 #1
AspiringPhysicist12
 16
 6
 Homework Statement:

In a concrete mixer, cement, gravel, and water are mixed by tumbling action in a slowly rotating drum. If the drum spins too fast, the ingredients stick to the drum wall instead of mixing.
Assume that the drum of a mixer has radius R = 0.5 m and that
it is mounted with its axle horizontal. What is the fastest the drum
can rotate without the ingredients sticking to the wall all the time?
Assume g = 9.8 m/s2.
 Relevant Equations:

Fn  Fc = 0
Fgcosθ  Fc=0 for object to stick to drum wall during circular motion
I attached my diagram/work done so far. I believe that as long as the angular velocity is always less than sqrt(2g), the object must fall downwards when it reaches the top of the drum wall. Also, the angular velocity cannot be greater than sqrt(2g), as ω=sqrt((g/r)cosθ) at all times, and can only have cosθ=1 at most. Therefore, we must have that ω<sqrt(2g) at all times to ensure the ingredient does not stick to the drum wall at all times.
My confusion is that in the answer key, it says ω ≤ sqrt(2g) rather than ω<sqrt(2g). I don't understand this because ω=sqrt(2g) is the angular velocity required for the ingredient to stick to the top of the drum wall. So if ω=sqrt(2g), the ingredient does not fall down after reaching the top of the drum wall, and the ingredient will thus will always stick to the wall. I would appreciate if it could be explained why the answer is ω ≤ sqrt(2g) rather than ω<sqrt(2g).
My confusion is that in the answer key, it says ω ≤ sqrt(2g) rather than ω<sqrt(2g). I don't understand this because ω=sqrt(2g) is the angular velocity required for the ingredient to stick to the top of the drum wall. So if ω=sqrt(2g), the ingredient does not fall down after reaching the top of the drum wall, and the ingredient will thus will always stick to the wall. I would appreciate if it could be explained why the answer is ω ≤ sqrt(2g) rather than ω<sqrt(2g).