You want to figure out the amount of friction that acts on a bicycle wheel. You

Question

You want to figure out the amount of friction that acts on a bicycle wheel. You give the am initial angular speed and find that it rotates 2 complete revolutions in first second, then takes an additional 59.0 seconds to come to rest. If the wheel has a mass 0.15 kg and a radius of 0.35 m, what is the moment of inertia, assuming all the mass is located at the radius? If the frictional torque is constant, calculate the magnitude of this torque. What is the radial acceleration of a point on the tire at t = 0 s? How many revolutions did the wheel make before coming to rest?

Explanation / Answer

here,

theta = 2 * 2 pi = 12.56 rad

time , t = 1 s

theta = u * t + 0.5 * a * t^2

12.56 = 0 + 0.5 * a * 1^2

a = 25.12 rad/s^2

the final angular speed , w = a * t = 25.12 * 1 = 25.12 rad/s

let the accelration due to friction be a'

0 = 25.12 + 59 * a

a = - 0.4258 rad/s^2

a)

mass ,m = 0.15 kg

R = 0.35 m

the moment of inertia of the wheel , I = m * r^2

I = 0.15 * 0.35^2 = 0.018 kg.m^2

b)

the torque due to friction , T = I * a'

T = - 7.66 * 10^-3 N.m

c)

at t= 0 , the radial accelration at t= 0 is 25.12 rad/s^2

d)

theta' = w^2 /2a' = 25.12^2 /( 2 * 0.4258)

theta' = 740.97 rad = 118 rev

number of rotation before coming rest , n = 2 + theta' = 120 rev

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