Velocity of a Car Skidding: Solving for Maximum Speed on a Circular Turn

  • Thread starter Thread starter mortho
  • Start date Start date
  • Tags Tags
    Car Velocity
AI Thread Summary
A discussion focused on calculating the maximum speed a car can achieve while rounding a circular turn without skidding, given a mass of 2.20 x 10^3 kg, a radius of 30.0 m, and a coefficient of static friction of 0.8. The initial attempt at solving the problem involved incorrect substitutions and confusion over the use of time in the equations. Participants clarified that the correct approach involves using centripetal acceleration and directly equating the frictional force to the centripetal force without needing to find time. After correcting the method, the final calculated maximum speed was determined to be 15.3 m/s, which was confirmed as reasonable. The discussion highlights the importance of correctly applying physics formulas in problem-solving.
mortho
Messages
99
Reaction score
0
[SOLVED] Velocity of a car skidding

Homework Statement


A 2.20 103 kg car rounds a circular turn of radius 30.0 m. If the road is flat and the coefficient of static friction between the tires and the road is 0.8, how fast can the car go without skidding?



Homework Equations


Fcentripital=ma
a=v2/t
v=2πr/t

The Attempt at a Solution



So i used F=ma and substituted F with Coefficient*mass*gravity and for ma i used
m 2πr/t and solved for t. Then i used the t for the velocity equation and solved it. I got 24.0 m/s but it was wrong. What did i do wrong?
 
Physics news on Phys.org
a=v2/t

I believe you mean this a = \frac{v^2}{r}

It's r for radius in the denominator, NOT t for time. This is centripetal acceleration here.

Try to solve it again.
 
Thank you soo much for pointing that out but I'm having algebra problems, so if i used
v2/r but i have no velocity so i use 2πr\t but then i have two divisions, i know this probably is a very stupid question but I'm really bad at math. so how would i put that equation to equal to t. Would it be (coefficient*m*g)/m(2πr2) ? The answer i got was completely ridiculous so I'm guessing it's wrong.
 
but i have no velocity so i use 2πr\t but then i have two divisions

But you are trying to find velocity, so there is no need to substitute something in for it. You don't need to find the time, either. Just equate the frictional force into the first equation listed in your "relevant equations" and the proper term for "a". Solve for v.
 
Thanks. i feel so retarded now that you said that. I got carried away trying to find time that i didn't even realize i was looking for velocity in the first place. So i solved for it and got 15.3 m/s correct?
 
Looks reasonable to me.
 
Thank you!
 
You're welcome. :smile:
 
Back
Top