Answer:
The effective coefficient of kinetic friction =0.3644
Explanation:
Given Data:
r=0.50 m
I=moment of inertia = 10.1 kg · m^2
ω=51 rev/min=0.885 rev/sec
t=4.0 sec
F=74 N
Required:
the effective coefficient of kinetic friction between the wheel and the wet rag=?
Solution:
Angular Acceleration=α=2π*ω/t
Angular Acceleration=α=2π*0.85/4
Angular Acceleration=α=1.3351 rad/s^2
Torque=τ=μ*F*r=I*α
Where:
μ is co-officient of friction
μ[tex]=\frac{I* \alpha}{F*r}[/tex]
μ=[tex]\frac{10.1*1.3351}{74*0.50}=0.3644[/tex]
The effective coefficient of kinetic friction =0.3644
Answer:
The effective coefficient of kinetic friction is 0.0682
Explanation:
The kinetic energy is equal to the frictional work:
Ek = W
[tex]E_{k} =\frac{1}{2} Iw^{2}[/tex]
Where
I = moment of inertia = 10.1 kg m²
w = angular velocity = 51 rev/min = 5.341 rad/s
[tex]E_{k} =\frac{1}{2} 10.1*(5.341)^{2}[/tex] (eq. 1)
The frictional work is:
[tex]W=\mu NX[/tex] (eq. 2)
The rotational expression of motion:
[tex]X=\frac{1}{2} Rwt[/tex] (eq. 3)
Where
R = radially inward force = 74 N
t = 4
Using eq. 1, 2 and 3:
[tex]\frac{1}{2}*10.1*(5.341)^{2} =\mu *74*0.5*\frac{1}{2} *(5.341)^{2} *4\\\mu =0.0682[/tex]
