Question

A child sits at a Ferris wheel car with 5.5 m from the axis of the Ferris wheel in an amusement park. If the Ferris wheel makes 1 revolution in 6 minutes, find the (a) tangential speed and (b) centripetal acceleration of the child.

Answer 1

The (a) tangential speed of the child is approximately
`\(0.96 \, \text{m/s}\)`, and the (b) centripetal acceleration of the child is approximately
`\(0.166 \, \text{m/s}^2\).`

To find the tangential speed (v) and centripetal acceleration
`(\(a_c\))` of the child sitting in the Ferris wheel car, we'll use the following formulas:

(a) Tangential speed (v) is given by:

`\[v = \frac{{2 \pi r}}{{T}}\]`

Where:

r = radius of the Ferris wheel (5.5 m)

T = time period for one revolution (6 minutes)

First, convert the time from minutes to seconds:

`\[T = 6 \, \text{minutes} * 60 \, \text{s/minute}\]`

`\[T = 360 \, \text{s}\]`

Now, use the formula for tangential speed:

`\[v = \frac{{2 \pi * 5.5 \, \text{m}}}{{360 \, \text{s}}}\]`

Calculating v:

`\[v \approx 0.96 \, \text{m/s}\]`

(b) Centripetal acceleration
`(\(a_c\))` is given by:

`\[a_c = \frac{{v^2}}{{r}}\]`

Where:

v = tangential speed (0.96 m/s)

r = radius of the Ferris wheel (5.5 m)

Using the formula for centripetal acceleration:

`\[a_c = \frac{{(0.96 \, \text{m/s})^2}}{{5.5 \, \text{m}}}\]`

Calculating
`\(a_c\):`

`\[a_c \approx 0.166 \, \text{m/s}^2\]`

Answer 2

(a) The tangential speed of the Ferris wheel is 0.096 m/s.

(b) The centripetal acceleration of the child is 1.68 x 10⁻³ m/s².

How to calculate the tangential speed?

(a) The tangential speed of the Ferris wheel is calculated by applying the following formula as follows;

v = ωr

where;

• ω is the angular speed
• r is the radius of the circular path

The angular speed is given as;

ω = 1 rev/ 6 min

ω = 1 rev/ 6 min x 2π rad/rev x 1 min / 60 s

ω = 0.0175 rad/s

v = 0.0175 rad/s x 5.5 m

v = 0.096 m/s

(b) The centripetal acceleration of the child is calculated as follows;

a = v²/r

a = ( 0.096 m/s )² / 5.5 m

a = 1.68 x 10⁻³ m/s²