Question

In an amusement park ride called The Roundup, passengers stand inside a 16.0m -diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane, as shown in the figure.

Part A) Suppose the ring rotates once every 3.60s. If a rider's mass is 55.0kg, with how much force does the ring push on her at the top of the ride?

______N

Part B) Suppose the ring rotates once every 3.60s. If a rider's mass is 55.0kg, with how much force does the ring push on her at the bottom of the ride?

_____N

Part C) What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?

_____seconds

Answer 1

Concepts and reason

The concepts required to solve the problems are the centripetal force and force equilibrium.

First use the period and the total circumference of the round up wheel to calculate the velocity of the rider. Then, use the velocity to calculate the centripetal force on the rider. Finally, use the force equilibrium to calculate the normal force.

Fundamentals

The radius of a circle is,

$r = \frac{d}{2}$

Here, $d$ is the diameter of the circle.

The circumference of a circle is,

$C = 2\pi r$

The speed is the total distance covered in a given time. The speed is,

$v = \frac{x}{t}$

Here, $x$ is the distance covered and $t$ is the time taken.

The centripetal force acting on a body moving in circular path is,

${F_C} = \frac{{m{v^2}}}{r}$

Here, $m$ is the mass of the body.

Weight of a body is,

$W = mg$

Here, $g$ is the acceleration due to gravity.

Part A

The radius of the wheel is,

$r = \frac{d}{2}$

Here, $d$ is the diameter of the wheel.

Substitute $16.0\,{\rm{m}}$ for $d$ . The radius of the wheel is,

$\begin{array}{c}\\r = \frac{{16.0\,{\rm{m}}}}{2}\\\\ = 8.00\,{\rm{m}}\\\end{array}$

The circumference of the wheel is,

$C = 2\pi r$

Substitute $8.00\,{\rm{m}}$ for $r$ and $3.14\,{\rm{rad}}$ for $\pi$ . The circumference of the wheel is,

$\begin{array}{c}\\C = 2\left( {3.14\,{\rm{rad}}} \right)\left( {8.00\,{\rm{m}}} \right)\\\\ = 50.3\,{\rm{m}}\\\end{array}$

The speed of rider is,

$v = \frac{C}{T}$

Here, $C$ is the total distance covered by the rider in one revolution and $T$ is the time period.

Substitute $50.3\,{\rm{m}}$ for $C$ and $3.60\,{\rm{s}}$ for $T$ . The speed of the rider is,

$\begin{array}{c}\\v = \frac{{50.3\,{\rm{m}}}}{{3.60\,{\rm{s}}}}\\\\ = 14.0\,{\rm{m/s}}\\\end{array}$

The centripetal force acting on the rider is,

${F_C} = \frac{{m{v^2}}}{r}$

Here, $m$ is the mass of the rider.

Substitute $55.0\,{\rm{kg}}$ for $m$ , $14.0\,{\rm{m/s}}$ for $v$ , and $8.00\,{\rm{m}}$ for $r$ . The centripetal force is,

$\begin{array}{c}\\{F_C} = \frac{{\left( {55.0\,{\rm{kg}}} \right){{\left( {14.0\,{\rm{m/s}}} \right)}^2}}}{{\left( {8.00\,{\rm{m}}} \right)}}\\\\ = 1.35 \times {10^3}\,{\rm{N}}\\\end{array}$

The weight of the rider is,

$W = mg$

Substitute $55.0\,{\rm{kg}}$ for $m$ and $9.81\,{\rm{m/}}{{\rm{s}}^2}$ for $g$ . The weight of the rider is,

$\begin{array}{c}\\W = \left( {55.0\,{\rm{kg}}} \right)\left( {9.81\,{\rm{m/}}{{\rm{s}}^2}} \right)\\\\ = 5.40 \times {10^2}\,{\rm{N}}\\\end{array}$

The force equilibrium at the top of the wheel is,

$- F - W = - {F_C}$

Here, $F$ is the force that the ring pushes the rider.

Rewrite the equation in terms of the force that the ring pushes on the rider at the top. The force is,

$F = {F_C} - W$

Substitute $1.35 \times {10^3}\,{\rm{N}}$ for ${F_C}$ and $5.40 \times {10^2}\,{\rm{N}}$ for $W$ . The force is,

$\begin{array}{c}\\F = \left( {1.35 \times {{10}^3}\,{\rm{N}}} \right) - \left( {5.40 \times {{10}^2}\,{\rm{N}}} \right)\\\\ = 801\,{\rm{N}}\\\end{array}$

Part B

The force equilibrium at the bottom of the wheel is,

$F - W = {F_C}$

Rewrite the equation in terms of the force that the ring pushes on the rider at the bottom. The force is,

$F = {F_C} + W$

Substitute $1.35 \times {10^3}\,{\rm{N}}$ for ${F_C}$ and $5.40 \times {10^2}\,{\rm{N}}$ for $W$ . The force is,

$\begin{array}{c}\\F = \left( {1.35 \times {{10}^3}\,{\rm{N}}} \right) + \left( {5.40 \times {{10}^2}\,{\rm{N}}} \right)\\\\ = 1.879 \times {10^3}\,{\rm{N}}\\\end{array}$

Part C

The force required to keep the rider in a circular motion without falling, at the top of the wheel is,

$\begin{array}{c}\\W = {F_C}\\\\mg = \frac{{m{v^2}}}{r}\\\end{array}$

Rewrite the equation in terms of speed of the rider. The speed is,

$v = \sqrt {rg}$

Substitute $8.00\,{\rm{m}}$ for $r$ and $9.81\,{\rm{m/}}{{\rm{s}}^2}$ for $g$ . The speed is,

$\begin{array}{c}\\v = \sqrt {\left( {8.00\,{\rm{m}}} \right)\left( {9.81\,{\rm{m/}}{{\rm{s}}^2}} \right)} \\\\ = 8.86\,{\rm{m/s}}\\\end{array}$

The time required for the rider to complete one revolution is,

$t = \frac{C}{v}$

Substitute $50.3\,{\rm{m}}$ for $C$ and $8.86\,{\rm{m/s}}$ for $v$ . The time is,

$\begin{array}{c}\\t = \frac{{50.3\,{\rm{m}}}}{{8.86\,{\rm{m/s}}}}\\\\ = 5.68\,{\rm{s}}\\\end{array}$

Ans: Part A

The force that the ring pushes the rider at the top is $801\,{\rm{N}}$ .