The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can function as a second hand (so that it makes one revolution every 60.0 s).

(a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. (b) A passenger weighs 882 Nat the weight-guessing booth on the ground. What is his apparent weight at the highest and at the lowest point on the Ferris wheel? (c) What would be the time for one revolution if the passenger’s apparent weight at the highest point were zero?

(d) What then would be the passenger’s apparent weight at the lowest point?

The given data can be listed below as,

• The diameter of the clock is, $D=100\text{\hspace{0.17em}m}$.
• The time to make one revolution is,$t=60\text{\hspace{0.17em}s}$ .
• The weight of the passenger is, $w=882\text{\hspace{0.17em}N}$.

The apparent weight of an object is a measure of downward force. The apparent weight is different from the actual weight of an object. The apparent weight is measured when the force of gravity acts on it.

Part (a)

The passenger is moving in a vertical circle. The expression for the speed can be expressed as,

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

Here $R$is the radius of the clock,$t$is the time taken for revolution.

Substitute 50 m for $R$, $60\text{\hspace{0.17em}s}$for $t$, 3.14 for $\pi$ in the above equation.

$\begin{array}{c}v=\frac{2\times 3.14\times 50\text{\hspace{0.17em}m}}{60\text{\hspace{0.17em}s}}\\ =5.23\text{\hspace{0.17em}m/s}\end{array}$

Hence, the required speed of the passenger is,$5.23\text{\hspace{0.17em}m/s}$ .

Part (b)

The expression for the mass of a passenger can be expressed as,

$m=\frac{W}{g}$

Here$W$is the weight of the passenger, and $g$is the acceleration due to gravity.

Substitute for $882\text{\hspace{0.17em}N}$, $W$and $9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$for $g$in the above equation.

$\begin{array}{c}m=\frac{882\text{\hspace{0.17em}N}}{9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}}\\ =90\text{\hspace{0.17em}kg}\end{array}$

Hence, the mass of the passenger is 90 kg.

The expression for the velocity of the Ferris wheel can be expressed as,

$\begin{array}{l}v=\frac{d}{t}\\ v=\frac{\pi D}{t}\end{array}$

Here$D$is the diameter of the wheel.

Substitute 3.14 for ,$\pi$and 100 m for $D$, 60 s for $t$in the above equation.

$\begin{array}{c}v=\frac{3.14\times 100\text{\hspace{0.17em}m}}{60\text{\hspace{0.17em}s}}\\ =5.23\text{\hspace{0.17em}m/s}\end{array}$

Hence, velocity is, $5.23\text{\hspace{0.17em}m/s}$.

The expression for the acceleration of the Ferris wheel can be expressed as,

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

Here $v$is the velocity of the Ferris wheel,$r$is the radius of the clock.

Substitute $5.23\text{\hspace{0.17em}m/s}$for $v$, for$r$in the above equation.

$\begin{array}{c}a=\frac{{\left(5.23\text{\hspace{0.17em}m/s}\right)}^2}{50\text{\hspace{0.17em}m}}\\ =0.547{\text{\hspace{0.17em}m/s}}^{\text{2}}\end{array}$

Hence, the acceleration is, .$0.547{\text{\hspace{0.17em}m/s}}^{\text{2}}$

The expression for the apparent weight at the lowest point can be expressed as,

$W=m(a+g)$

Here$m$is the mass of the passenger, $a$is the acceleration of the wheel,$g$ is the acceleration due to gravity.

Substitute$90\text{\hspace{0.17em}kg}$for m,$0.547{\text{\hspace{0.17em}m/s}}^{\text{2}}$for ,$a$and$9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$for$g$in the above equation.

role="math" localid="1659533883566" $\begin{array}{c}W=90\text{\hspace{0.17em}kg}({\text{0.547\hspace{0.17em}m/s}}^2+9.8{\text{\hspace{0.17em}m/s}}^{\text{2}})\\ =931.23\text{\hspace{0.17em}N}\end{array}$

Hence, the required apparent weight at the lowest point is, 931.23 N.

The expression for the apparent weight at the highest point can be expressed as,

$W=m(g-a)$

Substitute $90\text{\hspace{0.17em}kg}$for m,$0.547{\text{\hspace{0.17em}m/s}}^{\text{2}}$ for$a$, and $9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$for $g$in the above equation.

$\begin{array}{c}W=90\text{\hspace{0.17em}kg}(9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}-0.547{\text{\hspace{0.17em}m/s}}^{\text{2}})\\ =832.77\text{\hspace{0.17em}N}\end{array}$

Hence, the required apparent weight at the highest point is, 832.77 N.

Part (c)

The expression for the angular speed can be expressed as,

$\begin{array}{c}a=g={\omega }^{2}r\\ \omega =\sqrt{\frac{g}{r}}\end{array}$

Here $g$is the acceleration due to gravity,$r$is the radius of the clock wheel.

Substitute$9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$for $g$, 50 m for$r$in the above equation.

$\begin{array}{c}\omega =\sqrt{\frac{9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}}{50\text{\hspace{0.17em}m}}}\\ =0.442\text{\hspace{0.17em}rad/s}\end{array}$

Hence, angular speed is,$0.442\text{\hspace{0.17em}rad/s}$.

The expression for the time for one revolution can be expressed as,

$\begin{array}{l}\omega =\frac{2\pi }{T}\\ T=\frac{2\pi }{\omega }\end{array}$

Here $\omega$is angular speed.

Substitute 3.14 for $\pi$, $0.442\text{\hspace{0.17em}rad/s}$ for$\omega$ in the above equation.

$\begin{array}{c}T=\frac{2\times 3.14}{0.442\text{\hspace{0.17em}rad/s}}\\ =14.21\text{\hspace{0.17em}s}\end{array}$

Hence, the required time for one revolution is, 14.21 s.

Part (d)

The expression for the apparent weight at thelowest point can be expressed as,

$W=m(g+a)$

Here$m$ is the mass,$g$ is the acceleration due to gravity$a$, is the acceleration.

Substitute 90 kg for $m$, $9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$for$g$ ,$a=g=9.8{\text{\hspace{0.17em}m/s}}^{\text{2}}$ for in the above equation.

$\begin{array}{c}W=90\text{\hspace{0.17em}kg}({\text{9.8\hspace{0.17em}m/s}}^2+9.8{\text{\hspace{0.17em}m/s}}^{\text{2}})\\ =1764\text{\hspace{0.17em}N}\end{array}$

Hence, the required apparent weight of passenger at the lowest point is, 1764 N.