Calculating Tire Radius Given Mass, Friction & Velocity

In summary, the stone flew out of the tread of an automobile tire with a mass of 3.92 g due to a coefficient of static friction of 0.864 and a normal force of 2.37 N. Assuming only static friction supplies the centripetal force, the radius of the tire can be determined to be 0.324 m using the equation (mv^2)/r = 2mu*Fn.
  • #1
lalalah
18
0
1. A stone has a mass of 3.92 g and is wedged into the tread of an automobile tire, as the drawing shows. The coefficient of static friction between the stone and each side of the tread channel is 0.864. When the tire surface is rotating at 18.4 m/s, the stone flies out of the tread. The magnitude FN of the normal force that each side of the tread channel exerts on the stone is 2.37 N. Assume that only static friction supplies the centripetal force, and determine the radius r of the tire (in terms of m).



2. v = sq.root([tex]\mu[/tex]s*g*r), [tex]\Sigma[/tex]Fc = m(v2/r), ac = v2/r, v = [2[tex]\pi[/tex]r]/T



3. link to picture of the stone wedged in the tire.
http://www.flickr.com/photos/20949091@N00/2911847669/

as it is wedged, there are two normal forces acting directly opposite each to the other on either side of the stone. Even though the two normal forces should cancel out, I still want to use the 2.37 N in the v = sq.root(m_u*r*g) by substituting "g" for (Fn/m)

I am assuming that the velocity is 18.4 m/s. Is this correct?

So...

velocity = 18.4 m/s = sq.root((mu)*(Fn/m)*radiustire)
18.4 m/s = sq.root(0.864*(2.37N/.00392kg)*radius tire)
squaring both sides gets:
338.56 = 522.37 * radius
radius = .648 m

is there something I'm missing? my intuition tells me this answer is too big!
 
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  • #2


A simpler way to do it is to set the centrifugal force equal to the friction force on the stone. Remember there are two friction forces. That is [itex]F=2\mu N[/itex].
 
  • #3


ah i see.
so in this case, i can say there are two friction forces because they do not cancel each other out?

so it should be...
(mv^2)/r = 2 mu *mg
(.00392*18.4^2)/r = 2*0.864*.00392*9.8
1.327/r = 0.0664
r= 19.98 m
?
 
  • #4


wait let me try that again

Fn = 2.37
(mv^2)/r = 2 mu *Fn
(.00392*18.4^2)/r = 2*0.864*2.37
1.327/r = 4.095
r = .324

so it's the first answer divided by two! I'm starting to get the idea of how to use forces now... thank you so much for your help, i appreciate it!
 
  • #5


Your first answer was ok except you hadn't included the other friction force, hence why it was twice as much. The other method is a bit more intuitive.
 

What is the equation for calculating tire radius given mass, friction, and velocity?

The equation for calculating tire radius is r = (mv^2)/(f*g), where r is the radius of the tire, m is the mass of the vehicle, v is the velocity, f is the coefficient of friction, and g is the acceleration due to gravity.

How does friction affect the calculation of tire radius?

Friction is an important factor in the calculation of tire radius because it is directly related to the force that the tire must overcome to maintain its velocity. The higher the coefficient of friction, the smaller the tire radius must be to maintain the same velocity.

What units should be used for the variables in the tire radius equation?

The mass should be in kilograms (kg), velocity in meters per second (m/s), and the coefficient of friction and acceleration due to gravity should be unitless.

What is the significance of calculating tire radius?

Calculating tire radius is important for understanding the performance and capabilities of a vehicle. It can also help in determining the ideal tire size and pressure for optimal handling and fuel efficiency.

Are there any limitations to the tire radius equation?

Yes, the tire radius equation assumes that the tire is perfectly round, the mass is evenly distributed, and there are no other external forces at play. In reality, there may be variations in tire shape and weight distribution, as well as other factors such as air resistance, that can affect the accuracy of the calculation.

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