Chapter 5, Problem 2SP

A Ferris wheel with a radius of 15 m makes one complete rotation every 12 seconds.

1. a. Using the fact that the distance traveled by a rider in one rotation is 2πr, the circumference of the wheel, find the speed with which the riders are moving.
2. b. What is the magnitude of their centripetal acceleration?
3. c. For a rider with a mass of 50 kg, what is the magnitude of the centripetal force required to keep that rider moving in a circle? Is the weight of the rider large enough to provide this centripetal force at the top of the cycle?
4. d. What is the magnitude of the normal force exerted by the seat on the rider at the top of the cycle?
5. e. What will happen if the Ferris wheel is going so fast that the weight of the rider is not sufficient to provide the centripetal force at the top of the cycle?

To determine

The speed at which the riders are moving.

Answer to Problem 2SP

The speed at which the riders are moving is $7.85\text{ m/s}$.

Explanation of Solution

Given Info: The radius of the Ferris wheel is $15\text{ m}$ and the period of the rotation of the wheel is $12\text{ s}$.

Write the equation for the speed.

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

Here,

$v$ is the speed of the riders

$D$ is the distance travelled

$t$ is the time taken

Rewrite the above equation for the riders in the Ferris wheel.

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

Here,

$r$ is the radius of the wheel

$T$ is the time period of rotation of the wheel

Substitute $15\text{ m}$ for $r$ and $12\text{ s}$ for $T$ in the above equation to find $v$.

$\begin{array}{c}v=\frac{2\pi \left(15\text{ m}\right)}{12\text{ s}}\\ =7.85\text{ m/s}\end{array}$

Conclusion:

Thus the speed at which the riders are moving is $7.85\text{ m/s}$.

To determine

The magnitude of centripetal acceleration.

Answer to Problem 2SP

The magnitude of centripetal acceleration is $4.11{\text{ m/s}}^2$.

Explanation of Solution

Given Info: The radius of the Ferris wheel is $15\text{ m}$.

Write the equation for centripetal acceleration.

$a=\frac{v^2}{r}$

Here,

$a$ is the centripetal acceleration

Substitute $7.85\text{ m/s}$ for $v$ and $15\text{ m}$ for $r$ in the above equation to find $a$.

$\begin{array}{c}a=\frac{{\left(7.85\text{ m/s}\right)}^2}{15\text{ m}}\\ =4.11{\text{ m/s}}^2\end{array}$

Conclusion:

Thus the magnitude of centripetal acceleration is $4.11{\text{ m/s}}^2$.

To determine

The magnitude of the centripetal force required to keep the rider moving in a circle and whether the weight of the rider is large enough to provide the centripetal force at the top of the cycle.

Answer to Problem 2SP

The magnitude of the centripetal force required to keep the rider moving in a circle is $205.5\text{ N}$ and the weight of the rider is large enough to provide the centripetal force at the top of the cycle.

Explanation of Solution

Given Info: The mass of the rider is $50\text{ kg}$.

Write the equation for the centripetal force.

$F_c=ma$

Here,

$F_c$ is the magnitude of the centripetal force

Substitute $50\text{ kg}$ for $m$ and $4.11{\text{ m/s}}^2$ for $a$ in the above equation to find $F_c$.

$\begin{array}{c}{F}_c=\left(50\text{ kg}\right)\left(4.11{\text{ m/s}}^2\right)\\ =205.5\text{ N}\\ \approx 206\text{N}\end{array}$

Write the equation for the weight of the ride.

$W=mg$

Here,

$W$ is the weight of the rider

$g$ is the gravitational acceleration

The value of $g$ is $9.8{\text{ m/s}}^2$.

Substitute $50\text{ kg}$ for $m$ and $9.8{\text{ m/s}}^2$ for $g$ in the above equation to find $W$.

$\begin{array}{c}W=\left(50\text{ kg}\right)\left(9.8{\text{ m/s}}^2\right)\\ =490\text{ N}\end{array}$

$W>F_c$

Conclusion:

Thus the magnitude of the centripetal force required to keep the rider moving in a circle is $206\text{N}$ and the weight of the rider is large enough to provide the centripetal force at the top of the cycle.

To determine

The magnitude of the normal force exerted by the seat on the rider at the top of the cycle.

Answer to Problem 2SP

The magnitude of the normal force exerted by the seat on the rider at the top of the cycle is $284.5\text{ N}$.

Explanation of Solution

The forces acting on the rider at the top of the cycle are the normal force and the weight of the rider. The normal force acts upward and the weight of the rider acts downward. The net force is downward and it is equal to the centripetal force. This implies the difference between the weight of the rider and the centripetal force gives the magnitude of the normal force.

$F_N=W-F_c$

Here,

$F_N$ is the magnitude of the normal force

Substitute $490\text{ N}$ for $W$ and $206\text{ N}$ for $F_c$ in the above equation to find $F_N$.

$\begin{array}{c}{F}_N=490\text{ N}-206\text{ N}\\ =284\text{ N}\end{array}$

Conclusion:

Thus the magnitude of the normal force exerted by the seat on the rider at the top of the cycle is $284\text{ N}$.

To determine

What will happen if the Ferris wheel moves so fast that the weight of the rider is not sufficient to provide the centripetal force at the top of the cycle.

Answer to Problem 2SP

The rider will let go of the safety bar and he will fly out at a trajectory tangent to the Ferris wheel.

Explanation of Solution

On a Ferris wheel, the circular motion is vertical. The forces acting on the rider in the Ferris wheel are normal force and the weight of the rider.

At the bottom of the cycle the centripetal force needed for the vertical motion of the rider is provided by the weight of the rider. At the top of the cycle the centripetal acceleration is provided by the weight of the rider.

When the Ferris wheel moves so fast, the weight of the rider becomes not sufficient to provide the centripetal force at the top of the cycle. Then the rider will let go of the safety bar and he will fly out of at a trajectory tangent to the Ferris wheel. At this moment, gravity will also accelerate the rider to the ground.

Conclusion:

Thus the rider will let go of the safety bar and he will fly out at a trajectory tangent to the Ferris wheel.