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Q52P

Expert-verifiedFound in: Page 124

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is ${\mathbf{5}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\text{\hspace{0.17em}}}}{\mathbf{kN}}$****, and the circle’s radius is ${\mathbf{10}}{\mathbf{}}{\mathbf{m}}$****. At the top of the circle, what are the **

**(a) magnitude and **

**(b) direction (up or down) of the force on the car from the boom if the car’s speed is ${\mathbf{v}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{5}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\text{\hspace{0.17em}}}}{\mathbf{m}}{\mathbf{/}}{\mathbf{s}}$****? **

**What are (c)${{\mathbf{F}}}_{{\mathbf{B}}}$ *** ***and **

**(d) the direction if ${\mathbf{v}}{\mathbf{}}{\mathbf{=}}{\mathbf{12}}{\mathbf{}}{\mathbf{m}}{\mathbf{/}}{\mathbf{s}}$****?**

- The magnitude of the force on the car is $3.7\times {10}^{3}\text{\hspace{0.17em}N}$.
- The direction of the force is upward.
- The magnitude of ${F}_{B}$ is $2.3\times {10}^{3}\text{\hspace{0.17em}N}$.
- The direction of ${\mathrm{F}}_{\mathrm{B}}$ is downward.

- The combined weight of the car and riders is $\text{5.0\hspace{0.17em}kN}$.
- Circle’s radius is $\text{10\hspace{0.17em}m}$.

**Here, the car is moving in a vertical circle on the end of a rigid boom of negligible mass instead of a normal downward force, and we are dealing with the force of the boom ${{\mathbf{F}}}_{{\mathbf{B}}}$**** on the car, which is capable of pointing in any direction.**

** **

**Assume it to be upward and apply Newton’s second law to the car (of a total weight of 5000 N. Note that the centripetal acceleration is downward (our choice for negative direction) for a body at the top of its circular trajectory.**

Formula:

${\mathrm{F}}_{\mathrm{B}}-\mathrm{W}=\mathrm{ma}$

Applying Newton’s law,

${\mathrm{F}}_{\mathrm{B}}-\mathrm{W}=\mathrm{ma}$

Here, $\mathrm{m}=\frac{\mathrm{W}}{\mathrm{g}}$

And $\mathrm{a}=\frac{-{\mathrm{v}}^{2}}{\mathrm{r}}$

If $\mathrm{r}=10\text{\hspace{0.17em}m}$ and $\mathrm{v}=5.0\text{\hspace{0.17em}m/s}$,

$\begin{array}{l}{\mathrm{F}}_{\mathrm{B}}=\mathrm{W}\left(1+\frac{\mathrm{a}}{\mathrm{g}}\right)\\ {\mathrm{F}}_{\mathrm{B}}=\mathrm{W}\left(1+\frac{{\mathrm{v}}^{2}/\mathrm{r}}{\mathrm{g}}\right)\end{array}$

Substitute the values, and we get,

$\begin{array}{l}{\mathrm{F}}_{\mathrm{B}}=5000\left(1+\frac{{5}^{2}/10}{9.8}\right)\\ {\mathrm{F}}_{\mathrm{B}}=3.7\times {10}^{3}\text{\hspace{0.17em}N}\end{array}$

Thus, the magnitude of the force on the car is $3.7\times {10}^{3}\text{\hspace{0.17em}N}$ .

${\mathrm{F}}_{\mathrm{B}}=3.7\times {10}^{3}\text{\hspace{0.17em}N}$

Thus, the positive sign indicates that ${\mathrm{F}}_{\mathrm{B}}$ is upwards

If $\mathrm{r}=10\text{\hspace{0.17em}m}$ and $v=12.0\text{\hspace{0.17em}m/s}$,

${\mathrm{F}}_{\mathrm{B}}=\mathrm{W}\left(1-\frac{{\mathrm{v}}^{2}/\mathrm{r}}{\mathrm{g}}\right)$

Substitute the values, and we get,

$\begin{array}{l}{F}_{B}=5000\left(1-\frac{{12}^{2}/10}{9.8}\right)\\ {F}_{B}=-2.3\times {10}^{3}\text{\hspace{0.17em}N}\end{array}$

Thus, the magnitude of ${\mathrm{F}}_{\mathrm{B}}$ is $2.3\times {10}^{3}\text{\hspace{0.17em}N}$.

${\mathrm{F}}_{\mathrm{B}}=-2.3\times {10}^{3}\text{N}$

Thus, the minus sign indicates that ${\mathrm{F}}_{\mathrm{B}}$ is downward.

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