Chapter 5, Problem 2SP

(a)

To determine

The speed of the car with which the riders are moving in the Ferris wheel, If the radius of the wheel is $14\text{m}$ and it makes one complete rotation every $9\text{seconds}$.

Answer to Problem 2SP

The speed of the car with which the riders are moving is $9.77\text{m}/\text{s}$.

Explanation of Solution

Given Info: The radius of the curve is $14\text{m}$ and time taken for every rotation is $9\text{seconds}$.

Write the equation for the speed with which the riders are moving.

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

Here,

$v$ is the speed of the riders

$r$ is the radius of the wheel

$T$ is the time period of rotation

The time period is the time taken for an object to complete one rotation. Therefore the time period of rotation of the riders is $9\text{s}$.

Substitute $14\text{m}$ for $r$ and $9\text{s}$ for $T$ in the above equation to find $v$.

$\begin{array}{c}v=\frac{2\times 3.14\times \left(14\text{m}\right)}{9\text{s}}\\ =9.77\text{m}/\text{s}\end{array}$

Conclusion:

Thus the magnitude of the centripetal acceleration of the car is $9.77\text{m}/\text{s}$.

(b)

To determine

The magnitude of the centripetal acceleration for the circular motion of the Ferris wheel.

Answer to Problem 2SP

The magnitude of the centripetal acceleration for the circular motion of the Ferris wheel is $6.82\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Given Info: The radius of the Ferris wheel is $14\text{m}$.

Write the equation for centripetal acceleration.

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

Here,

$a_c$ is the centripetal acceleration

$v$ is the speed of the motion

$r$ is the radius of the Ferris wheel

Substitute $9.77\text{m}/\text{s}$ for $v$ and $14\text{m}$ for $r$ in the above equation to find $a_c$.

$\begin{array}{c}{a}_c=\frac{{\left(9.77\text{m}/\text{s}\right)}^2}{14\text{m}}\\ =6.82\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Thus, the centripetal acceleration of the circular motion is $6.82\text{m}/{\text{s}}^{\text{2}}$.

(c)

To determine

The magnitude of the centripetal force required to keep a rider of mass $35\text{kg}$ moving in a circle and to check whether the weight of the rider is large enough to provide this centripetal force at the top of the cycle.

Answer to Problem 2SP

The magnitude of the centripetal force required to keep a rider moving in a circle is $239\text{N}$.

Explanation of Solution

Given Info: The mass of the rider is $35\text{kg}$.

Write the expression for the magnitude of centripetal force.

$F_c=m{a}_c$

Here,

$F_c$ is the centripetal force

$m$ is the mass of the rider

$a_c$ is the centripetal acceleration

Substitute $35\text{kg}$ for $m$ and $6.82\text{m}/{\text{s}}^{\text{2}}$ for $a_c$ in  the above equation to get $F_c$.

$\begin{array}{c}{F}_c=\left(35\text{kg}\right)\left(6.82\text{m}/{\text{s}}^{\text{2}}\right)\\ =238.7\text{N}\\ =239\text{N}\end{array}$

The centripetal force required to keep the rider in the circular path is $239\text{N}$.

At the top of the cycle the weight of the body can give this centripetal force, if the weight is equal to or greater than the centripetal force. If the weight of the body is less than centripetal force the motion, the weight cannot provide centripetal force.

Write the expression for the weight of the body.

$W\text{= }mg$

Here,

$W$ is the weight of the rider

$m$ is the mass of the rider

$g$ is the acceleration due to gravity of the rider

Substitute $35\text{kg}$ for $m$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ in the above equation to get $\text{W}$.

$\begin{array}{c}W\text{=}\left(35\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =343\text{N}\end{array}$

Since the weight of the rider is greater than the centripetal force, weight can give provide centripetal force required for the circular motion.

Conclusion:

Thus, the magnitude of the centripetal force required to keep a rider moving in a circle is $239\text{N}$ and the weight of the rider is large enough to provide this centripetal force.

(d)

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 $104\text{N}$.

Explanation of Solution

Given Info: The mass of the rider is $35\text{kg}$.

The weight of the rider is $343\text{N}$ and the centripetal force required to keep in circular motion is $239\text{N}$

Since the rider is accelerates towards the center of the circle, the net force is the centripetal force acting towards the center.

Write the expression for the net force on the rider.

$F_{\text{net }}=F_c=W-F_N$

Here,

$F_{\text{net}}$ is the net force acting towards the center of the wheel

$F_N$ is the normal force

The negative sign in the above equation indicates that the weight of the rider and the normal force are opposite in direction.

Rearrange the above equation to get $F_N$.

$F_N=W-F_c$

Substitute $343\text{N}$ for $\text{W}$ and $239\text{N}$ for $F_c$ in the above equation to get $F_N$.

$\begin{array}{c}{F}_N=343\text{N}-239\text{N}\\ \text{=104}\text{N}\end{array}$

Therefore the normal force acting on the rider by the seat at the top of the wheel is $104\text{N}$.

Conclusion:

Thus, the magnitude of the normal force exerted by the seat on the rider at the top of the cycle is $104\text{N}$.

(e)

To determine

State of the rider 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.

Answer to Problem 2SP

The rider will fly out at a trajectory tangent to the Ferris wheel because, the weight of the rider not sufficient to provide the centripetal force.

Explanation of Solution

The centripetal force required to hold the rider in circular motion in Ferris wheel of radius $14\text{m}$ is $239\text{N}$. During the ride the weight of the rider supplies this centripetal force. Since the weight of the rider is $343\text{N}$, rider can easily move in circular motion.

Write the expression for the centripetal force.

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

The above expression implies that the centripetal acceleration is directly proportional to the square of the speed of the rider.

Therefore, if the speed of the rider is very fast then centripetal force will be also very large. In such case weight of rider cannot supply enough centripetal force to the rider. This will cause imbalance in the net force and the rider will fly out from the seat.

If there are safety measures then it will give enough force to avoid outward push. If safety measures are not available then the rider will fly out in tangential direction.

Conclusion:

Thus, if the weight of the rider is less compared to the centripetal force, then the rider will fly out in the tangential direction because the absence of any force to provide the centripetal force.