Solve Equation for Min/Max Speed on Glare Ice

[junlee]

Homework Statement

A circular curve of radius R in a new highway is designed so that a car traveling at speed v can negotiate the turn safely on glare ice (zero friction). If a car travels too slowly, then it will slip toward the center of the circle. If it travels too fast, then it will slip away from the center of the circle. If the coefficient of static friction increases, a car can stay on the road while traveling at any speed within a range from vmin to vmax. Derive formulas for vmin and vmax using mu for µ, and v, R, and g as appropriate.

Homework Equations

I think it is Summation of Fr=mAr=mV^2/r

The Attempt at a Solution

I tried various formula but it turn out to be wrong. I can't seem to fit in the coeffient of friction.

 

[anathema]

Can someone help me with this?

Hello,

In this scenario, we can use the equation for centripetal force to derive formulas for vmin and vmax. The centripetal force is given by Fc = mv^2/R, where m is the mass of the car, v is its speed, and R is the radius of the circular curve.

Since there is no friction on the ice, the only force acting on the car is the centripetal force. Therefore, the maximum speed at which the car can travel without slipping is when the centripetal force is equal to the maximum static friction force, which is given by μmg, where μ is the coefficient of static friction and g is the acceleration due to gravity.

Setting these two forces equal to each other, we get:

Fc = μmg

mv^2/R = μmg

vmax = √(μgR)

Similarly, the minimum speed at which the car can travel without slipping is when the centripetal force is equal to the minimum static friction force, which is given by μmgcosθ, where θ is the angle between the direction of motion and the normal force.

Setting these two forces equal to each other, we get:

Fc = μmgcosθ

mv^2/R = μmgcosθ

vmin = √(μgRcosθ)

Since the angle θ is not given in the problem, we can assume it to be 0 degrees, which gives us vmin = √(μgR).

I hope this helps! Let me know if you have any further questions.

 

[ogb p]

I would approach this problem by first identifying the relevant physical principles involved. In this case, we are dealing with the concept of centripetal force, which is necessary for a car to safely negotiate a circular curve. The centripetal force is provided by the friction between the car's tires and the road.

To solve for the minimum and maximum speeds on glare ice, we can use the formula for centripetal force:

F = mv^2 / r

where F is the centripetal force, m is the mass of the car, v is the velocity, and r is the radius of the curve.

On glare ice, the coefficient of friction (µ) is equal to zero, meaning there is no friction between the car's tires and the road. This means that the maximum possible centripetal force is also equal to zero.

In order to stay on the road, the centripetal force must be greater than or equal to the gravitational force, which can be represented by mg.

Therefore, we can set up the following inequality:

mv^2 / r ≥ mg

Solving for v, we get:

v ≥ √(rg)

This is the minimum speed (vmin) that a car must travel in order to safely negotiate the curve on glare ice.

To find the maximum speed (vmax), we can use the same formula, but instead of setting the centripetal force equal to zero, we set it equal to the maximum possible value, which is provided by the coefficient of static friction:

mv^2 / r = µmg

Solving for v, we get:

vmax = √(µrg)

In conclusion, the formulas for the minimum and maximum speeds on glare ice are:

vmin = √(rg)

vmax = √(µrg)

where r is the radius of the curve, g is the acceleration due to gravity, and µ is the coefficient of static friction.