(m * g) For the car to remain at the same vertical position, the sum of the vertical forces must equal 0 N. N * cos ? – µ * N * sin ? – (m * g) = 0 m * g = N * cos ? – µ * N * sin ? Now we have 2 equations Eq .#1 m * (v^2 ÷ r) = N * sin ? – µ * N * cos ? Eq. #2 m * g = N * cos ? + µ * N * sin ? When you divide Eq. #1 by Eq. #2, m on the left side of the equal sign will cancel, and N on the right side of the equal sign will cancel. Eq. #3 (v^2 ÷ r) ÷ g = (sin ? + µ * cos ?) ÷ (cos ? – µ * sin ?) Multiply both sides by g (v^2 ÷ r) = g * (sin ? + µ * cos ?) ÷ (cos ? – µ * sin ?) Multiply both sides by r v^2 = r * g * (sin ? + µ * cos ?) ÷ (cos ? – µ * sin ?) v = [r * g * (sin ? + µ * cos ?) ÷ (cos ? – µ * sin ?)]^0.5 For minimum velocity The friction force is in opposite direction Eq .#1 m * (v^2 ÷ r) = N * sin ? – µ * N * cos ? Eq. #2 m * g = N * cos ? – µ * N * sin ? Eq. #3 (v^2 ÷ r) ÷ g = (sin ? – µ * cos ?) ÷ (cos ? + µ * sin ?) v = [r * g * (sin ? – µ * cos ?) ÷ (cos ? + µ * sin ?)]^0.5

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.