Question

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 $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{kN}$, and the circle’s radius is$\mathbf{10}\mathbf{}\mathbf{m}$. At the top of the circle, what are the

(a) magnitudeand

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

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

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

Short answer

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

Step-by-step solution

Given data

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

To understand the concept

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:

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

(a) To calculate the magnitude and direction of the force on the car from the boom if the car’s speed is v = 5.0 m/s 

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$3.7\times {10}^3\text{\hspace{0.17em}N}$ .

(b) To calculate the magnitude and direction of the force on the car from the boom if the car’s speed is v = 5.0 m/s

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

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

(c), (d) To calculate the magnitude of the force on the car from the boom if the car’s speed is v = 12 m/s

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 $2.3\times {10}^3\text{\hspace{0.17em}N}$.

(d) To calculate the direction of the force on the car from the boom if the car’s speed is v = 12 m/s

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

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