Centripetal Force Homework: Minibus, Mass 1.6 Tonnes, 25m Radius, 72km/h Speed

[raphile]

Homework Statement

A minibus of mass 1.6 tonnes is rounding a flat bend of constant radius 25m at a steady speed of 72km/h. Find the centripetal force acting on the vehicle. What is the minimum co-efficient of sideways friction between the tyres and the road for this motion to take place without skidding?

The Attempt at a Solution

I thought I knew how to do this. Using mv^2/r, I got the centripetal force to be 25,600N (changing the speed into from km/h to m/s). Then, for the co-efficient of friction (denote as u), I used uR = u(mg) = mv^2/r and got, by taking g=9.81, an answer of u=1.631.

But I thought co-efficients of friction were supposed to be less than 1. An answer of 1.631 seems a bit high. Is it right?

 

[konthelion]

Your answers are correct. $$\mu_$$ is not always <1. Since $$\mu>1$$, what do you think is happening?

 

[Moetasim]

I would like to clarify that the calculation for the centripetal force is correct, assuming that the minibus is traveling in a circular motion at a constant speed of 72km/h. However, the calculation for the coefficient of friction is not accurate. The coefficient of friction is a dimensionless quantity and cannot be equal to the value of 1.631.

To calculate the coefficient of friction, we can use the formula u = v^2/(rg), where v is the linear speed, r is the radius of the circle, and g is the acceleration due to gravity. Substituting the values given in the problem, we get u = (20m/s)^2 / (25m)(9.81m/s^2) = 1.63.

This value is within the range of typical coefficients of friction for rubber tires on dry pavement, which can range from 0.7 to 1.0. Therefore, the minimum coefficient of friction needed for the minibus to travel around the bend without skidding is approximately 1.63.

It is important to note that this calculation assumes ideal conditions and does not take into account any external factors that may affect the motion of the minibus, such as road conditions or the condition of the tires. Further analysis and experimentation may be needed to accurately determine the coefficient of friction in real-world scenarios.