Answer:
The minimum speed is 14.53 m/s.
Explanation:
Given that,
r = 11 m
Friction coefficient = 0.51
Suppose we need to find the minimum speed, that the cylinder must make a person move at to ensure they will stick to the wall
When frictional force becomes equal to or greater than the weight of person
Then, he sticks to the wall
We need to calculate the minimum speed
Using formula for speed
[tex]f_{s}=\mu N\geq mg[/tex]
Where, [tex]N =\dfrac{mv^2}{r}[/tex]
[tex]\mu\times\dfrac{mv^2}{r}\gep mg[/tex]
[tex]v^2\geq\dfrac{gr}{\mu}[/tex]
Put the value into the formula
[tex]v=\sqrt{\dfrac{9.8\times11}{0.51}}[/tex]
[tex]v=14.53\ m/s[/tex]
Hence, The minimum speed is 14.53 m/s.
This question is incomplete. The complete question is to find out the minimum speed of the cylinder. This question involves the concept of the frictional force.
The minimum speed of the cylinder is "14.55 m/s".
In order for the people to remain in contact with the wall all the time, the weight of the people must be equal to the frictional force acting on the people.
[tex]Weight = frictional\ force = \mu Normal\ Force\\Weight = \mu Centripetal\ Force\\\\mg = \mu \frac{mv^2}{r}\\\\v= \sqrt{\frac{gr}{\mu}}[/tex]
where,
v = minimum speed = ?
g = acceleration due to gravity = 9.81 m/s²
r = radius = 11 m
μ = coefficient of friction = 0.51
Therefore,
[tex]v = \sqrt{\frac{(9.81\ m/s^2)(11\ m)}{0.51}}[/tex]
v = 14.55 m/s
Learn more about frictional force here:
https://brainly.com/question/1714663?referrer=searchResults
The attached picture shows the free-body diagram of this scenario.
