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[tex]\tau[/tex] = r * F

1. The problem statement, all variables and given/known data

youve got our bicycle upside-down for repairs with its 66 cm diameter wheel spinning freely at 230 rpm. the mass of the wheel is 1.9 kg and is concentrated mostly at the rim. you hold a

wrench against the tire for 3.1s with a normal force of 2.7 N. if the coefficient of friction between the wrench and the tire is 0.46 what is theFINAL ANGULAR SPEED OF THE WHEEL?

2. Relevant equations

diameter = 66 cm

radius = .033m

mass = 1.9 kg

Norm Force = 2.7 N

friction coef = .46

[tex]\omega[/tex]0 = 230 rpm

[tex]\omega[/tex]0 = 24.1 rad/s

w = ?

eqns

I = Mr^2

[tex]\omega[/tex] = [tex]\omega[/tex]0 + [tex]\alpha[/tex] * t

[tex]\tau[/tex] = r * F * sin(theta)

[tex]\tau[/tex] = I [tex]\alpha[/tex]

v=[tex]\omega[/tex] * r

v=[tex]\omega[/tex] * r

3. The attempt at a solution

x

[tex]\tau[/tex] = r * F

where F is the Friction Force

r*F = M * r^2 * [tex]\alpha[/tex]

we can solve for [tex]\alpha[/tex] and it = 19.08

[tex]\omega[/tex] = ( [tex]\omega[/tex]0 + [tex]\alpha[/tex] *t)

[tex]\omega[/tex] = 85.48 rad/s

that means it sped up...

idk how to do this problem keep getting confused with the answers im getting

any help?

thanks in advance for your time and effort.

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# Homework Help: Angular Speed of a bicycle

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