Banking curve problem (find maximum speed)

AI Thread Summary
A car traveling around a circular curve with a radius of 42 meters can achieve a maximum speed determined by the frictional force, which is 20% of the car's weight. The relevant equation for maximum speed is v = √(rg), where g is the acceleration due to gravity. Calculating this gives a maximum speed of approximately 20.29 m/s. The discussion also touches on the relationship between centripetal force and friction, noting that static friction plays a crucial role in maintaining the car's motion around the curve. Understanding the dynamics of friction and centripetal force is essential for solving such physics problems effectively.
5ymmetrica1
Messages
88
Reaction score
0

Homework Statement


A car travels around a circular curve on a flat, horizontal road at a radius of 42meters.
the maximum frictional force between the tyres and the road is equal to 20% of the weight of the car
calculate the maximum speed at which the car can travel around the curve at a constant radius of 42 meters


Homework Equations


I think I am supposed to use v = √rg
But I don't have a set of answers for these problems so I am not sure as this doesn't include the part of the question, the maximum frictional force between the tyres and the road is equal to 20% of the weight of the car


The Attempt at a Solution



v = √42*9.8
∴ v = 20.29 m/s
 
Physics news on Phys.org
5ymmetrica1 said:

Homework Statement


A car travels around a circular curve on a flat, horizontal road at a radius of 42meters.
the maximum frictional force between the tyres and the road is equal to 20% of the weight of the car
calculate the maximum speed at which the car can travel around the curve at a constant radius of 42 meters


Homework Equations


I think I am supposed to use v = √rg
But I don't have a set of answers for these problems so I am not sure as this doesn't include the part of the question, the maximum frictional force between the tyres and the road is equal to 20% of the weight of the car


The Attempt at a Solution



v = √42*9.8
∴ v = 20.29 m/s

The centripetal force is mv^2/R but what force is in play to produce that force ? It's not mg since that's the weight of the entire vehicle. What's the static friction force?
 
I haven't studied any formula for static friction at this time in my physics course so I'm not quite sure what you mean sorry.

I know Fw = Fg (and Fv is equal but opposite in direction)

and that when split into vector components Fh = Fc

so is the frictional force Ff = FN (Normal force)?
 
5ymmetrica1 said:
I haven't studied any formula for static friction at this time in my physics course so I'm not quite sure what you mean sorry.

I know Fw = Fg (and Fv is equal but opposite in direction)

and that when split into vector components Fh = Fc

so is the frictional force Ff = FN (Normal force)?

You don't need the actual formula of friction since you know Fμ=0.2*G
 
5ymmetrica1 said:
I haven't studied any formula for static friction at this time in my physics course so I'm not quite sure what you mean sorry.

I know Fw = Fg (and Fv is equal but opposite in direction)

and that when split into vector components Fh = Fc

so is the frictional force Ff = FN (Normal force)?


If yo have a weight W sitting on the table and the coefficient of static friction is μ, how much horizontal force can you apply before it starts to slip?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Back
Top