Respuesta :
Answer:
The maximum period of rotation of the cylinder is 2.84 seconds
Explanation:
Radius of the cylinder, r = 5 m
Coefficient of static friction, [tex]\mu = 0.4[/tex]
The frictional force, [tex]F = \mu R[/tex]
The normal reaction, R is a centripetal force:
[tex]R = \frac{mv^{2} }{r}[/tex]
[tex]F = \frac{\mu mv^{2} }{r}[/tex]
The force due to gravity, F = mg
For equilibrium, the force due to gravity must equal the frictional force
[tex]mg = \frac{\mu mv^{2} }{r}[/tex]
[tex]gr = \mu v^{2}[/tex]
[tex]9.8 * 5 = 0.4v^{2} \\v^{2} = \frac{9.8*5}{0.4} \\v = \sqrt{\frac{9.8*5}{0.4}}[/tex]
v = 11.07 m/s
The angular speed, w = v/r
w = 11.07/5
w = 2.214 rad/s
For a complete revolution through a circle, θ = 2π rad
w = θ/t
t = θ/w
t = 2π/2.214
t = 2.84 seconds
