91影视

• 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

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.

(a)

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

(b)

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

$$\begin{gathered} \underset{}{鈭�}{F}_x=m{a}_x \\ {F}_n=mr{蝇}^2 \\ =m\left(2.5m\right){\left(3.769rad/s\right)}^2 \\ =35.51m \end{gathered}$$

Here,m of personr is the radius, and$蝇$ is angular velocity.

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

$$\begin{gathered} \underset{}{鈭�}{F}_y=m{a}_y \\ f=mg \\ {渭}_s{F}_n=mg \\ {渭}_s=\frac{mg}{F_n} \end{gathered}$$

Here,${渭}_s$ 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,

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

Therefore, the required static friction is 0.28.

(c)

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

Therefore, the answer does not depend on the person鈥檚 mass.