• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon

Suggested languages for you:

Americas

Europe

Q47P

Expert-verified
Found in: Page 143

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# A circular-motion addict of mass${}^{\mathbf{80}\mathbf{k}\mathbf{g}}$ rides a Ferris wheel around in a vertical circle of radius${}^{\mathbf{10}\mathbf{}\mathbf{m}}$ at a constant speed of${}^{\mathbf{6}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}}$. (a) What is the period of the motion? What is the magnitude of the normal force on the addict from the seat when both go through (b) the highest point of the circular path and (c) the lowest point?

1. The period of the motion is${}^{10\mathrm{s}}$.

2. The magnitude of the normal force on the addict from the seat when both go through the highest point of the circular path is${}^{4.9×{10}^{2}\mathrm{N}}$

3. The magnitude of the normal force on the addict from the seat when both go through the lowest point is${}^{1.1\mathrm{kN}}$.

See the step by step solution

## Step 1: Given

$\mathrm{Mass}=80\mathrm{kg},\mathrm{circle}\mathrm{of}\mathrm{radius}=10\mathrm{m},\mathrm{speed}=6.1\mathrm{m}/\mathrm{s}$

## Step 2: Determining the concept

This problem is based on the concept of uniform circular motion. Uniform circular motion is a motion in which an object moves in a circular path with constant velocity. Also, it involves Newton’s second law of motion.

Formula:

The velocity in uniform circular motion is given by,

$\mathrm{v}=\frac{2\mathrm{\pi R}}{\mathrm{T}}$

where, v is the velocity, R is the radius and T is the time period

According to Newton’s second law of motion

${\mathrm{F}}_{\mathrm{N}}-\mathrm{mg}={\mathrm{ma}}_{\mathrm{c}}$

where,${}^{{\mathrm{a}}_{\mathrm{c}}}$ is an acceleration, g is an acceleration due to gravity, m is mass,${}^{{\mathrm{F}}_{\mathrm{N}}}$ is the normal force and R is the radius.

## Step 3: (a) Determining the period of the motion

Using equation (i) the time can be written as,

$\mathrm{T}=2\mathrm{\pi R}/\mathrm{V}\phantom{\rule{0ex}{0ex}}-2\mathrm{\pi }\left(10\mathrm{M}\right)/\left(6.1\mathrm{m}/\mathrm{s}\right)\phantom{\rule{0ex}{0ex}}=10\mathrm{s}$

Hence, the period of the motion is 10 s.

## Step 4: (b)Determining the magnitude of the normal force on the addict from the seat when both go through the highest point of the circular path

In this case, Normal force ${}^{{\mathrm{F}}_{\mathrm{N}}}$ is directed upward, gravitational force mg is to downward and acceleration is also directed to the down. Thus, by using Newton’s 2nd law,

${\mathrm{F}}_{\mathrm{N}}-\mathrm{mg}={\mathrm{ma}}_{\mathrm{c}}\phantom{\rule{0ex}{0ex}}$(Centripetal acceleration is due to the circular motion)

${\text{F}}_{\mathrm{N}}=\mathrm{m}\left(\mathrm{g}-{\mathrm{v}}^{2}/\mathrm{R}\right)\phantom{\rule{0ex}{0ex}}=486\mathrm{N}\phantom{\rule{0ex}{0ex}}\approx 4.9×{10}^{2}\mathrm{N}$

Hence, the magnitude of the normal force on the addict from the seat when both go through the highest point of the circular path is${}^{4.9×{10}^{2}N}$.

## Step 5: (c) Determining the magnitude of the normal force on the addict from the seat when both go through the lowest point

Now, reverse both the normal force direction and the acceleration direction

Thus,

$\mathrm{F}=\mathrm{m}\left(\mathrm{g}+{\mathrm{v}}^{2}/\mathrm{R}\right)\phantom{\rule{0ex}{0ex}}-1081\mathrm{N}\phantom{\rule{0ex}{0ex}}\approx 1.1\mathrm{kN}$

Hence, the magnitude of the normal force on the addict from the seat when both go through the lowest point is${}^{1.1\mathrm{kN}}$

## Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.