An amusement park ride consists of a large vertical cylinder that spins around its axis fast enough such that any person inside is held up against the wall when the floor drops away. The coefficient of static friction between person and wall is µs, and the radius of the cylinder is R.
Show that the maximum period of revolution necessary to keep the person from falling is T = (4ð2Rµs/g)^1/2
12
To find the maximum period of revolution necessary to keep the person from falling, we need to analyze the forces acting on the person when they are stuck against the wall of the spinning cylinder.
Let's consider a free-body diagram of the person. The two forces acting on them are the gravitational force (mg) pulling them downward and the static friction force (fs) pushing them towards the center of the cylinder.
Since the person is not falling, the static friction force exactly balances the gravitational force. Therefore, we can write:
fs = mg
Now, let's analyze the static friction force. The maximum static friction force (fs = µs*N) is proportional to the normal force (N) between the person and the wall. In this case, the normal force is equal to the gravitational force (N = mg).
So, the maximum static friction force can be written as:
fs = µs * mg
Now, let's consider the forces acting on the person in the radial direction. The force causing the person to move towards the center of the cylinder is the net radial force (Fr) and is given by:
Fr = ma
Where m is the mass of the person and a is the acceleration towards the center of the cylinder.
Since the person is pushed against the wall, the net radial force is equal to the static friction force in magnitude:
Fr = fs
We can rewrite this equation using the relationship between centripetal acceleration (a) and angular velocity (ω) as a = Rω², where R is the radius of the cylinder.
So, we have:
fs = mRω²
Now, let's substitute the maximum static friction force (fs = µs * mg) into the equation:
µs * mg = mRω²
We can cancel out the mass (m) from both sides:
µs * g = Rω²
Finally, solve for the angular velocity (ω) by rearranging the equation:
ω² = µs * g / R
ω = √(µs * g / R)
The period of revolution (T) is the time taken for one complete revolution and is given by T = 2π / ω. Substituting the value of ω in terms of the variables, we get:
T = 2π / √(µs * g / R)
To simplify this further, multiply the numerator and denominator by R:
T = (2πR) / √(µs * g * R / R²)
T = (2πR) / √(µs * g * R²)
T = (2πR) / √(µs * g * R² / 1)
T = (2πR) / √(µs * g * R² * 1)
T = (2πR) / √(µs * g * R² * 1)
T = (2πR) / √(µs * g * R²)
T = (2πR / R) * √(1 / (µs * g))
T = (2π / 1) * √(1 / (µs * g))
T = 2π * √(1 / (µs * g))
T = 2π * √(1 / µs) * √(1 / g)
T = 2π * √(1 / µs) * √(1 / g)
T = (2π / √µs) * √(1 / g)
Finally, simplify the expression:
T = (4π / 2) * √(2 / µs) * √(1 / g) = (4π / 2) * √(2 * µs / g)
T = (4π / 2) * √(2µs / g)
T = (4π / 2) * √(2µs / g)
T = (2π) * √(2µs / g)
Hence, the maximum period of revolution necessary to keep the person from falling is T = (4ð2Rµs/g)^1/2.
To show that the maximum period of revolution necessary to keep the person from falling is T = (4π^2Rµs/g)^(1/2), we can use the concept of equilibrium.
In this case, the maximum period of revolution occurs when the static friction between the person and the wall is at its maximum, which means the person is just about to slip or fall.
Let's consider the forces acting on the person when they are at the highest point on the vertical cylinder:
1. Gravity (mg) acts downward.
2. Normal force (N) acts perpendicular to the wall, directed towards the center of the cylinder.
3. Frictional force (F friction) acts horizontally towards the center of the cylinder, opposing the tendency of the person to fall.
At the highest point, the frictional force F friction provides the centripetal force required to keep the person in circular motion. The centripetal force is given by:
F centripetal = m × a centripetal
Where:
m is the mass of the person
a centripetal is the acceleration towards the center of the cylinder
The centripetal acceleration is given by:
a centripetal = v^2 / R
Where:
v is the velocity of the person in circular motion
R is the radius of the cylinder
Now, we can equate the frictional force to the centripetal force:
F friction = m × a centripetal
Since the person is just about to slip or fall, the frictional force is at its maximum and is equal to the maximum static friction:
F friction = µs × N
The normal force N can be related to the weight of the person:
N = mg
Substituting the expressions for F friction, a centripetal, N, and rearranging the terms, we get:
µs × mg = m × v^2 / R
Canceling out the mass, we have:
µs × g = v^2 / R
Now, solving for v, we have:
v = √(µs × g × R)
The period of revolution T is given by the time taken for one complete revolution (2πR) divided by the velocity:
T = (2πR) / v
Substituting the expression for v, we get:
T = (2πR) / √(µs × g × R)
Simplifying further, we have:
T = (2π√R) / √(µs × g)
Finally, we can rearrange the terms to get the desired result:
T = (4π^2Rµs/g)^(1/2)
Thus, we have shown that the maximum period of revolution necessary to keep the person from falling is T = (4π^2Rµs/g)^(1/2).