Q105P - University Physics with Modern Physics

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 dropped about 0.5 m. The people remained pinned against the wall without touching the floor. 

(a) Draw a force diagram for a person on this ride after the floor has dropped. (b) What minimum coefficient of static friction was required for the person not to slide downward to the new position of the floor? 

(c) Does your answer in part (b) depend on the person’s mass? (Note: When such a ride is over, the cylinder is slowly brought to rest. As it slows down, people slide down the walls to the floor.)

Step-by-Step Solution

Answer

(a) The force diagram is as follows.

(b) The required static friction is 0.28.

 

(c) No

Step 1: Identify the given information

• The radius of the inner wall, $\mathrm{r}=2.5 \mathrm{m}$. 
• The rotation rate is $0.60\mathrm{rev}/\mathrm{s}$.
• The drop I floor is 0.5 E  

Step 2: Describe the rotational speed and acceleration

The speed at which a particle rotates about an axis is known as rotational speed, and the rotational acceleration is the change in the change in rotational speed per unit of time.

Step 3: Draw a force diagram for a person

(a)

 

The force diagram for a person on the ride is shown below.

Step 4: Find the minimum coefficient of static friction

(b)

 

The net force acting on the person along the x-axis is given by,

$$\begin{gathered} \sum _{ }{F}_x=m{a}_x \\ {F}_n=mr{\omega }^2 \\ =m\left(2.5m\right){\left(3.769rad/s\right)}^2 \\ =35.51m \end{gathered}$$ 

 

Here, m of person r is the radius, and $\omega$ is angular velocity.

 

The net force acting on the person along the y-axis is given by,

$$\begin{gathered} \sum _{ }{F}_y=m{a}_y \\ f=mg \\ {\mu }_s{F}_n=mg \\ {\mu }_s=\frac{mg}{F_n} \end{gathered}$$ 

 

Here, ${\mu }_s$ is the coefficient of static friction, $F_n$ is the normal force, and g is the acceleration due to gravity.

 

Substitute $9.8m/s^2$ for g, and 35.51m for $F_n$ in the above equation, and we get,

$$\begin{gathered} {\mu }_s=\frac{m\left(9.8 m/s^2\right)}{35.51m} \\ =0.28 \end{gathered}$$ 

 

Therefore, the required static friction is 0.28.

Step 5: Find whether the answer in part (b) depends on the person’s mass

(c)

 

From the above calculation, it can be observed that the person’s mass cancels out, and the final answer does not depend on the person’s mass. It depends on the radius of the cylinder and its rotation frequency.

 

Therefore, the answer does not depend on the person’s mass.