Answer:
0.52378
Explanation:
V = Velocity = 50 m/s
r = Radius = 1.25 m
I = Moment of inertia = 110 kgm²
N = Weight supported by wheels = 14000 N
The angular velocity
[tex]\omega=\dfrac{V}{r}\\\Rightarrow \omega=\dfrac{50}{1.25}\\\Rightarrow \omega=40\ rad/s[/tex]
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{40-0}{0.48}\\\Rightarrow \alpha=83.33\ rad/s^2[/tex]
Frictional force is given by
[tex]f=\dfrac{I\alpha}{r}\\\Rightarrow f=\dfrac{110\times 83.33}{1.25}\\\Rightarrow f=7333.04\ N[/tex]
The coefficient of friction is given by
[tex]\mu=\dfrac{f}{N}\\\Rightarrow \mu=\dfrac{7333.039}{14000}\\\Rightarrow \mu=0.52378[/tex]
The coefficient of friction is 0.52378
In this given case, the coefficient of kinetic friction between the wheels and the runway - 0.524.
Kinetic friction is a force that is applied between two moving surfaces.
Given:
moment of inertia = 110kgm[tex]^2[/tex]
speed = 50 m/s
radius - 1.25 m
weight = 14000
angular speed = 0.480 s
solution:
The final angular speed of the wheel
ωf = V /r
= 50/1.25
= 40 rad/s
the angular acceleration
α =( ωf -ωi) / Δt
= (40-0) / 0.48
= 83.3rad/[tex]S^2[/tex]
The frictional force is
Fk = I α / r
= 110(83.3) / 1.25
=7.33* N
coefficient of friction
μ = Fk / N
= 7.33*103 /14000
= 0.524.
Thus, the coefficient of kinetic friction between the wheels and the runway - 0.524.
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