If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.72 {\rm rev}/{\rm s}, friction in the bearings causes the wheel to stop
in just 12 {\rm s}.
0.11 N*m
That may be true for this specific bicycle, but you have not asked a question.
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If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.72rev/s, friction in the bearings causes the wheel to stop
in just 12s.
Part A
If the moment of inertia of the wheel about its axle is 0.30kg\m2 , what is the magnitude of the frictional torque?
To find the coefficient of friction in the bearings of the bicycle wheel, we can use the equation that relates the deceleration of the rotation to the coefficient of friction. The deceleration (α) is related to the angular velocity (ω) and the time taken (t) by the formula:
α = (ω_f - ω_i) / t
Where:
- α is the deceleration (angular acceleration)
- ω_f is the final angular velocity (0 rad/s, as the wheel stops rotating)
- ω_i is the initial angular velocity (0.72 rev/s, given in the problem)
- t is the time taken (12 s)
First, we need to convert the initial angular velocity from rev/s to rad/s since the SI unit for angular velocity is radian/second. One revolution is equal to 2π radians, so we have:
ω_i = 0.72 rev/s * 2π rad/rev ≈ 4.52 rad/s
Now, we can calculate the deceleration:
α = (0 - 4.52 rad/s) / 12 s ≈ -0.38 rad/s^2
Since the deceleration is caused by the friction in the bearings of the wheel, we can apply the following equation:
α = -μ * g * r
Where:
- μ is the coefficient of friction
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- r is the radius of the wheel
We can rearrange the equation to solve for the coefficient of friction:
μ = -α / (g * r)
Let's assume the radius of the wheel is 0.5 meters, then:
μ = -(-0.38 rad/s^2) / (9.8 m/s^2 * 0.5 m)
μ ≈ 0.039
Therefore, the coefficient of friction in the bearings of the poorly maintained bicycle wheel is approximately 0.039.