Respuesta :
For the position of button we can use force balance as
Friction Force = Centripetal force
so here we will have
[tex]\mu_s mg = m\omega^2 R[/tex]
here we know that
R = radius of circle where button is placed
[tex]\omega = 2\pi f[/tex]
f = 35 rev/min
[tex]f = \frac{35}{60} rev/s = 0.58 rev/s[/tex]
[tex]\omega = 2\pi (0.58) = 3.66 rad/s[/tex]
now from above equation
[tex]\mu_s(m)(9.81) = (3.66)^2(0.240)[/tex]
[tex]\mu_s = 0.33[/tex]
so friction coefficient will be 0.33
The coefficient of friction is a dimensionless number that represents the ratio between friction force and normal force.
The friction coefficient between the button and the platform is 0.329.
What is the coefficient of friction?
The coefficient of friction can be defined as the ratio of the frictional force resisting the motion of an object to the normal force applied to an object.
Given that the diameter of the button is 0.520 m and the speed of revolution is 35.0 rev/min, the button is no more than 0.240 m from the axis. It means that this is the radius R of the circle where the button is placed.
The angular velocity is calculated as given below.
[tex]\omega = 2\pi f[/tex]
Where f is the speed of revolution.
[tex]\omega = 2\times 3.14\times \dfrac{35}{60}[/tex]
[tex]\omega = 3.67\;\rm rad/s[/tex]
The friction coefficient is calculated as given below.
The centripetal force acting on the button will be equivalent to the friction force acting between the button and the platform.
[tex]\mu mg = m\omega^2 R[/tex]
Where m is the mass of the button, and g is the gravitational acceleration.
[tex]\mu\times m\times 9.8 = m\times 3.67^3 \times 0.240[/tex]
[tex]\mu = 0.329[/tex]
Hence we can conclude that the friction coefficient between the button and the platform is 0.329.
To know more about the friction coefficient, follow the link given below.
https://brainly.com/question/11808898
