If the coefficient of friction is 0.41 and the cylinder has a radius of 2.5 m, what is the minimum angular speed of the cylinder so that the people don't fall out? (Normally the operator runs it considerably faster as a safety measure.)
I am not seeing the physical setup here.
its like the amusement park ride where people stand against a wall and the floor drops out..
OK, the force of friction is mu*normal component.
The normal component is mass*w*radius where w is in radians/sec. The force of friction needs to be greater than mg, so
mu*mass*w radius > mg
solve for w.
To find the minimum angular speed of the cylinder so that the people don't fall out, we need to use the equation mu*mass*w*radius > mg, where mu is the coefficient of friction, mass is the total mass of the people, w is the angular speed in radians per second, radius is the radius of the cylinder, and g is the acceleration due to gravity.
1. Start by multiplying both sides of the equation by radius:
mu*mass*w*radius^2 > mg*radius
2. Divide both sides of the equation by mu*mass*radius^2:
w > (mg*radius) / (mu*mass*radius^2)
3. Simplify the equation:
w > g / (mu*radius)
Now, you can substitute the given values and calculate the minimum angular speed:
w > (9.8 m/s^2) / (0.41 * 2.5 m)
w > 9.8 / 1.025
w > 9.56 radians/sec
Therefore, the minimum angular speed of the cylinder, so that the people don't fall out, is 9.56 radians per second.