Solve for Tire Radius: Mass, Friction Coefficient, Normal Force, and Speed Given

[rockmorg]

Hey guys -
Here is a problem I have been working on... I've tried several things, I think I'm pretty close...
Stone has a mass of 7e-3 kg and is wedged into the tread of a tire. Coefficient of static friction between each side of tread channel is 0.74. When the tire surface is rotating at 18 m/s the stone flies out. The magnitude of the normal force that each side of tread exerts is 1.8 N. Assume only static friction supplies centripetal force and determine radius of the tire.

So I have mass, coefficient of static friction, normal force, and speed.

I've tried...

Fc = ma_c = m(v^2/r)

F_c = f_s = u_sF_n
= 2(.74)2(1.8 N)

F_c = 5.3 N

r = F_c/(v2m)

5.3 N/(18 m/s)^2(7e-3) = 2.3 m

I've tried this and it came out wrong, of course thinking realistically it would be wrong anyways because usually tires (much less their radius) would be 2.3 m.

I could use F_c = 1.3 N (from not multiplying .74 by 2 and 1.8 N by 2)...


Any thoughts? Thanks!

 

[Fermat]

A couple of comments.

The friction force on one side is = µ*NR = 0.74*1.8
So, the friction from both sides is 2*µ*NR = 2*0.74*1.8 = 2.664 N.

Also, Fc = mv²/r

giving,

r = mv²/Fc rather than Fc/mv²

 

[quicknote]

Hi there,

It seems like you're on the right track in solving for the radius of the tire. However, there are a few things to consider in your calculations:

1. The normal force is not doubled when you multiply it by 2. The normal force is the force perpendicular to the surface, so it remains the same regardless of the number of sides of the tread channel.

2. The coefficient of static friction may not necessarily be multiplied by 2 either. The coefficient of friction is a property of the two surfaces in contact, so it would remain the same regardless of the number of sides of the tread channel.

3. You are correct in using the formula Fc = ma_c = m(v^2/r), but remember that the centripetal force is equal to the frictional force, not the other way around. So, it should be written as Fc = F_s = m(v^2/r).

With these considerations in mind, your final calculation should look like this:

r = F_s/(v^2m) = (0.74)(1.8 N)/(18 m/s)^2(7e-3 kg) = 0.037 m = 3.7 cm

Therefore, the radius of the tire is approximately 3.7 cm. I hope this helps! Keep up the good work.