Chapter 13.3, Problem 10MCQ

a.

To determine

To find: the angle of inclination made by the child, if the sine of the angle of inclination of the child is $\frac{1}{5}$ .

Answer to Problem 10MCQ

  $\theta \approx {11.54}^{\circ }.$

Explanation of Solution

Given information: Suppose a child on a merry −go −round is seated on an outside

horse. The diameter of the merry −go- round is 16 meters. The angle of inclination is represented by the equation $\tan \theta =\frac{v^2}{gR}$ , where R is the radius of the circular path, v is the speed in meters per second and g is 9.8 meters per second squared.

Calculation:

  $\begin{array}{c}\tan \theta =\frac{v^2}{gR}\\ g=9.8\\ D=16\\ \sin \theta =\frac{1}{5}\\ \theta ={\sin }^{-1}(\frac{1}{5})\approx 11.54\\ \theta \approx {11.54}^{\circ }.\end{array}$

Therefore, $\theta \approx {11.54}^{\circ }.$

b.

To determine

To find: the velocity of the merry-go-round.

Answer to Problem 10MCQ

  $v\approx 4\text{meters/second}$ .

Explanation of Solution

Given information: Suppose a child on a merry −go −round is seated on an outside

horse. The diameter of the merry −go- round is 16 meters. The angle of inclination is represented by the equation $\tan \theta =\frac{v^2}{gR}$ , where R is the radius of the circular path, v is the speed in meters per second and g is 9.8 meters per second squared.

Calculation:

  $\begin{array}{c}\tan \theta =\frac{v^2}{gR}\\ g=9.8\\ D=16\\ \sin \theta =\frac{1}{5}\\ {\cos }^2\theta =1-{\sin }^2\theta \\ \cos \theta =\sqrt{1-{\sin }^2\theta }=\sqrt{1-{(\frac{1}{5})}^2}=\sqrt{\frac{24}{25}}=\frac{2\sqrt{6}}{5}\\ v^2=gR\tan \theta \\ v=\sqrt{\frac{gD}{2}\times \frac{\sin \theta }{\cos \theta }}\\ v=\sqrt{\frac{9.8\times 16}{2}\times \frac{\frac{1}{5}}{\frac{2\sqrt{6}}{5}}}\\ v=\sqrt{78.4\times \frac{1}{2\sqrt{6}}}\\ v\approx 4\text{meters/second}\end{array}$

Therefore, $v\approx 4\text{meters/second}$ .

c.

To determine

To find: the value of the angle of inclination if the speed of the merry-go-round is 3.6 meters per second.

Answer to Problem 10MCQ

  $\theta \approx {9.39}^{\circ }.$

Explanation of Solution

Given information: Suppose a child on a merry −go −round is seated on an outside

horse. The diameter of the merry −go- round is 16 meters. The angle of inclination is represented by the equation $\tan \theta =\frac{v^2}{gR}$ , where R is the radius of the circular path, v is the speed in meters per second and g is 9.8 meters per second squared.

Calculation:

  $\begin{array}{c}\tan \theta =\frac{v^2}{gR}\\ g=9.8\\ D=16\\ v=3.6\\ \tan \theta =\frac{v^2}{\frac{gD}{2}}=\frac{{3.6}^2}{\frac{9.8\times 16}{2}}\\ \tan \theta =\frac{12.96}{78.4}=\mathrm{0.16530612.}\\ \theta ={\tan }^{-1}(0.16530612)\approx {9.39}^{\circ }\\ \theta \approx {9.39}^{\circ }.\end{array}$

Therefore , $\theta \approx {9.39}^{\circ }.$