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Q47P

Expert-verifiedFound in: Page 143

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?**

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

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

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

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

**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.

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

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

Hence, the period of the motion is 10 s.

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)

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

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${}^{4.9\times {10}^{2}N}$.

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

Thus,

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

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

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