Answer:
a) [tex]a_w=4.331m/s^2[/tex]
b) [tex]F=53.7N[/tex]
c)[tex]\mu=96.5N[/tex]
Step-by-step explanation:
From the question we are told that
mass of unbalanced wheel [tex]M=15kg[/tex]
Radius of gyration [tex]G=64m[/tex]
Mass center [tex]M_c =40mm=[/tex][tex]4*10^{-2} m[/tex]
Angular velocity [tex]\omega=4rads/s[/tex]
a)
Generally the equation for moment at point A is mathematically given as
[tex]\sum M_a=I \alpha+\sum M \=ad[/tex]
Where [tex]a_w=wheel\ accelaration[/tex]
[tex]15*9.81*(0.04)=0.06144*\frac{a_w}{0.1}+15(0.04)(0.04)\frac{a_w}{0.1}+(15a_w-15(0.04)(4^2)0.1)[/tex]
[tex]5,87=(a_w*0.8544+15a_0-0.96)[/tex]
[tex]a_w=4.331m/s^2[/tex]
b)
Generally the equation for equilibrium of system on the horizontal axis is mathematically given as
[tex]\sum f_x=ma_w[/tex]
[tex]F=ma_w-m \=r w^2[/tex]
[tex]F=(25*4.221)-(15*0.04*4^2)[/tex]
[tex]F=63.3-9.6[/tex]
[tex]F=53.7N[/tex]
c)
Generally the equation for equilibrium of system on the vertical axis is mathematically given as
[tex]\sum f_y=ma_w[/tex]
Where
[tex]F=\mu-mg+m \=r \alpha[/tex]
[tex]\alpha=\frac{a_0}{0.1}[/tex]
[tex]\mu-mg+m \=r \alpha=0[/tex]
[tex]\mu-15*9.81+15*0.04*\frac{4.221}{0.1}[/tex]
[tex]\mu-96.5N=0[/tex]
[tex]\mu=96.5[/tex]
