Question

A Ferris wheel with radius 14.0 m is turning about a horizontal axis through its center. The linear speed of a passenger on the rim is constant and equal to 6.00 m/s. What are the magnitude and direction of the passenger’s acceleration as she passes through (a) the lowest point in her circular motion and (b) the highest point in her circular motion? (c) How much time does it take the Ferris wheel to make one revolution?

Answer 1

Answer:

a) The acceleration experimented by the passenger when she passes through the lowest point of her circular motion is: $a_{R} = 2.571\,\frac{m}{s^{2}}$, $\angle = 90^{\circ}$.

b) Hence, the acceleration experimented by the passenger when she passes through the highest point of her circular motion is: $a_{R} = 2.571\,\frac{m}{s^{2}}$, $\angle = 270^{\circ}$.

c) The Ferris wheel takes 14.646 seconds to make a revolution.

Explanation:

a) An object that rotates at constant angular velocity reports a centripetal acceleration and no tangential acceleration. When passenger passes through the lowest point in her circular motion, centripetal acceleration goes up to the center.

In addition, centripetal acceleration is determined by the following expression:

$a_{R} = \frac{v^{2}}{R}$ (Eq. 1)

Where:

$a_{R}$ - Centripetal acceleration, measured in meters per square second.

$v$ - Linear speed, measured in meters per second.

$R$ - Radius of the Ferris wheel, measured in meters.

If we know that $v = 6\,\frac{m}{s}$ and $R = 14\,m$, the magnitude of radial acceleration is:

$a_{R} = \frac{\left(6\,\frac{m}{s} \right)^{2}}{14\,m}$

$a_{R} = 2.571\,\frac{m}{s^{2}}$

The acceleration experimented by the passenger when she passes through the lowest point of her circular motion is: $a_{R} = 2.571\,\frac{m}{s^{2}}$, $\angle = 90^{\circ}$.

b) In the highest point the magnitude of radial acceleration is the same but direction is the opposed to that at lowest point. That is, centripetal acceleration goes down to the center.

Hence, the acceleration experimented by the passenger when she passes through the highest point of her circular motion is: $a_{R} = 2.571\,\frac{m}{s^{2}}$, $\angle = 270^{\circ}$.

c) At first we need to calculate the angular velocity of the Ferris wheel ($\omega$), measured in radians per second, by using the following expression:

$\omega = \frac{v}{R}$ (Eq. 2)

If we know that $v = 6\,\frac{m}{s}$ and $R = 14\,m$, then the angular velocity of the Ferris wheel is:

$\omega = \frac{6\,\frac{m}{s} }{14\,m}$

$\omega = 0.429\,\frac{rad}{s}$

Now we proceed to obtain the period of the Ferris wheel ($T$), measured in seconds, which is the time needed by that wheel to make on revolution:

$T = \frac{2\pi}{\omega}$ (Eq. 3)

Where $\omega$ is the angular velocity of the Ferris wheel, measured in radians per second.

If we get that $\omega = 0.429\,\frac{rad}{s}$, then:

$T = \frac{2\pi}{0.429\,\frac{rad}{s} }$

$T = 14.646\,s$

The Ferris wheel takes 14.646 seconds to make a revolution.