The rotor is an amusement park ride where people stand against the inside of a cylinder. Once the cylinder is spinning fast enough, the floor drops out.
A-What force keeps the people from falling out the bottowm of the ride?
I think that it's the Normal force?
B-If the coefficient of friction is 0.40 and the cylinder has a radius of 2.5m, what is the minimum angular speed the of the cylinder so the people don't fall out?
Do I use NL #2 Sum F=mv2/r
which=ma(sub)c ??
Also, how do I set up the math to solve for the angular speed?
Thank you
Yes, normal force = m v^2/r
max friction force up = mu m v^2/r
weight down = mg
slip when m g = m v^2/r
v^2 = r g
Yes, normal force = m v^2/r
max friction force up = mu m v^2/r
weight down = mg
slip when m g = mu m v^2/r
v^2 = r g / mu
A- Yes, you are correct. The force that keeps the people from falling out of the bottom of the ride is the normal force. The normal force is the force exerted by a surface to support the weight of an object resting on it.
B- To determine the minimum angular speed (ω) required for the people to not fall out, you can use the equation:
Frictional force (Ff) = μN
where:
Ff = frictional force
μ = coefficient of friction
N = normal force
In this case, since the people are standing against the inside of the cylinder, the normal force (N) is equal to the weight (mg) of the people. Thus, the frictional force can be calculated as:
Ff = μmg
Additionally, the frictional force can be expressed as the centripetal force required to keep the people from falling out of the ride:
Ff = mass (m) * radial acceleration (ac) =
m * (ω^2 * r)
where:
m = mass of the people
ac = radial acceleration
ω = angular speed
r = radius of the cylinder
Setting the two expressions for Ff equal to each other, we get:
μmg = mω^2r
Rearranging the equation, we can solve for the minimum angular speed (ω):
ω = sqrt(μg/r)
where:
g = acceleration due to gravity (approximately 9.8 m/s^2)
Substituting the given values, the minimum angular speed can be calculated using the equation above.
A- Yes, you are correct. The force that keeps the people from falling out of the bottom of the ride is the normal force. The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the normal force is the force exerted by the floor of the cylinder on the people standing inside.
B- To determine the minimum angular speed the cylinder must have so that the people don't fall out, we can use the concept of centripetal force. The centripetal force is the force that keeps an object moving in a circular path and is given by the formula:
Fc = m * v^2 / r
Where Fc is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius of the circular path.
In this case, the force providing the centripetal force is the friction force between the people and the cylinder. The friction force can be calculated as Ff = μ * N, where μ is the coefficient of friction and N is the normal force.
Setting the centripetal force equal to the friction force:
m * v^2 / r = μ * N
Since N is equal to the weight of the person standing on the ride, N = m * g, where g is the acceleration due to gravity.
m * v^2 / r = μ * m * g
Simplifying the equation, we can cancel out the mass (m) on both sides:
v^2 / r = μ * g
To solve for the minimum angular speed (ω), we can use the relationship: v = r * ω
Plugging in this value for v, we get:
(ω * r)^2 / r = μ * g
Simplifying further:
ω^2 * r = μ * g
Finally, solving for ω:
ω^2 = (μ * g) / r
ω = sqrt((μ * g) / r)
Using the given values of μ = 0.40 and r = 2.5 m, and using the standard acceleration due to gravity, g = 9.8 m/s^2, we can substitute these values into the equation to find the minimum angular speed.