A car of mass = 1200 kg traveling at 50.0 km/hour enters a banked turn covered w

Question

A car of mass = 1200 kg traveling at 50.0 km/hour enters a banked turn covered with ice. The road is banked at an angle , and there is no friction between the road and the car's tires.
Part A
What is the radius of the turn if = 20.0 degrees (assuming the car continues in uniform circular motion around the turn)?
I found the answer is 54.0 m for part A.

Need help on this-Part B
Now, suppose that the curve is level (=0) and that the ice has melted, so that there is a coefficient of static friction between the road and the car's tires. What is min, the minimum value of the coefficient of static friction between the tires and the road required to prevent the car from slipping? Assume that the car's speed is still 50.0 km/hour and that the radius of the curve is given by the value you found for r in Part A.

Explanation / Answer

To set part B up, you should draw a force-body diagram. In this particular problem, we are dealing with 2 forces affecting motion - we have the centripetal force (Fc) and the static-friction force (Ff). Our equation then looks like: Fc-Ff=0. So, Fc=Ff.

Plugging in the defining terms of both forces yield: m(v2/r)=mg. Of course, we can eliminate (m) from the equation by dividing it out. This leaves us with: (v2/r)=g. And because we are concerned with evaluating (), we simply solve for () by dividing through by (g).

Thus: (v2/rg)=.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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