posted this question this morning but I'm not sure of the response Damon provided. I did it once but then I updated it later in this at the bottom- please check booth of my answers.
If you were in a rotor style ride and the riders accelerate until the speed of ride is reached- if the radius of the cylinder is 5.0m and the coefficient of friction between clothes and wall is 0.5, how do you find minimum speed you would need to stick to the wall of the ride? Is he saying that I take 2*5(r) *9.8(g)? If not, please explain.Thank you-please scroll all the way for the next way I tried to solve it after Damon's response-Thank you very much
Physics-Please help with formula - Damon, Tuesday, December 7, 2010 at 11:11am
you need a centripetal acceleration equal to 2g if the mu is 1/2
Force down = m g
friction force up = mu m v^2/r
g = mu v^2/r
v^2/r = g/mu = 2 g
v^2 = 2 r g
Physics-I think I have it-Please check - Joey, Tuesday, December 7, 2010 at 5:29pm
Is this correct? I published something else earlier-that has to be wrong but I think this is correct.
Force friction balances weight
Force friction comes from force normal to create centripetal force
mv^2/r = Force normal
Force friction = mu x Force normal = mg
mu x mv^2/r = mg
9.8/05 = 19.6 m/s^2
19.6/5=3.92
sqrt 3.92 = 1.98 rads/s
Damon's response is correct. In order for you to stick to the wall of the ride, you need a centripetal acceleration equal to 2g, where g is the acceleration due to gravity and mu (μ) is the coefficient of friction between your clothes and the wall.
Damon derived the formula v^2 = 2rg, where v is the minimum speed you would need to stick to the wall and r is the radius of the cylinder.
Let's go through your attempted solution and see if it matches Damon's formula:
Force friction balances weight:
mv^2/r = Force normal
Force friction comes from force normal to create centripetal force:
Force friction = mu x Force normal = mg
mu x mv^2/r = mg
9.8/0.5 = 19.6 m/s^2
Here, you made a mistake. The value of mu should be 0.5, but you used 0.05 instead. So, 0.5/0.5 = 1, not 19.6.
19.6/5 = 3.92
This step is incorrect. You divided 19.6 by 5, which is not necessary in this case.
sqrt(3.92) = 1.98 rad/s
Here, you took the square root of 3.92, which gives you 1.98. However, this is the angular velocity, not the linear speed, which is what we are trying to find.
To find the linear speed, you need to take the square root of 2rg, as Damon mentioned. So, the correct calculation would be:
v = sqrt(2rg) = sqrt(2*5*9.8) ≈ 9.9 m/s
Therefore, the correct answer is approximately 9.9 m/s, which is the minimum speed you would need to stick to the wall of the ride.