Chapter 6, Problem 47AP

(a)

To determine

The force of static friction exerted by the carousel on the bag.

Answer to Problem 47AP

The static friction force exerted by the carousel on the bag is $\underset{\_}{106\text{N}}$.

Explanation of Solution

Write the expression for the angular speed of the luggage as.

  $\omega =\frac{2\pi }{T}$                                                                                                         (I)

Here, $\omega$ is the angular speed of the luggage and $T$ is time period to complete one rotation by the luggage.

Write the expression for the angular speed of the luggage in terms on linear speed as.

  $\omega =\frac{v}{r}$                                                                                                          (II)

Here, $v$ is the linear speed of the luggage and $r$ is the radius of rotation.

Substitute $\frac{v}{r}$ for $\omega$ in equation (I)

  $\frac{v}{r}=\frac{2\pi }{T}$

Simplify the above expression for $v$ as.

  $v=\frac{2\pi r}{T}$                                                                                                       (III)

Write the expression for centripetal acceleration for luggage as.

  $a_c=\frac{v^2}{r}$                                                                                                     (IV)

Here, $a_c$ is centripetal acceleration for luggage.

If the luggage carousel is steadily rotating about its vertical axis then three forces act on the luggage. The static friction force acts along the inclined metallic surface in an outward direction, normal reaction force exerts by the metallic surface on the luggage in normal to the metallic plane and force due to weight acts downward.

As the luggage is rotating in a horizontal plane so centripetal acceleration will acts in $x$-direction.

The forces act on the luggage on luggage carousel as shown below.

Write the expression corresponding to Newton’s second law of motion in $x$-direction as.

  $f\cos 20°-n\sin 20°=m{a}_c$                                                                          (V)

Substitute $\frac{v^2}{r}$ for $a_c$ in equation (V).

  $f\cos 20°-n\sin 20°=\frac{m{v}^2}{r}$                                                                        (VI)

Multiply $\cos 20°$ on both sides in equation (VI)

  $f{\cos }^{2}20°-n\cos 20°\sin 20°=\frac{m{v}^2}{r}\cos 20°$                                            (VII)

Simplify the equation (VII).

  $n\cos 20°\sin 20°=f{\cos }^{2}20°-\frac{m{v}^2}{r}\cos 20°$                                            (VIII)

Here, $f$ is static friction force, $n$ is normal reaction force exerted by the metallic surface on the luggage and $m$ is mass of the luggage.

Write the expression corresponding to the condition of equilibrium in $y$-direction as.

  $f\sin 20°+n\cos 20°-mg=0$

Re-arrange the terms.

  $f\sin 20°+n\cos 20°=mg$                                                                          (IX)

Multiply $\sin 20°$ on both sides in the equation (IX).

  $f{\sin }^{2}20°+n\sin 20°\cos 20°=mg\sin 20°$                                                   (X)

Substitute $f{\cos }^{2}20°-\frac{m{v}^2}{r}\cos 20°$ for $n\cos 20°\sin 20°$ in equation (X).

  $\begin{array}{c}f{\sin }^{2}20°+f{\cos }^{2}20°-\frac{m{v}^2}{r}\cos 20°=mg\sin 20°\\ f\left({\sin }^{2}20°+{\cos }^{2}20°\right)=\frac{m{v}^2}{r}\cos 20°+mg\sin 20°\end{array}$

Simplify the above expression for $f$ as.

  $f=\frac{m{v}^2}{r}\cos 20°+mg\sin 20°$                                                                     (XI)

Conclusion:

Substitute $7.46\text{m}$ for $r$ and $38.0\text{s}$ for $T$ in equation (III).

  $\begin{array}{c}v=\frac{2\pi \left(7.46\text{m}\right)}{\left(38.0\text{s}\right)}\\ =1.233\text{m}/\text{s}\\ \approx 1.23\text{m}/\text{s}\end{array}$

Substitute $30\text{kg}$ for $m$, $7.46\text{m}$ for $r$, $1.23\text{m}/\text{s}$ for $v$ and $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (XI).

  $\begin{array}{c}f=\frac{\left(30\text{kg}\right){\left(1.23\text{m}/\text{s}\right)}^2}{\left(7.46\text{m}\right)}\cos 20°+\left(30\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\sin 20°\\ =5.71\text{N}+\text{100}\text{.55}\text{N}\\ \text{=106}\text{.271}\text{N}\\ \approx \text{106}\text{N}\end{array}$

Thus, the static friction force exerted by the carousel on the bag is $\underset{\_}{106\text{N}}$.

(b)

To determine

The coefficient of static friction between the bag and the carousel.

Answer to Problem 47AP

The coefficient of static friction between the bag and the carousel is $\underset{\_}{0.39}$.

Explanation of Solution

Multiply equation (VI) by $-\sin 20°$ on the both sides.

  $-f\cos 20°\sin 20°+n{\sin }^{2}20°=-\frac{m{v}^2}{r}\sin 20°$

Simplify the above expression for $f\cos 20°\sin 20°$

  $f\cos 20°\sin 20°=\frac{m{v}^2}{r}\sin 20°+n{\sin }^{2}20°$                                              (XII)

Multiply $\cos 20°$ on both sides in the equation (IX).

  $f\cos 20°\sin 20°+n{\cos }^{2}20°=mg\cos 20°$                                              (XIII)

Substitute $\frac{m{v}^2}{r}\sin 20°+n{\sin }^{2}20°$ for $f\cos 20°\sin 20°$ in equation (XIII).

  $\begin{array}{l}\frac{m{v}^2}{r}\sin 20°+n{\sin }^{2}20°+n{\cos }^{2}20°=mg\cos 20°\\ n\left({\sin }^{2}20°+{\cos }^{2}20°\right)=mg\cos 20°-\frac{m{v}^2}{r}\sin 20°\end{array}$

Simplify the above expression for $n$ as.

  $n=mg\cos 20°-\frac{m{v}^2}{r}\sin 20°$                                                                 (XIV).

Write the expression for the coefficient of static friction as.

  ${\mu }_s=\frac{f}{n}$                                                                                                     (XV)

Here, ${\mu }_s$ is static friction coefficient.

Conclusion:

Substitute $7.94\text{m}$ for $r$ and $34.0\text{s}$ for $T$ in equation (III).

  $\begin{array}{c}v=\frac{2\pi \left(7.94\text{m}\right)}{\left(34.0\text{s}\right)}\\ =1.4673\text{m}/\text{s}\\ \approx 1.47\text{m}/\text{s}\end{array}$

Substitute $30\text{kg}$ for $m$, $7.94\text{m}$ for $r$, $1.47\text{m}/\text{s}$ for $v$ and $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (XIV).

  $\begin{array}{c}n=\left(30\text{kg}\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\cos 20°-\frac{\left(30\text{kg}\right){\left(1.47\text{m}/\text{s}\right)}^2}{\left(7.94\text{m}\right)}\sin 20°\\ =\text{276}\text{.2696}\text{N}-2.792\text{N}\\ \text{=273}\text{.4}\text{N}\\ \approx \text{273}\text{N}\end{array}$

Substitute $\text{273}\text{N}$ for $n$ and $106\text{N}$ for $n$  in equation (XV).

  $\begin{array}{c}{\mu }_s=\frac{\left(106\text{N}\right)}{\text{273}\text{N}}\\ =0.3882\\ \approx 0.39\end{array}$

Thus, the coefficient of static friction between the bag and the carousel is $\underset{\_}{0.39}$.