(15 Marks) An amusement park ride consists of a large vertical cylinder
that spins about its axis fast enough that any person inside is
held up against the wall when the floor drops away. The coefficient
of static friction between the person and the wall is µs, and the radius
of the cylinder is R.
a. Show that the maximum period of revolution necessary to
keep the person from falling is T = (4πRµs/g)
1/2.
well, the friction force has to equal weigh
mv^2/r * µ=mg
or v^2=g*r/µ
but v= 2PIr/T
2PI^2 r^2/T^2=g*r/µ
T= sqrt(4πR*µ/g)
To derive the expression for the maximum period of revolution necessary to keep the person from falling, we can use the following steps:
Step 1: Determine the force acting on the person towards the center of the cylinder when at the maximum period of revolution.
When the floor drops away, the person is held up against the wall of the cylinder by the force of static friction. This frictional force provides the centripetal force that keeps the person moving in a circular path. The force of friction (Ff) can be expressed as:
Ff = μs * N
where μs is the coefficient of static friction and N is the normal force exerted by the wall on the person.
Step 2: Calculate the normal force (N) acting on the person.
The normal force is equal to the weight of the person, which is given by:
N = mg
where m is the mass of the person, and g is the acceleration due to gravity.
Step 3: Express the frictional force (Ff) in terms of the mass of the person.
Substituting the expression for N into the equation for Ff, we get:
Ff = μs * mg
Step 4: Equate the frictional force (Ff) to the centripetal force (Fc).
The centripetal force required to keep the person moving in a circular path is given by:
Fc = m * (v^2 / R)
where v is the velocity of the person and R is the radius of the cylinder.
Setting Ff equal to Fc, we have:
μs * mg = m * (v^2 / R)
Step 5: Solve for the maximum velocity (v) of the person.
Rearranging the equation, we find:
v^2 = μs * g * R
Taking the square root of both sides, we get:
v = sqrt(μs * g * R)
Step 6: Express the maximum velocity (v) in terms of the time period (T).
The maximum velocity occurs when the person is at the lowest point of the circular path. At this point, the change in position is equal to the circumference of the circle, 2πR, divided by the period of revolution (T).
Therefore, v = 2πR / T
Step 7: Substitute the expression for v from Step 5 into the equation from Step 6.
sqrt(μs * g * R) = 2πR / T
Squaring both sides of the equation, we have:
μs * g * R = (2πR / T)^2
Simplifying the right side, we get:
μs * g * R = 4π^2 * R^2 / T^2
Step 8: Rearrange the equation to solve for T.
T^2 = (4π^2 * R^2) / (μs * g)
Taking the square root of both sides, we finally obtain:
T = sqrt((4π^2 * R^2) / (μs * g))
Therefore, the maximum period of revolution necessary to keep the person from falling is T = sqrt((4π^2 * R^2) / (μs * g)).
To determine the maximum period of revolution (T) necessary to keep the person from falling, we need to consider the balance of forces acting on the person when they are pressed against the wall of the ride.
Assuming there is no slipping between the person and the wall, the frictional force provides the necessary centripetal force to keep the person in circular motion. The maximum value of static friction can be determined using the equation:
fs(max) = µs * N,
Where fs(max) is the maximum frictional force, µs is the coefficient of static friction, and N is the normal force acting on the person.
The normal force is equal to the weight of the person, which can be calculated as:
N = mg,
Where m is the mass of the person and g is the acceleration due to gravity.
Thus, the maximum frictional force is given by:
fs(max) = µs * mg.
The maximum frictional force must provide the centripetal force required for circular motion, which can be expressed as:
fs(max) = m * (v^2 / R),
Where v is the linear velocity of the person and R is the radius of the cylinder.
Since v = (2πR) / T (where T is the period of revolution), we can substitute this into the equation:
µs * mg = m * ((2πR / T)^2 / R).
Cancelling out the mass of the person (m) on both sides and rearranging the equation, we get:
µs * g = ((4π^2R) / T^2).
Solving for T, we find:
T^2 = (4πRµs) / g.
And taking the square root of both sides, we obtain the expression for the maximum period of revolution:
T = √((4πRµs) / g).
Hence, the maximum period of revolution necessary to keep the person from falling is T = √((4πRµs) / g).