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

AI Thread Summary
The discussion centers on solving for the radius of a tire given a mass of 7e-3 kg for a stone, a static friction coefficient of 0.74, and a normal force of 1.8 N at a tire speed of 18 m/s. The initial calculations attempted to equate centripetal force with static friction but yielded an unrealistic radius of 2.3 m. A correction was suggested, noting that the total friction force from both sides of the tread is 2.664 N. The formula for radius should be r = mv²/Fc, not Fc/mv². The conversation emphasizes the importance of correctly calculating the forces involved to determine an accurate tire radius.
rockmorg
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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!
 
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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
 
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