A cylindrical carnival ride is spinning with a frequency of 0.5 Hz in a circle of radius 3 m which allows a 60 kg person to “stick” to the side of the ride as it spins.

T = 2s v = 9.42 m/s
A) Find the normal force exerted on the person.
B) What minimum µs is necessary between the rider and the wall for the rider to stick to the wall?

A)The normal force exerted by the cylinder must equal the centripetal force, M v^2/R

B) The static coefficient of friction. µs, must be large enough so that
µs*M v^2/R >or= M g

µs >or= g*R/v^2

To solve this problem, we can start by finding the acceleration of the person due to the circular motion.

Given:
Frequency (f) = 0.5 Hz
Radius (r) = 3 m

Step 1: Calculate the angular velocity (ω):
ω = 2πf
= 2π(0.5)
= π rad/s

Step 2: Calculate the linear velocity (v):
v = ωr
= π(3)
= 3π m/s

Step 3: Calculate the centripetal acceleration (ac):
ac = v²/r
= (3π)²/3
= 3π² m/s²

Now, let's move on to answering the questions.

A) Find the normal force exerted on the person:
The normal force (N) balances the gravitational force (mg) acting on the person.

Given:
Mass of the person (m) = 60 kg
Acceleration due to gravity (g) ≈ 9.8 m/s²

Step 4: Calculate the normal force (N):
N = mg + mac
= mg + m(3π²)
= 60(9.8) + 60(3π²)
≈ 588 + 563π² N
≈ 1891 N (approximately)

The normal force exerted on the person is approximately 1891 N.

B) What minimum µs is necessary between the rider and the wall for the rider to stick to the wall?
The rider will stick to the wall if the static frictional force (fs) between the rider and the wall is greater than or equal to the maximum possible static frictional force (µsN), where µs is the coefficient of static friction.

Step 5: Calculate the static frictional force:
fs = µsN

Given:
Normal force (N) = 1891 N

We can rearrange the equation to solve for µs:
µs = fs/N

In this case, the maximum possible static frictional force is equal to the centripetal force acting on the person (mac):

fs = mac

Step 6: Calculate the minimum µs required:
µs = mac / N
= (3π²) / 1891
≈ 0.00505 (approximately)

The minimum coefficient of static friction (µs) required between the rider and the wall for the rider to stick to the wall is approximately 0.00505 (or 0.5%).

To solve this problem, we need to use the concepts of centripetal force, normal force, and frictional force.

A) Find the normal force exerted on the person:
The normal force is the force exerted by a surface perpendicular to the surface. In this case, it is the force that prevents the person from falling off the ride.

We know the mass of the person is 60 kg.

To find the normal force, we need to consider the forces acting on the person. Since the person is "sticking" to the side of the ride, the normal force and gravity are the only forces acting on the person in the vertical direction.

The gravitational force (weight) is given by: Fg = mg, where m is the mass of the person and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Fg = 60 kg * 9.8 m/s^2 = 588 N

In this case, the normal force (N) is equal in magnitude and opposite in direction to the gravitational force. So the normal force exerted on the person is 588 N.

B) What minimum µs is necessary between the rider and the wall for the rider to stick to the wall:
To determine the minimum coefficient of static friction (µs), we need to consider the net horizontal force acting on the person.

The net horizontal force in this scenario is provided by the centripetal force, which is given by:
Fc = (mv^2)/r, where m is the mass of the person, v is the velocity, and r is the radius of the ride.

Fc = (60 kg * (9.42 m/s)^2) / 3 m = 582.34 N

The frictional force (Ff) is the force needed to maintain static equilibrium or prevent the person from sliding off the ride. It acts in the opposite direction of the net horizontal force.

By applying Newton's second law in the horizontal direction, we can write:
Fnet = Ff
Fc = Ff
(60 kg * (9.42 m/s)^2) / 3 m = µs * N

Now, substituting the value of N (which is equal to 588 N) into the equation, we can solve for µs:
(60 kg * (9.42 m/s)^2) / 3 m = µs * 588 N

µs = [(60 kg * (9.42 m/s)^2) / 3 m] / 588 N

Evaluating this expression, we find that the minimum µs necessary between the rider and the wall for the rider to stick to the wall is approximately 0.831.