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

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SUMMARY

The centripetal force acting on a minibus with a mass of 1.6 tonnes, rounding a bend with a radius of 25 meters at a speed of 72 km/h, is calculated to be 25,600 N using the formula F = mv²/r. The minimum coefficient of friction required to prevent skidding is determined to be 1.631, which exceeds the typical maximum value of 1 for friction coefficients. This indicates that the vehicle's tires may have enhanced grip or that external factors, such as road conditions, are contributing to the high friction requirement.

PREREQUISITES
  • Understanding of centripetal force calculations
  • Familiarity with the formula F = mv²/r
  • Knowledge of friction coefficients and their significance
  • Basic physics concepts related to motion and forces
NEXT STEPS
  • Research advanced friction models in vehicle dynamics
  • Explore the effects of tire materials on friction coefficients
  • Learn about centripetal acceleration and its applications in real-world scenarios
  • Investigate the impact of road conditions on vehicle handling
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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?
 
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Your answers are correct. \mu_ is not always <1. Since \mu&gt;1, what do you think is happening?
 

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