91Ó°ÊÓ

• The combined weight of the car and riders is$\text{5.0 k±·}$.
• Circle’s radius is $\text{10″¾}$.

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 $\mathrm{r}=10\text{″¾}$and $\mathrm{v}=5.0\text{″¾/s}$,

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

Substitute the values, and we get,

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

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

${\mathrm{F}}_{\mathrm{B}}=3.7×{10}^3\text{ N}$

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

If$\mathrm{r}=10\text{″¾}$and $v=12.0\text{″¾/s}$,

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

Substitute the values, and we get,

$\begin{array}{l}{F}_B=5000\left(1−\frac{{12}^2/10}{9.8}\right)\\ F_B=−2.3×{10}^3\text{ N}\end{array}$

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

${\mathrm{F}}_{\mathrm{B}}=−2.3×{10}^3\text{N}$

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