Question

On the ride "spindletop" at the amusement park six flags over texas, people stood against the inner wall of a hollow vertical cylinder with radius 2.5 m. the cylinder started to rotate, and when it reached a constant rotation rate of 0.60 rev/s, the floor on which people were standing dropped about 0.5 m. the people remained pinned against the wall.

Answer 1

The minimum value of coefficient of static friction for which the person does not slide is $\boxed{0.276}$.

Further Explanation:

As the cylinder start to rotate, the people standing against the wall remain stick to the wall. So if we draw the free body diagram of the person, then we can see the balanced forces acting on person's body.

Given:

The rate of rotation of the cylinder is $0.60\text{ rev}/\text{s}$.

The radius of the cylinder is $2.50\text{ m}$.

Concept:

The friction force between the person’s body and the inner surface is balanced by the gravitational firce acting on the person's body.

$\boxed{f_\text{friction}=f_\tect{gravity}}$                                                   ...... (1)

The force of friction acting on the person's body is:

$f_\text{friction}=\mu N$

Here, $N$ is the normal reaction,  $\mu$ is coefficient of static friction.

The gravitational force acting on the person's body is:

$f_\text{gravity}=mg$

Here, $m$ is mass of the body, and $g$ is the acceleration due to gravity.

Substitute $\mu N$ for $f_\text{friction}$ and $mg$ for $f_\text{gravity}$ in equation (1).

${\begin{aligned}\mu N&=mg\\\mu&=\dfrac{mg}{N}\end{aligned}$

The Normal force acting on the person's body is balanced by the centrifugal force acting on the persion due to the rotation of the cylinder.

$\boxed{f_\text{normal}=f_\text{centrifugal}}$                                                  ...... (2)

The centrifugal force acting on the person's body is:

$f_\text{centrifugal}=m\omega^2r$

Here, $\omega$ is the angular speed of the cylinder, $r$ is the radius of the cylinder.

Substitute $m\omega^2r$ for $f_\text{centrifugal}$ in equation (2).

$N=m\omega^2r$

Substitute the value of $N$ in equation of static friction coefficient.

$\begin{aligned}\mu&=\dfrac{mg}{m\omega^2r}\\&=\dfrac{g}{\omega^2r}\end{aligned}$

The acceleration due to gravity on the surface of Earth is $9.81\text{ m}/\text{s}^2$.

The angular speed of rotation of the cylinder is given as:

$\begin{aligned}\omega&=2\pi\times\text{Revolutions per second}\\&=2\pi\times0.6\\&=3.77\text{ rad}/\text{s}\end{aligned}$

Substitute the values of $\omega$, $g$ and $r$ in above expression.

$\begin{aligned}\mu&=\dfrac{9.81}{(3.77)^2\times2.5}\\&=\dfrac{9.81}{35.53}\\&=0.276\end{aligned}$

Thus, the minimum value of coefficient of static friction for which the person does not slide is $\boxed{0.276}$.

Learn More:

1. For typical rubber-on-concrete friction, what is the shortest time brainly.com/question/3017271

2. Motion in two dimensions brainly.com/question/1913142

3. the time taken by a car to accelerate on a friction surface brainly.com/question/7174363

Answer Details:

Grade: Senior school

Subject: Physics

Chapter: Circular motion

Keywords:

Amusement park, ride, spindle top, six flags, texas, vertical cylinder, hollow, 2.5m, 0.60 rev/s, dropped about, minimum static friction coefficient, rotation rate 0.60.

Answer 2

a) In this case the forces are the centrifugal force Fcp, which is directed horizontally toward the wall; the force of static friction Ff with the wall, directed upward; the normal force Fn by the wall, which is directed away the wall; the force of gravity Fg, directed downwards. Then we have that the horizontal forces are all equal in magnitude; similarly the vertical forces are also all equal in magnitude.

 

b) The minimum coefficient s occurs when force of gravity is equals the max friction force, that is

Fg = Ff,max

m g = s Fn

 

Also, the normal force has equal magnitude to the centrifugal force:

m g = s Fcp

m g = s m w^2 r

g = s w^2 r

s = g / (r w^2)

 

With values: g = 9.81 m/s^2; r = 2.5 m; and w = 2pi * 0.60 = 3.77 rad/s; we find

 

s = 9.81 / (2.5 * 3.77^2) = 0.276