Chapter 8, Problem 8.49AP

A skateboarder with his board can be modeled as a particle of mass 76.0 kg, located at his center of mass (which we will study in Chapter 9). As shown in Figure P8.49, the skateboarder starts from rest in a crouch-ing position at one lip of a half-pipe (point Ⓐ). The half-pipe is one half of a cylinder of radius 6.80 m with its axis horizontal. On his descent, the skateboarder moves without friction so that his center of mass moves through one quarter of a circle of radius 630 m. (a) Find his speed at the bottom of the half-pipe (point Ⓑ (b) Immediately after passing point Ⓑhe stands up and raises his arms, lifting his center of mass from 0.500 in to 0.950 m above the concrete (point ©). Next, the skateboarder glides upward with his center of mass moving in a quarter circle of radius 5.85 m. His body is horizontal when he passes point Ⓓ, the far lip of the half-pipe. As he passes through point Ⓓ, the speed of the skateboarder is 5.14 m/s. How much chemical potential energy in the body of the skateboarder was converted to mechanical energy in the skateboarder—Earth system when he stood up at point Ⓑ? (c) How high above point Ⓓ does he rise? Caution: Do not try this stunt yourself without the required skill and protective equipment.

Figure P8.49

To determine

The speed at the bottom of the half pipe.

Answer to Problem 8.49AP

The speed at the bottom of the half pipe is $11.1\text{m}/\text{s}$.

Explanation of Solution

Given info: The mass of the particle is $76.0\text{kg}$, radius of the cylinder is $6.80\text{m}$, radius of circle is $6.30\text{m}$, speed of the skateboarder is $5.14\text{m}/\text{s}$.

The formula to calculate the initial gravitational potential energy of the particle at point $A$ is,

$U_{\text{gA}}=mg{x}_{\text{A}}$

Here,

$m$ is the mass of the particle.

$g$ is the acceleration due to gravity.

$x_{\text{A}}$ is the initial height of the particle at point $A$.

Thus, the initial gravitational potential energy of the particle at point $A$ is $mg{x}_{\text{A}}$.

The formula to calculate the gravitational potential energy is,

$U_{\text{gB}}=mg{x}_{\text{B}}$

Here,

$m$ is the mass of the particle.

$g$ is the acceleration due to gravity.

$x_{\text{B}}$ is the height of the particle at point $B$.

The height of the particle at point $B$ is 0.

Substitute 0 for $x_{\text{2B}}$ in the above formula to find $U_{\text{2B}}$.

$\begin{array}{c}{U}_{\text{gB}}=mg{x}_{\text{B}}\\ =mg\left(0\right)\\ =0\text{J}\end{array}$

Thus, the gravitational potential energy at point $B$ is $0\text{J}$.

The formula to calculate the initial kinetic energy of the particle is,

$K.E_{\text{1A}}=\frac{1}{2}m{v}_{1\text{A}}{}^2$

Here,

$m$ is the mass of the particle.

$v_{\text{1A}}$ is the velocity of the particle at point $A$.

The initial velocity of the particle is 0 as the particle is at rest then the kinetic energy at point $A$ is 0.

Substitute 0 for $v_{\text{1A}}$ in the above formula to find $K.E_{\text{1A}}$.

$\begin{array}{r}K.E_{\text{1A}}=\frac{1}{2}m{v}_{1\text{A}}{}^2\\ =\frac{1}{2}m{\left(0\right)}^2\\ =0\text{J}\end{array}$

Thus, the kinetic energy of the particle at point $A$ is $0\text{J}$.

The formula to calculate the kinetic energy of the particle at point $B$ is,

$K.E_{\text{2B}}=\frac{1}{2}m{v}_{\text{2B}}{}^2$

Here,

$m$ is the mass of the particle.

$v_{\text{1A}}$ is the velocity of the particle at point $A$.

Thus, the kinetic energy at point $B$ is $\frac{1}{2}m{v}_{\text{2B}}{}^2$.

The formula to calculate the law of conservation of energy is,

$K.E_{\text{1A}}+U_{\text{gA}}=K.E_{\text{2B}}+U_{\text{gB}}$

Here,

$K.E_{\text{1A}}$ is the kinetic energy at point $A$.

$U_{\text{gA}}$ is the gravitational potential energy at point $A$.

$K.E_{\text{2B}}$ is the kinetic energy at point $B$.

$U_{\text{gB}}$ is the gravitational potential energy at point $B$.

Substitute $\frac{1}{2}m{v}_{\text{2B}}{}^2$ for $K.E_{\text{2B}}$ , $\frac{1}{2}m{v}_{1\text{A}}{}^2$ for $K.E_{\text{1A}}$, $mg{x}_{\text{1A}}$ for $U_{\text{1A}}$ , $mg{x}_{\text{2B}}$ for $U_{\text{2B}}$  in the above formula to find $v_{\text{2B}}$.

$\begin{array}{c}K.E_{\text{1A}}+U_{\text{gA}}=K.E_{\text{2B}}+U_{\text{gB}}\\ \frac{1}{2}m{v}_{\text{1A}}{}^2+mg{x}_{\text{1A}}=\frac{1}{2}m{v}_{\text{2B}}{}^2+mg{x}_{\text{2B}}\end{array}$

Substitute $0$ for $v_{\text{1A}}$, 0 for $x_{\text{2B}}$ in the above formula to find $v_{\text{2B}}$.

$\begin{array}{c}\frac{1}{2}m{v}_{\text{1A}}{}^2+mg{x}_{\text{1A}}=\frac{1}{2}m{v}_{\text{2B}}{}^2+mg{x}_{\text{2B}}\\ 0+mg{x}_{\text{1A}}=\frac{1}{2}m{v}_{\text{2B}}{}^2+0\end{array}$

Rearrange the above formula for $v_{\text{B}}$.

$\begin{array}{c}mg{x}_{\text{1A}}=\frac{1}{2}m{v}_{\text{2B}}{}^2\\ v_{\text{2B}}{}^2=2g{x}_{\text{1A}}\\ v_{\text{2B}}=\sqrt{2g{x}_{\text{1A}}}\end{array}$

Substitute $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $6.30\text{}\text{m}$ for $x_{\text{1A}}$ in the above formula to find $v_{\text{2B}}$.

$\begin{array}{c}{v}_{\text{2B}}=\sqrt{2g{x}_{\text{1A}}}\\ =\sqrt{2\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left(6.30\text{m}\right)}\\ =\sqrt{123.48{\text{m}}^2/{\text{s}}^{\text{2}}}\\ =11.1\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed at the bottom of the half pipe is $11.1\text{m}/\text{s}$.

To determine

The amount of chemical potential energy converted into mechanical energy in the skateboarder- Earth system when he stood up at point $B$.

Answer to Problem 8.49AP

The amount of chemical potential energy converted into mechanical energy in the skateboarder- Earth system when he stood up at point $B$ is $1.00\times {10}^3\text{J}$.

Explanation of Solution

Given info: The mass of the particle is $76.0\text{kg}$ , radius of the cylinder is $6.80\text{m}$,  radius of circle is $6.30\text{m}$, speed of the skateboarder is $5.14\text{m}/\text{s}$.

The formula to calculate the centripetal acceleration of the particle  at the point $B$ is,

$a=\frac{v_{\text{2B}}{}^2}{r}$

Here,

$v_{\text{2B}}{}^2$ is the speed of the particle at point $B$.

$r$ is the radius of the circle.

Substitute $11.1\text{m}/\text{s}$ for $v_{\text{2B}}$ , $6.30\text{m}$ for $r$ in the above formula to find $a$.

$\begin{array}{c}a=\frac{v_{\text{2B}}{}^2}{r}\\ =\frac{{\left(11.1\text{m}/\text{s}\right)}^2}{6.30\text{m}}\\ =19.55{\text{m}/\text{s}}^2\\ \approx 19.6{\text{m}/\text{s}}^2\end{array}$

Thus, the centripetal acceleration of the particle at point $B$  is $19.6{\text{m}/\text{s}}^2$.

The formula to calculate the normal force acting on the particle at point $B$ is,

$N_{\text{B}}=ma+mg$

Here,

$m$ is the mass of the particle.

$a$ is the Centripetal acceleration of the particle.

$g$ is the acceleration due to gravity.

Substitute $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $19.6\text{m}/{\text{s}}^{\text{2}}$ for $a$, $76.00\text{kg}$ in the above formula to find $N_{\text{B}}$.

$\begin{array}{c}{N}_{\text{B}}=ma+mg\\ =\left(76.00\text{kg}\right)\left(19.6\text{m}/{\text{s}}^{\text{2}}\right)+\left(76.00\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =1489.6\text{}\text{N+744}\text{.8}\text{N}\\ \text{=2234}\text{.4}\text{}\text{N}\end{array}$

Thus, the value of normal force acting on the particle at point $B$  is $\text{2234}\text{.4}\text{}\text{N}$.

The formula to calculate the chemical energy of the skateboarder converted into mechanical energy at point $B$ is,

$W=N_{\text{B}}\left(r_2-r_1\right)\cos \theta$

Here,

$N_{\text{B}}$ is the normal force at point $B$.

$r_2$ final position of center of mass.

$r_1$ initial position of center of mass.

$\theta$ is the angle between the normal force and center of mass.

Substitute $\text{2234}\text{.4}\text{}\text{N}$ for $N_{\text{B}}$ , $0.950\text{m}$ for $r_2$ , $0.500\text{m}$ for $r_1$ in the above formula to find $W$.

$\begin{array}{c}W=\left(\text{2234}\text{.4}\text{}\text{N}\right)\left(0.950\text{}\text{m-0}\text{.500}\text{m}\right)\cos 0°\\ =1005.48\text{J}\\ \cong 1.00\times {10}^3\text{J}\end{array}$

Conclusion:

Therefore, the amount of chemical potential energy converted into mechanical energy in the skateboarder- Earth system when he stood up at point $B$ is $1.00\times {10}^3\text{J}$.

To determine

The height above point $D$ does the skateboarder rise.

Answer to Problem 8.49AP

The height above point $D$ does the skateboarder rise is $\text{1}\text{.35}\text{m}$.

Explanation of Solution

Given info: The mass of the particle is $76.0\text{kg}$ , radius of the cylinder is $6.80\text{m}$,  radius of circle is $6.30\text{m}$, speed of the skateboarder is $5.14\text{m}/\text{s}$.

The formula to calculate the initial gravitational potential energy of the particle at point $D$ is,

$U_{\text{gD}}=mg{x}_D$

Here,

$m$ is the mass of the particle.

$g$ is the acceleration due to gravity.

$x_{\text{D}}$ is the initial height of the particle at point $D$.

Thus, the initial gravitational potential energy of the particle at point $D$ is $mg{x}_{\text{D}}$.

The formula to calculate the gravitational potential energy is,

$U_{\text{gE}}=mg{x}_{\text{E}}$

Here,

$m$ is the mass of the particle.

$g$ is the acceleration due to gravity.

$x_{\text{E}}$ is the height of the particle at point $E$.

Thus, the gravitational potential energy at point $E$ is $mg{x}_{\text{E}}$

The formula to calculate the initial kinetic energy of the particle is,

$K.E_{\text{1D}}=\frac{1}{2}m{v}_{1\text{D}}{}^2$

Here,

$m$ is the mass of the particle.

$v_{\text{1D}}$ is the velocity of the particle at point $D$.

Thus, the kinetic energy of the particle at point $D$ is $\frac{1}{2}m{v}_{1\text{D}}{}^2$.

The formula to calculate the kinetic energy of the particle at point $E$ is,

$K.E_{\text{2E}}=\frac{1}{2}m{v}_{\text{2E}}{}^2$

Here,

$m$ is the mass of the particle.

$v_{\text{2E}}$ is the velocity of the particle at point $E$.

Thus, the kinetic energy at point $E$ is $\frac{1}{2}m{v}_{\text{2B}}{}^2$.

The formula to calculate the law of conservation of energy is,

$K.E_{\text{1D}}+U_{\text{gD}}=K.E_{\text{2E}}+U_{\text{gD}}$

Here,

$K.E_{\text{1D}}$ is the kinetic energy at point $D$.

$U_{\text{gD}}$ is the gravitational potential energy at point $D$.

$K.E_{\text{2E}}$ is the kinetic energy at point $E$.

$U_{\text{gE}}$ is the gravitational potential energy at point $E$.

Substitute $\frac{1}{2}m{v}_{\text{2E}}{}^2$ for $K.E_{\text{2E}}$ , $\frac{1}{2}m{v}_{1\text{D}}{}^2$ for $K.E_{\text{1D}}$, $mg{x}_{\text{1D}}$ for $U_{\text{1D}}$ , $mg{x}_{\text{2E}}$ for $U_{\text{2E}}$  in the above formula to find $x_{\text{1D}}-x_{\text{2E}}$.

$\begin{array}{c}K.E_{\text{1D}}+U_{\text{gD}}=K.E_{\text{2E}}+U_{\text{gE}}\\ \frac{1}{2}m{v}_{\text{1D}}{}^2+mg{x}_{\text{1D}}=\frac{1}{2}m{v}_{\text{2E}}{}^2+mg{x}_{\text{2E}}\end{array}$

Substitute $0$ for $v_{2\text{E}}$ in the above formula to find $x_{\text{1D}}-x_{\text{2E}}$.

$\begin{array}{c}\frac{1}{2}m{v}_{\text{1D}}{}^2+mg{x}_{\text{1D}}=\frac{1}{2}m{v}_{\text{2E}}{}^2+mg{x}_{\text{2E}}\\ mg\left(x_{\text{2E}}-x_{\text{1D}}\right)=\frac{1}{2}m{v}_{\text{1D}}{}^2\\ x_{\text{2E}}-x_{\text{1D}}=\frac{v_{\text{D}}{}^2}{2g}\end{array}$

Substitute $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $5.14\text{m}/\text{s}$ for $v_{\text{D}}$ in the above formula to find $x_{\text{1D}}-x_{\text{2E}}$.

$\begin{array}{c}{x}_{\text{2E}}-x_{\text{1D}}=\frac{v_{\text{D}}{}^2}{2g}\\ =\frac{{\left(5.14\text{m}/\text{s}\right)}^2}{2\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}\\ =1.3479\text{m}\\ \text{=1}\text{.35}\text{m}\end{array}$

Conclusion:

Therefore, the height above point $D$ does the skateboarder rise is $\text{1}\text{.35}\text{m}$.