Question

An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away (Fig. P6.59). The coefficient of static friction between person and wall is ${\mathit{\mu }}_{\mathbf{s}}$, and the radius of the cylinder is$\mathit{R}$. (a) Show that the maximum period of revolution necessary to keep the person from falling is$\mathit{T}\mathbf{=}{\left(4{\pi }^{2}R{\mu }_s/g\right)}^{\mathbf{1}\mathbf{/}\mathbf{2}}$. (b) If the rate of revolution of the cylinder is made to be somewhat larger, what happens to the magnitude of each one of the forces acting on the person? What happens in the motion of the person? (c) If the rate of revolution of the cylinder is instead made to be somewhat smaller, what happens to the magnitude of each one of the forces acting on the person? How does the motion of the person change?

Short answer

a) The maximum period of revolution necessary to keep the person from falling is $T=\sqrt{\frac{4{\pi }^2{R}^2{\mu }_s}{g}}$.

b) If the rate of revolution of the cylinder is made to be somewhat larger, then the centrifugal and normal forces increase while frictional force stays equal to the person's weight.

c) If the rate of revolution of the cylinder is made to be somewhat smaller, all the force decreases, including centrifugal, normal, and frictional forces.

Step-by-step solution

Defining centripetal force

The force required to move the particle in circular motion is known as centripetal force.

The expression for the centripetal force is given by

$F=m\frac{v^2}{R}$

Here $m$is the mass of the object,$v$ is the speed and$R$ is the radius,$F$ is the centrifugal force. Centrifugal force is also known by the name of centripetal force.

Obtaining the expression for the maximum period of revolution

The expression for the speed is given by

$v=\frac{2\pi R}{T}$

Here is the speed, is the radius of the circular path and is the period.

As, here normal force is equal to the centripetal force.

$F=m\frac{v^2}{R}$

$\Rightarrow n=m\frac{v^2}{R}$ ....... (1)

The static force is given by

$f_s=mg$ ........ (2)

Also,

$f_s={\mu }_{s}n$ ...... (3)

From equation (2) and (3)

$mg={\mu }_{s}n$

$\Rightarrow n=\frac{mg}{{\mu }_s}$ ...... (4)

From equation (1) and (4)

$$\begin{gathered} m\frac{v^2}{R}=\frac{mg}{{\mu }_s} \\ {v}^2=\frac{gR}{{\mu }_s} \end{gathered}$$

Substitute$\frac{2\pi R}{T}$for$v$in the above equation.

role="math" localid="1663673016026" $$\begin{gathered} {\left(\frac{2\pi r}{T}\right)}^2=\frac{gR}{{\mu }_s} \\ \frac{4{\pi }^2{R}^2}{T^2}=\frac{gR}{{\mu }_s} \\ {T}^2=\frac{4{\pi }^2{R}^2{\mu }_s}{g} \\ T=\sqrt{\frac{4{\pi }^2{R}^2{\mu }_s}{g}} \end{gathered}$$

Therefore, the maximum period of revolution necessary to keep the person from falling is $T=\sqrt{\frac{4{\pi }^2{R}^2{\mu }_s}{g}}$.

Explanation for (b)

(b)

If the rate of revolution of the cylinder is made to be somewhat larger, then the centrifugal and normal forces increases, while frictional force stays equal to person weight.

Explanation for (c)

(c)

If the rate of revolution of the cylinder is made to be somewhat smaller, all the forces decrease, including centrifugal force, normal force, and friction force.

The period of the motion in the presence of frictional force depends upon the coefficient of friction and the radius of the circular path.