Maximum speed of car round banked bend

AI Thread Summary
The discussion revolves around calculating the maximum speed of a car navigating a banked curve with a radius of 44 meters, a banking angle where tan(α) = 3/4, and a coefficient of friction of 0.6. The initial calculations suggest a maximum speed of 27.2 m/s, but further analysis using force components leads to a revised maximum speed of 32.5 m/s. Participants emphasize the importance of correctly identifying force components and suggest drawing free body diagrams for clarity. Despite initial confidence in the first answer, the final conclusion indicates that the revised speed is more accurate. The conversation highlights common challenges in solving physics problems related to circular motion and friction.
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Homework Statement


Car mass m drives round a bend in a circular arc radius 44m with the road banked at an angle \alpha where tan\alpha=3/4. The coefficient of friction \mu=0.6 of the car with the road. What is the maximum velocity the car can travel at without sliding up the banked road?

Homework Equations


F=ma F=mv^2/r Friction(max)=\mur
3^2 + 4^2= 5^2

The Attempt at a Solution


The normal reaction force R perpendicular to the bank will be mgcos\alpha
R=0.8mg

Component of R towards teh center of horizontal circle = R/sin\alpha = mg4/3

mv^2/r= 0.6x0.8mgxcos\alpha + mg4/3
v^2/44=0.384g +g4/3
V=27.2m/s

I've always struggled with this type of question due to taking the wrong components for reaction forces etc so would greatly appreciate any help to see if this is right.

cheers
 
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Take the components of the forces along and normal to the slope. It'll seem easier. Draw the freebody diagram. At the maximum speed, the static frcition force along the slope is max.

(Don't plug in numbers to make it messy -- use symbols.)
 
Yeah i know...this was actually in a test i did this morning so i did draw out everything. Basically this was the answer i did in the test and I still stand by after working through it again but was just wanting to see if it was actually right
 
After simplification,

v^2/r
= g[sin(theta) + kcos(theta)]/[cos(theta) - ksin(theta)]
= g[tan(theta) + k]/[1 - ktan(theta)],

which gives me v = 32.5 m/s.

(Sorry...)
 
Last edited:
ah thanks alot! shame i didnt get it right tho lol
 

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