In a popular amusement park ride, a rotating cylinder of radius R = 3.10 m is set in rotation at an angular speed of 4.20 rad/s, as in the figure shown below. The floor then drops away, leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction between a rider's clothing and the wall is needed to keep the rider from slipping? Hint: Recall that the magnitude of the maximum force of static friction is equal to μsn, where n is the normal force—in this case, the force causing the centripetal acceleration.
μs =
Well, this is quite the sticky situation we've got here! We need to find the minimum coefficient of friction between the rider's clothing and the wall to keep them from slipping.
First things first, let's find the centripetal acceleration of the rider. The centripetal acceleration is given by the formula:
ac = rω^2
Where r is the radius of the cylinder (3.10 m) and ω is the angular speed (4.20 rad/s).
Plugging in the values, we get:
ac = (3.10 m)(4.20 rad/s)^2
That comes out to a centripetal acceleration of 54.996 m/s^2. Quite the dizzying speed!
Now, the force of friction is what keeps the rider from slipping down the wall. The maximum force of static friction can be found using the formula:
Ff = μsn
Where Ff is the force of friction, μs is the coefficient of friction, and n is the normal force.
In this case, the normal force is what causes the centripetal acceleration, so we can find it using the formula:
n = m * g
Where m is the mass of the rider and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now, we don't have the mass of the rider, so we'll have to leave that as a variable for now. Let's plug in the known values:
Ff = μs * (m * g)
The rider is just barely not slipping, so the force of friction is equal to the centripetal force:
Ff = m * ac
Setting the two equations equal to each other, we get:
μs * (m * g) = m * ac
The mass cancels out, leaving us with:
μs * g = ac
Now we can solve for the coefficient of friction (μs):
μs = ac / g
Plugging in the values we found earlier:
μs = (54.996 m/s^2) / (9.8 m/s^2)
Doing the math, we find that the minimum coefficient of friction needed to keep the rider from slipping is approximately 5.61.
So, we need a coefficient of friction "μs" that's greater than 5.61 to keep the rider stuck to the wall. That's one sticky situation indeed! But let's not slip up on our calculations - remember, safety is no joke!
To determine the minimum coefficient of friction (μs) needed to keep the rider from slipping, we can start by analyzing the forces acting on the rider.
1. Centripetal Force (Fc):
The rider is essentially moving in a circular path due to the rotation of the cylinder. The centripetal force required to keep the rider moving in this circular path is given by:
Fc = m * a = m * (v^2 / R),
where m is the mass of the rider, a is the centripetal acceleration, v is the linear velocity of the rider, and R is the radius of the cylinder.
2. Normal Force (Fn):
The normal force (Fn) acting on the rider is the force that is perpendicular to the wall. In this case, it is equal to the gravitational force acting on the rider:
Fn = m * g,
where g is the acceleration due to gravity.
3. Frictional Force (Ff):
The frictional force (Ff) acting between the rider's clothing and the wall opposes the motion and prevents slipping. It is given by:
Ff = μs * Fn.
To find the minimum coefficient of friction (μs), we can equate the magnitudes of the centripetal force and the maximum static frictional force:
Fc = Ff.
Substituting the values and solving for μs, we get:
m * (v^2 / R) = μs * m * g.
Canceling out the mass (m) on both sides of the equation, we have:
(v^2 / R) = μs * g.
Simplifying further, we get:
μs = (v^2 / (R * g)).
Substituting the given values, we have:
μs = (4.20 rad/s)^2 / (3.10 m * 9.81 m/s^2).
Calculating this expression will give you the minimum coefficient of friction required to keep the rider from slipping.
To find the minimum coefficient of friction between a rider's clothing and the wall, we need to determine the centripetal force acting on the rider and then calculate the force of static friction required to keep the rider from slipping.
First, let's find the centripetal force acting on the rider. The centripetal force is given by the equation:
Fc = m * ac
Where Fc is the centripetal force, m is the mass of the rider, and ac is the centripetal acceleration.
The centripetal acceleration is given by:
ac = R * ω²
Where R is the radius of the rotating cylinder and ω is the angular speed.
Given: R = 3.10 m, ω = 4.20 rad/s
Now we can calculate the centripetal acceleration:
ac = (3.10 m) * (4.20 rad/s)²
= 55.08 m/s²
Next, we need to determine the centripetal force:
Fc = m * ac
To find the normal force n, we can use the gravitational force acting on the rider:
mg = n
Where m is the mass of the rider and g is the acceleration due to gravity. Since the riders are in a vertical position, the normal force is equal to the weight of the riders.
Now, let's find the normal force:
n = mg
Given that g is approximately 9.8 m/s², we can substitute this value into the equation to find the normal force.
Finally, we can calculate the minimum coefficient of friction using the equation for the maximum force of static friction:
Fs,max = μs * n
We want to find the minimum coefficient of friction, so the maximum force of static friction should be equal to the centripetal force:
Fs,max = Fc
Substituting the values, we have:
μs * n = Fc
Now we can solve for the minimum coefficient of friction (μs):
μs = Fc / n
Note: Make sure to properly convert units, mass should be in kilograms (kg), and force in newtons (N).
So μs = (Fc / n)
mu * normal force = m g
normal force = m omega^2 R
so
m g = mu m omega^2 R
mu = g/(omega^2 R)
or as we could have guessed
mu = gravity acceleration/centripetal acceleration
mu = 9.81/[ 4.2^2 * 3.1 ]