A playground slide is in the form of an arc of a circle that has a radius of 12 m. The maximum height of the slide ish = 4.0 m, and the ground is tangent to the circle (Fig. 8-70). A 25 kgchild starts from rest at the top of the slide and has the speed of 6.2m/sat the bottom. (a) What is the length of the slide? (b) What average frictional force acts on the child over this distance? If, instead of the ground, a vertical line through the top of the slide is tangent to the circle, what is (c) the length of the slide and (d) the average friction on the child?

The radius of the circle is,$\mathrm{R}=12\mathrm{m}$

The maximum height of the slide is,$\mathrm{h}=4.0\mathrm{m}$

The ground is tangent to the circle.

The mass of the child is,$\mathrm{m}=24\mathrm{kg}$

The initial speed of the child is,${\mathrm{v}}_{\mathrm{i}}=0\mathrm{m}/\mathrm{s}$

The final velocity of the child at the end of the slide is,${\mathrm{v}}_{\mathrm{f}}=6.2\mathrm{m}/\mathrm{s}$

We can find the angle subtended by the slide from the analogy of the given system with the swinging pendulum. Then using the formula for arc length, we can find the length of the slide if the ground is tangent to the circle. Using the formula for work done on a system by external force we can find the average frictional force acting on the child if the ground is tangent to the circle. Similarly, we can find answers for parts c and d.

Formulae:

The work done by the body, $\mathrm{W}=∆{\mathrm{E}}_{\mathrm{mech}}+∆{\mathrm{E}}_{\mathrm{th}}$ (1)

The length of the arc $\left(\mathrm{\theta }\mathrm{in}\mathrm{radians}\right)$, $\mathrm{S}=\mathrm{R\theta }$ (2)

The height is analogous to the swinging pendulum, $\mathrm{h}=\mathrm{R}\left(1-\mathrm{cos\theta }\right)$ (3)

The potential energy at a height, $\mathrm{PE}=\mathrm{mgh}$ (4)

The kinetic energy of the body, $\mathrm{KE}=\frac{1}{2}{\mathrm{mv}}^2$ (5)

The thermal energy of the body due to friction, $∆{\mathrm{E}}_{\mathrm{th}}={\mathrm{f}}_{\mathrm{k}}\mathrm{S}$ (6)

The ground is tangent to the circle.

The system ofslidesgiven in the problem is analogous to the swinging pendulum. So, using equation (3), we can say that the angle of inclination is given as:

$$\begin{gathered} \mathrm{cos\theta }=\frac{\mathrm{R}-\mathrm{h}}{\mathrm{R}} \\ \mathrm{\theta }={\cos }^{-1}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right) \\ \mathrm{\theta }={\cos }^{-1}\left(1-\frac{4}{12}\right) \\ \mathrm{\theta }=0.84\mathrm{rad} \end{gathered}$$

The slide is in the form of an arc of the circle. So the length of the slide is the arc length. Thus, the length of the slide is given using equation (2):

$$\begin{gathered} \mathrm{S}=12\mathrm{m}\left(0.84\mathrm{rad}\right) \\ =10.1\mathrm{m} \end{gathered}$$

Therefore, the length of the slide is 10.1 m

Work done on a system by an external force is given by equation (1). In this case,$\mathrm{W}=0$

Then, the frictional force using equations (4), (5), and (6) is given as:

$$\begin{gathered} 0\mathrm{J}=∆\mathrm{K}.\mathrm{E}+∆\mathrm{P}.\mathrm{E}+∆{\mathrm{E}}_{\mathrm{th}} \\ 0\mathrm{J}=\frac{1}{2}{\mathrm{mv}}^2-\mathrm{mgh}+{\mathrm{f}}_{\mathrm{k}}\mathrm{S} \\ 0\mathrm{J}=\frac{1}{2}\left(25\mathrm{kg}\right){\left(6.2\mathrm{m}/\mathrm{s}\right)}^2-\left(25\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(4\mathrm{m}\right){\mathrm{f}}_{\mathrm{k}}\left(10.1\mathrm{m}\right) \\ {\mathrm{f}}_{\mathrm{k}}=49\mathrm{N} \end{gathered}$$

Therefore, the average frictional force acting on the child is 49 N .

The vertical line through the top of the slide is tangent to the circle.

If θ₁and θ₂ is the initial angle made by the child with horizontal respectively then using the analogy with a swinging pendulum we can write that using equation (3) as:

$∆\mathrm{h}=\mathrm{R}\left(1-{\mathrm{cos\theta }}_2\right)-\mathrm{R}\left(1-{\mathrm{cos\theta }}_1\right)$

But,${\mathrm{\theta }}_1=90°$

Thus, the angle value is given as:

$$\begin{gathered} ∆\mathrm{h}=-\mathrm{R}\left({\mathrm{cos\theta }}_2\right) \\ {\mathrm{\theta }}_2={\cos }^{-1}\left(-\frac{∆\mathrm{h}}{\mathrm{R}}\right) \\ ={\cos }^{-1}\left(-\frac{\left(-4\right)}{12}\right) \\ =72.5° \end{gathered}$$

The angle subtended by the arc is given as:

$$\begin{gathered} ∆\mathrm{\theta }=\left(90°-72.5°\right) \\ =19.5° \\ =0.34\mathrm{rad} \end{gathered}$$

The length of the slide using equation (2) is given as:

$$\begin{gathered} \mathrm{S}\text{'}=12\mathrm{m}\left(0.34\mathrm{rad}\right) \\ =4.1\mathrm{m} \end{gathered}$$

Therefore, the length of the slide if a vertical line through the top of the slide is tangent to the circle is 4.1 m

From parts b) and c) we can get the value of average frictional force as follows using equations (4), (5), and (6) as follows:

$$\begin{gathered} 0\mathrm{J}=\frac{1}{2}{\mathrm{mv}}^2-\mathrm{mgh}+{\mathrm{f}}_{\mathrm{k}}\mathrm{S} \\ \frac{1}{2}\left(25\mathrm{kg}\right){\left(6.2\mathrm{m}/\mathrm{s}\right)}^2-\left(25\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(4\mathrm{m}\right)+{\mathrm{f}}_{\mathrm{k}}\text{'}\left(4.1\mathrm{m}\right) \\ {\mathrm{f}}_{\mathrm{k}}\text{'}=1.2\times {10}^2\mathrm{N} \end{gathered}$$

Therefore, the average frictional force acting on the child if a vertical line through the top of the slide is tangent to the circle is $1.2\times {10}^2\mathrm{N}$