Question

Question: 47. (a) A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of 20.0^owith the horizontal. A piece of luggage having mass 30.0 kgis placed on the carousel at a position 7.46 mmeasured horizontally from the axis of rotation. The travel bag goes around once in 38.0 s. Calculate the force of static friction exerted by the carousel on the bag. (b) The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to another position, 7.94 mfrom the axis of rotation. Now going around once in every 34.0 s, the bag is on the verge of slipping down the sloped surface. Calculate the coefficient of static friction between the bag and the carousel.

Short answer

(a) The force of static friction exerted by the carousel on the bag is 106 N.

(b) The coefficient of static friction between the bag and the carousel is 0.396.

Step-by-step solution

Expression for centripetal force

If a particle is moving in circular path there must be some centripetal force acting towards the centre.

The expression for the centripetal force is given by

$F_c=m\frac{v^2}{r}$

Here F_c is the centripetal force,m is the mass of the body andr is the radius of the circular path.

Step 2: Calculating the speed of the bag

The below figure shows the free body diagram of the luggage.

The speed of the bag is given by

$v=\frac{2\pi r}{T}$

Here vis the speed of the bag, ris the radius of the circular path and T is the time period.

Substitute 7.46 m for r and 38.0 s for T in the above equation.

$\begin{array}{c}v=\frac{2\pi \left(7.46\right)}{38}\\ =1.23\text{m}/\text{s}\end{array}$

Apply particle in uniform circular motion model to the bag in the horizontal direction (taking inward as positive).

$\sum F_x=f_s\cos \theta -n\sin \theta =m{a}_c=\frac{m{v}^2}{r}.................\left(1\right)$

Apply particle in equilibrium model to the bag in the vertical direction (taking upward as positive).

$$\begin{gathered} \sum F_y=f_s\text{sinθ}+n\text{cosθ}-mg=\text{0} \\ {f}_s\text{sinθ}+n\text{cosθ}=mg......................\left(2\right) \end{gathered}$$

Eliminate n from Eq.(1) and Eq.(2) by multiplying equation (1) by cosθ and equation (2) by sinθ .

$\begin{array}{c}{f}_s{\cos }^2\theta -n\sin \theta \cos \theta +f_s{\sin }^2\theta +n\cos \theta \sin \theta =\frac{m{v}^2}{r}\cos \theta +mg\sin \theta \\ f_s{\cos }^2\theta +f_s{\sin }^2\theta =\frac{m{v}^2}{r}\cos \theta +mg\sin \theta \end{array}$

Take f_s as a common factor on the LHS.

$f_s\left({\sin }^2\theta +{\cos }^2\theta \right)=\frac{m{v}^2}{r}\cos \theta +mg\sin \theta$

Where, $\left({\sin }^2\theta +{\cos }^2\theta =1\right)$

$\begin{array}{l}{f}_s=\frac{m{v}^2}{r}\cos \theta +mg\sin \theta \\ f_s=m\left(\frac{v^2}{r}\cos \theta +g\sin \theta \right)\end{array}$

Substitute 30 kg for m , 9.8 m/s² for g , 20^ofor θ, 1.23 m/s for v and 7.46 m for r in the above equation.

$\begin{array}{c}{f}_s=30\left[\frac{{\left(1.23\right)}^2}{7.46}\cos {20}^{°}+\left(9.8\right)\sin {20}^{°}\right]\\ =106\text{N}\end{array}$

Therefore, the force of static friction exerted by the carousel on the bag is 106 N.

Step 3: Finding the speed of the bag from the following relation equation (b)

The speed of the bag is given by,

$v=\frac{2\pi r}{T}$

Here vis the speed of the bag,r is the radius of the circular path andT is the time period.

Substitute 7.46m for r and 38.0 s for T in the above equation.

$\begin{array}{c}v=\frac{2\pi \left(7.94\right)}{34}\\ =1.47\text{m}/\text{s}\end{array}$

Eliminate f_s from equation (1) and equation (2) by multiplying equation (1) by $-\sin \theta$and Equation (2) by $\cos \theta$.

$\begin{array}{l}-f_s\cos \theta \sin \theta +n{\sin }^2\theta +f_s\sin \theta \cos \theta +n{\cos }^2\theta =mg\cos \theta -\frac{m{v}^2}{r}\sin \theta \\ n{\sin }^2\theta +n{\cos }^2\theta =mg\cos \theta -\frac{m{v}^2}{r}\sin \theta \end{array}$

Take n as a common factor on the LHS.

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

Substitute 30 kg for m, for g, 20^ofor θ, 1.47 m/s forv and 7.49m forr in the above equation.

$\begin{array}{c}n=30\left[\left(9.8\right)\cos {20}^{°}-\left(\frac{{1.47}^2}{7.94}\sin {20}^{°}\right)\right]\\ =273\text{N}\end{array}$

Therefore, the coefficient of static friction is found by

$\begin{array}{l}{f}_s={\mu }_{s}n\\ {\mu }_s=\frac{f_s}{n}\end{array}$

Substitute 108 N for f_s and 273 N for n in the above equation.

$\begin{array}{c}{\mu }_s=\frac{108}{273}\\ =0.396\end{array}$

Therefore, the coefficient of static friction between the bag and carousel is 0.396.