An amusement park ride consists of a large vertical cylinder that spins about

Figure P6.61
its axis fast enough such that any person inside is held up against the wall
persons who walk around on the inner wall of the outer rim. Find the rate of rotation of the wheel (in revolutions
when the floor drops away. The coefficient of static friction between person
per minute) that will produce this effect.
and wall is μ , and the radius of the cylinder is R. (a) Draw the free body s 63. An amusement park ride consists of a rotating circular
diagram of the woman inside the cylinder. (b)platform 8.00 m in diameter from which 10.0-kg seats re suspended at the end of 2.50-m massless chains
aWhat is the minimum velocity
(Fig. P6.63). When the system rotates, the chains make an s angle 􏰃 􏰀 28.0° with the vertical. (a) What is the speed of
(v) of the woman, in terms of R, μ , and g, to keep her from falling. each seat? (b) Draw a free-body diagram of a 40.0-kg child

Well, this is quite a physics problem! Let's see if I can provide a humorous response.

Why did the amusement park ride start spinning so fast? Because it wanted to make sure no one could escape and go on a bathroom break! It's all about keeping everyone on their toes, or in this case, on the walls!

But back to the question at hand. To find the rate of rotation needed for the effect, we need to consider the minimum velocity (v) of the woman inside the cylinder. Let's assume she wants to stay put against the wall, just like that stubborn piece of spinach between your teeth that refuses to leave.

Now, to prevent her from falling, we need to take into account the coefficient of static friction between the person and the wall (μ), the radius of the cylinder (R), and good ol' gravity (g). It's like a delicate dance between the forces!

So, the minimum velocity (v) can be calculated using the equation v = μRg. Just remember, it's not about how fast you spin, but about how fast you spin to keep everyone glued to the walls. Safety first, my friends!

I hope that brings a smile to your face while tackling this physics problem. Now, let's get back to clowning around! 🤡

To find the rate of rotation of the amusement park ride to produce the effect of people being held up against the wall, we need to consider the forces acting on a person inside the cylinder.

(a) Drawing a free body diagram of the woman inside the cylinder:
- Draw a circle to represent the woman inside the cylinder.
- Draw arrows representing the forces acting on the woman.
- The force of gravity acts vertically downward from the woman's center of mass.
- The normal force acts radially inward towards the center of the cylinder.
- The static friction force acts tangentially and radially outward, preventing the woman from slipping down the wall.
- Label the forces accordingly.

(b) Finding the minimum velocity (v) of the woman:
- The net force acting radially inward should equal the centripetal force required to keep the woman against the wall.
- The centripetal force is given by the product of the woman's mass and the radial acceleration.
- The radial acceleration can be calculated using the rotational speed (ω) of the wheel and the radius (R) of the cylinder.
- The normal force minus the force of gravity should equal the static friction force.
- Set up equations using the force components and solve for the minimum velocity (v), in terms of R, μ, and g.

Once you have determined the minimum velocity, you can convert it to the rate of rotation in revolutions per minute by considering how many times the woman would travel around the circumference of the cylinder in a minute, based on the given diameter of the platform.