Respuesta :
Answer:
Part a)
[tex]F_f = 107.8 N[/tex]
Part b)
[tex]\mu = 0.415[/tex]
Explanation:
Part a)
Time period of one revolution is given as
T = 42 s
now the angular speed of the belt is given as
[tex]\omega = \frac{2\pi}{T}[/tex]
[tex]\omega = \frac{2\pi}{42}[/tex]
[tex]\omega = 0.15 rad/s[/tex]
Now at rest position the force along the surface of carousel must be constant
so we will have
[tex]mgsin\theta + m\omega^2 r cos\theta = F_f[/tex]
[tex]30(9.81)sin20.5 + 30(0.15^2)(7.46)cos20.5 = F_f[/tex]
[tex]F_f = 107.8 N[/tex]
Part b)
New Time period of one revolution is given as
T = 30 s
now the angular speed of the belt is given as
[tex]\omega = \frac{2\pi}{T}[/tex]
[tex]\omega = \frac{2\pi}{30}[/tex]
[tex]\omega = 0.21 rad/s[/tex]
Now at rest position the force along the surface of carousel must be constant
so we will have
[tex]mgsin\theta + m\omega^2 r cos\theta = F_f[/tex]
[tex]30(9.81)sin20.5 + 30(0.21^2)(7.94)cos20.5 = F_f[/tex]
[tex]F_f = 112.9 N[/tex]
Also we know that in perpendicular direction also force is balanced
[tex]F_n + m\omega^2 r sin\theta = mgcos\theta[/tex]
[tex]F_n = mgcos\theta - m\omega^2 r sin\theta[/tex]
[tex]F_n = 30(9.81)cos20.5 - 30(0.21)^2(7.94)sin20.5[/tex]
[tex]F_n = 272 N[/tex]
now for friction coefficient we will have
[tex]F_f = \mu F_n[/tex]
[tex]112.9 = \mu 272[/tex]
[tex]\mu = 0.415[/tex]