Question

P49 The Ferris wheel in Figure 5.80is a vertical, circular amusement ride with radius 10m . Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 10.5s. Consider a rider whose mass is 56kg .

(a) At the bottom of the ride, what is the rate of change of the rider's momentum? (b) At the bottom of the ride, what is the vector gravitational force exerted by the Earth on the rider?

(c) At the bottom of the ride, what is the vector force exerted by the seat on the rider?

(d) Next consider the situation at the top of the ride. At the top of the ride, what is the rate of change of the rider's momentum?

(e) At the top of the ride, what is the vector gravitational force exerted by the Earth on the rider?

(f) At the top of the ride, what is the vector force exerted by the seat on the rider?

A rider feels heavier if the electric, interatomic contact force of the seat on the rider is larger than the rider's weight mg (and the rider sinks more deeply into the seat cushion). A rider feels lighter if the contact force of the seat is smaller than the rider's weight (and the rider does not sink as far into the seat cushion).

(g) Does a rider feel heavier or lighter at the bottom of a Ferris wheel ride?

(h) Does a rider feel heavier or lighter at the top of a Ferris wheel ride?

Short answer

(a) The rate of change of the rider's momentum at the bottom of the ride is$200\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2$

(b) The vector gravitational force exerted by the Earth on the rider at the bottom of the ride is$-549\hat{\mathrm{j}}\mathrm{N}$

(C) The vector force exerted by the seat on the rider at the bottom of the ride is$749\hat{\mathrm{j}}\mathrm{N}$

(d) The rate of change of the rider's momentum at the top of the ride is$200\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2$

(e) The vector gravitational force exerted by the Earth on the rider is:${\overrightarrow{\mathrm{F}}}_{\mathrm{g}}=-549\stackrel{⏜}{\mathrm{j}}\mathrm{N}$

(f)The vector force exerted by the seat on the rider at the top is $-349\stackrel{⏜}{\mathrm{j}}\mathrm{N}$

(g) A rider feel heavier at the bottom of a Ferris wheel ride.

(h) A rider feel lighter at the top of a Ferris wheel ride.

Step-by-step solution

Given

The Ferris wheel in is a vertical, circular amusement ride with radius R=10m . Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in t=10.5s . Consider a rider whose mass is m=56kg .

The expression to get the velocity and distance

The net force on an object is equal to the rate of change of momentum and can be written as the sum of two parts. The parallel rate of change of momentum${\mathbf{(}\mathbf{d}\overrightarrow{\mathbf{p}}\mathbf{/}\mathbf{dt}\mathbf{)}}_{\mathbf{\left|}\mathbf{\right|}}$and the perpendicular rate of change of momentum role="math" localid="1656857746837" ${\mathbf{(}\mathbf{d}\overrightarrow{\mathbf{p}}\mathbf{/}\mathbf{dt}\mathbf{)}}_{⟂}$are the two parts that we are concerned with.

Change in momentum is given by,

$\frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{\mathbf{d}{\mathbf{t}}_{\mathbf{\left|}\mathbf{\right|}}}\mathbf{+}\frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{\mathbf{d}{\mathbf{t}}_{\mathbf{⟂}}}$ ……………………… (1)

The object's speed is affected by the parallel rate of change of momentum, and because the speed is constant, the parallel rate is zero and equal to the rate of change of the magnitude of the momentum.

$\frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{{\mathbf{dt}}_{\mathbf{\left|}\mathbf{\right|}}}\mathbf{=}\frac{\mathbf{d}\mathbf{\left|}\overrightarrow{\mathbf{p}}\mathbf{\right|}}{\mathbf{dt}}\stackrel{\mathbf{⏜}}{\mathbf{p}}\mathbf{=}\mathbf{0}$

The direction shift generated by the perpendicular rate of change is known as the rate change. The quantity of the perpendicular rate change matches the rate change of the direction of the momentum at speeds significantly slower than the speed of light. .

$\frac{\mathbf{d}\overrightarrow{\mathbf{p}}}{{\mathbf{dt}}_{\mathbf{⟂}}}\mathbf{=}\left|\overrightarrow{p}\right|\frac{\mathbf{d}\stackrel{\mathbf{⏜}}{\mathbf{p}}}{\mathbf{dt}}\mathbf{=}\frac{{\mathbf{mv}}^{\mathbf{2}}}{\mathbf{R}}$ ………………………. (2)

It will also be equivalent to the rate of change of momentum$\mathbf{}\mathbf{d}\overrightarrow{\mathbf{p}}\mathbf{/}\mathbf{dt}$.

We know the values of mand R and want to know how fast v is. The change in distance over time is the speed. As a result, it is provided by

$\mathbf{v}\mathbf{=}\frac{\mathbf{d}}{\mathbf{t}}$

(a) The value of the change of momentumat the bottom of the ride

When the rider completes one round, it moves above the circle's circumference. The circumference is calculated using $2\mathrm{\pi R}$.Here R is the radius of the wheel and the distance is $d=2\mathrm{\pi R}$.

To get the value of v,the value of d has to be substituted in $v=\frac{d}{t}$

$v=\frac{2\mathrm{\pi R}}{t}$ …………………………. (3)

The speed of the rider in the wheel is:

$\mathrm{v}=\frac{2\mathrm{\pi R}}{\mathrm{t}}=\frac{2\mathrm{\pi }\left(10\mathrm{m}\right)}{10.5\mathrm{s}}=5.98\mathrm{m}/\mathrm{s}$

The value of$\frac{d\overrightarrow{p}}{d{t}_{⟂}}$is:

$$\begin{gathered} \frac{\mathrm{d}\overrightarrow{\mathrm{p}}}{{\mathrm{dt}}_{⟂}}=\frac{{\mathrm{mv}}^2}{\mathrm{R}} \\ =\frac{\left(56\mathrm{kg}\right){\left(5.98\mathrm{m}/\mathrm{s}\right)}^2}{10\mathrm{m}} \\ =200\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

So,the change in momentum is:

$$\begin{gathered} \frac{\mathrm{d}\overrightarrow{\mathrm{p}}}{\mathrm{dt}}=\frac{\mathrm{d}\overrightarrow{\mathrm{p}}}{\mathrm{dt}}+\frac{\mathrm{d}\overrightarrow{\mathrm{p}}}{\mathrm{dt}} \\ =0+200\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =200\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

And the net force is:

localid="1656859363787" $F_{net}=\frac{d\overrightarrow{p}}{dt}$

The rate of change of the rider's momentum at the bottom of the ride is $200\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2$.

(b) The value of the vector gravitational force at the bottom of the ride

The variable m is the rider's mass and g is the gravitational acceleration, and $9.8\mathrm{m}/{\mathrm{s}}^2$is the result substitute and g numbers into F=mg to calculate the gravitational force acting on the rider.

$$\begin{gathered} \mathrm{F}=\mathrm{mg} \\ =\left(56\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =-549\mathrm{N} \end{gathered}$$

The component of the gravitational force ${\mathrm{F}}_{\mathrm{g}}=-549\mathrm{N}$was calculated. For the gravitational force, there is no force component in the z-direction and no force in the x- direction, therefore ${\mathrm{F}}_{\mathrm{z}}=0,{\mathrm{F}}_{\mathrm{x}}=0$.The three-dimensional force vector is given by

${\overrightarrow{\mathrm{F}}}_{\mathrm{g}}={\mathrm{F}}_{\mathrm{x}}\stackrel{⏜}{\mathrm{i}}+{\mathrm{F}}_{\mathrm{y}}\stackrel{⏜}{\mathrm{j}}+{\mathrm{F}}_{\mathrm{z}}\stackrel{⏜}{\mathrm{k}}$

The vector of the gravitational force is:

$$\begin{gathered} {\overrightarrow{\mathrm{F}}}_{\mathrm{g}}={\mathrm{F}}_{\mathrm{x}}\stackrel{⏜}{\mathrm{i}}+{\mathrm{F}}_{\mathrm{y}}\stackrel{⏜}{\mathrm{j}}+{\mathrm{F}}_{\mathrm{z}}\stackrel{⏜}{\mathrm{k}} \\ =\left(0\stackrel{⏜}{\mathrm{i}}+\left(-549\stackrel{⏜}{\mathrm{j}}\right)+0\stackrel{⏜}{\mathrm{k}}\right)\mathrm{N} \\ =-549\stackrel{⏜}{\mathrm{j}}\mathrm{N} \end{gathered}$$

The vector gravitational force exerted by the Earth on the rider at the bottom of the ride is $-549\stackrel{⏜}{\mathrm{j}}\mathrm{N}$

(c)The vector force exerted on the rider at the bottom of the ride

The rider is under two forces at the bottom: the seat's normal force ${\mathrm{F}}_{\mathrm{N}}$(upward) and its weight (downward), therefore the net force of both forces is

${\mathrm{F}}_{\mathrm{net}}={\mathrm{F}}_{\mathrm{N}}-\mathrm{mg}$

The negative sign is owing to the fact that both forces are moving in the opposing direction. Derive the following form for $F_N$.

${\mathrm{F}}_{\mathrm{net}}={\mathrm{F}}_{\mathrm{N}}-\mathrm{mg}$

The value of normal force is:

role="math" localid="1656863293802" $$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}={\mathrm{F}}_{\mathrm{N}}-\mathrm{mg} \\ =200\mathrm{N}+549\mathrm{N} \\ =749\mathrm{N} \end{gathered}$$

The y-component of the normal force was determined.${\mathrm{F}}_{\mathrm{N}}=749\mathrm{N}$For the normal force, there is no force component in the z-direction and no force in the x-direction, hence $F_z=0F_{x}0$. The three-dimensional force vector is given by

${\overrightarrow{\mathrm{F}}}_{\mathrm{g}}={\mathrm{F}}_{\mathrm{x}}\stackrel{⏜}{\mathrm{i}}+{\mathrm{F}}_{\mathrm{y}}\stackrel{⏜}{\mathrm{j}}+{\mathrm{F}}_{\mathrm{z}}\stackrel{⏜}{\mathrm{k}}$

The vector of the normal force is

$$\begin{gathered} {\overrightarrow{\mathrm{F}}}_{\mathrm{N}}={\mathrm{F}}_{\mathrm{x}}\stackrel{⏜}{\mathrm{i}}+{\mathrm{F}}_{\mathrm{y}}\stackrel{⏜}{\mathrm{j}}+{\mathrm{F}}_{\mathrm{z}}\stackrel{⏜}{\mathrm{k}} \\ =\left(0\stackrel{⏜}{\mathrm{i}}+\left(749\stackrel{⏜}{\mathrm{j}}\right)+0\stackrel{⏜}{\mathrm{k}}\right)\mathrm{N} \\ =749\stackrel{⏜}{\mathrm{j}}\mathrm{N} \end{gathered}$$

The vector force exerted by the seat on the rider is $749\stackrel{⏜}{\mathrm{j}}\mathrm{N}$

(d) The vector force exerted by the seat on the rider at the top

The rate of change of momentum $\frac{\mathrm{d}\overrightarrow{\mathrm{p}}}{\mathrm{dt}}$at the top will be the same as the rate of change at the bottom. As portion (a) shows, the rate of change of momentum $\frac{\mathrm{d}\overrightarrow{\mathrm{p}}}{\mathrm{dt}}$will berole="math" localid="1656863749645" $\frac{d\overrightarrow{p}}{dt}=200\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2$

The rate of change of the rider's momentum at the top of the ride is $-200\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2$

(e) The vector force exerted by the seat on the rider at the top

According to the part ,the vector gravitational force exerted by the Earth on the rider is:${\mathrm{F}}_{\mathrm{g}}=-549\stackrel{⏜}{\mathrm{j}}\mathrm{N}$

(f) The vector force exerted by the seat on the rider at the top

Consider the equation (2) of step (3) ,the vector of the normal force is:

$$\begin{gathered} {\overrightarrow{\mathrm{F}}}_{\mathrm{N}}={\mathrm{F}}_{\mathrm{x}}\stackrel{⏜}{\mathrm{i}}+{\mathrm{F}}_{\mathrm{y}}\stackrel{⏜}{\mathrm{j}}+{\mathrm{F}}_{\mathrm{z}}\stackrel{⏜}{\mathrm{k}} \\ =\left(0\stackrel{⏜}{\mathrm{i}}+\left(-349\stackrel{⏜}{\mathrm{j}}\right)+0\stackrel{⏜}{\mathrm{k}}\right)\mathrm{N} \\ =349\stackrel{⏜}{\mathrm{j}}\mathrm{N} \end{gathered}$$

The vector force exerted by the seat on the rider at the top is -$349\stackrel{⏜}{\mathrm{j}}\mathrm{N}$

(g) Feeling of the rider at the bottom of a Ferris wheel ride

When the rider is at the bottom, the contact force of the seat is ${\overrightarrow{\mathrm{F}}}_{\mathrm{N}}=749\stackrel{⏜}{\mathrm{j}}\mathrm{N}$, and the gravitational force is ${\overrightarrow{\mathrm{F}}}_{\mathrm{g}}=-549\stackrel{⏜}{\mathrm{j}}\mathrm{N}$, as determined in part (c). As a result, the contact force exceeds the gravitational force. As a result, when the rider sinks further into the seat, he feels heavier.

A rider feel heavier at the bottom of a Ferris wheel ride

(h) Feeling of the rider at the top of a Ferris wheel ride

When the rider is at the top, the contact force of the seat is${\overrightarrow{\mathrm{F}}}_{\mathrm{N}}=-349\stackrel{⏜}{\mathrm{j}}\mathrm{N}$ , and the gravitational force is${\overrightarrow{\mathrm{F}}}_{\mathrm{g}}=-549\stackrel{⏜}{\mathrm{j}}\mathrm{N}$ , as calculated in part (f). As a result, the gravitational force is smaller than the contact force.

Hence, when the rider does not sink into the seat, he feels lighter.