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Max spped for a car to go around a curve track without skidding...confusing 
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#1
Aug2205, 10:39 AM

P: 3

here's my question,
a car of mass 1000kg moves along the corner of a level road having a radius of curvature 35.0m. If the frictional force between the tyres and the road is 4kN, the maximum speed of the car without skidding at the corner is A. 4.0 m/s B. 8.8 m/s C. 11.8 m/s D. 140.0 m/s ok, i knew there's a formula for the max spped of a car go around the curve track without skidding is v=(rag/h)^1/2, but is it possible to find the height form centre of gravity of car in the above case? I've tried usin' v=(Fr/m)^1/2 which is derived from the centripetal force, F=(mv^2)/r, nonetheless, if we using v=(Fr/m)^1/2, is this could be the maximum spped of car instead of using v=(rag/h)^1/2???? 


#2
Aug2205, 11:33 AM

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#3
Aug2205, 12:59 PM

HW Helper
P: 507

The difference between skidding and toppling (overturning) is creating the confusion.
If the car is toppling before skidding then we think of height on Center of mass of the car to calculate the torques. 


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