Chapter 11, Problem 11.48AP

(a)

To determine

The speed at the bottom of the half-pipe.

Answer to Problem 11.48AP

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

Explanation of Solution

Given info: The mass of particle is $76.0\text{kg}$, the distance from the ground of its center of mass is $0.500\text{m}$. The radius of the half of the pipe is $6.80\text{m}$.

Write the expression for conservation of work and energy law.

$KE=U$ (1)

Here,

$KE$ is the kinetic energy which is at point B.

$U$ is the potential energy when the skateboarder is at rest at point A.

The skateboarder is at rest at point A, so there is a potential energy at point A,

$U=mgx$

Here,

$x$ is the radius of the circle.

$m$ is the mass of the particle

$g$ is the acceleration due to gravity.

The center of mass moves through one quarter of the circle.

The radius of the circle is,

$\begin{array}{c}x=6.80\text{m}-0.500\text{m}\\ =6.30\text{m}\end{array}$

The skateboarder is in motion so it acquires the kinetic energy at point B,

$KE=\frac{1}{2}m{v}^2$

Here,

$v$ is the speed of skateboarder at the bottom of the half pipe.

Substitute $\frac{1}{2}m{v}^2$ for $KE$ and $mgx$ for $U$ in equation (1).

$\begin{array}{c}\frac{1}{2}m{v}^2=mgx\\ v^2=2gx\\ v=\sqrt{2gx}\end{array}$

Substitute $6.30\text{m}$ for $x$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in above equation.

$\begin{array}{c}v=\sqrt{2\left(9.8\text{m}/{\text{s}}^2\right)\left(6.30\text{m}\right)}\\ =11.1\text{m}/\text{s}\end{array}$

Conclusion:

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

(b)

To determine

The angular momentum of him about the center of curvature at the point B.

Answer to Problem 11.48AP

The angular momentum of him about the center of curvature at the point B is $5.314\times {10}^3\text{kg}\cdot {\text{m}}^2/\text{s}$.

Explanation of Solution

Given info: The mass of particle is $76.0\text{kg}$, the distance from the ground of its center of mass is $0.500\text{m}$. The radius of the half of the pipe is $6.80\text{m}$.

Write the expression for the angular momentum about the center of curvature.

$L=mvx$

Here,

$L$ is the angular momentum about the center of curvature.

Substitute $6.30\text{m}$ for $x$, $76.0\text{kg}$ for $m$ and $11.1\text{m}/\text{s}$ for $v$ in above equation.

$\begin{array}{c}L=\left(76.0\text{kg}\right)\left(11.1\text{m}/\text{s}\right)\left(6.30\text{m}\right)\\ =5.314\times {10}^3\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Conclusion:

Therefore, the angular momentum of him about the center of curvature at the point B is $5.314\times {10}^3\text{kg}\cdot {\text{m}}^2/\text{s}$.

(c)

To determine

To explain: The angular momentum of him is constant in this maneuver, whereas the kinetic energy of his body is not constant.

Answer to Problem 11.48AP

After the passing point B, there is no torque about the axis of the channel act on him so; the angular momentum will be constant, but his legs convert the chemical energy into mechanical energy and the kinetic energy of his body is not constant.

Explanation of Solution

Given info: The mass of particle is $76.0\text{kg}$, the distance from the ground of its center of mass is $0.500\text{m}$. The radius of the half of the pipe is $6.80\text{m}$.

A skateboarder passes the point B, so there is no tangential force acts on him because the wheels on the skate prevent this force. The torque is zero due to no tangential force, so the angular momentum will be constant.

The kinetic energy increase because his legs convert chemical energy into mechanical energy and the kinetic energy will not be constant. While the normal force rises trajectory to enhance his linear momentum.

Conclusion:

Therefore, after the passing point B, there is no torque about the axis of the channel act on him so; the angular momentum will be constant, but his legs convert the chemical energy into mechanical energy and the kinetic energy of his body is not constant.

(d)

To determine

The speed immediately after the skateboarder stands up.

Answer to Problem 11.48AP

The speed of skateboarder after he stands up is $12.0\text{m}/\text{s}$.

Explanation of Solution

Given info: The mass of particle is $76.0\text{kg}$, the distance from the ground of its center of mass is $0.500\text{m}$. The radius of the half of the pipe is $6.80\text{m}$.

The skateboarder stands up, so the distance is,

$\begin{array}{c}{x}^{\prime }=6.8\text{m}-0.950\text{m}\\ \text{=5}\text{.85}\text{m}\end{array}$

Write the expression for angular momentum.

$L=m{v}^{\prime }{x}^{\prime }$

Here,

$v^{\prime }$ is the speed of skateboarder at point C.

Substitute $76.0\text{kg}$ for $m$, $\text{5}\text{.85}\text{m}$ for $x^{\prime }$ and $5.314\times {10}^3\text{kg}\cdot {\text{m}}^2/\text{s}$ for $L$ in above equation.

$\begin{array}{c}5.314\times {10}^3\text{kg}\cdot {\text{m}}^2/\text{s}=\left(76.0\text{kg}\right)v^{\prime }\left(\text{5}\text{.85}\text{m}\right)\\ v^{\prime }=\frac{5.314\times {10}^3\text{kg}\cdot {\text{m}}^2/\text{s}}{\left(76.0\text{kg}\right)\left(\text{5}\text{.85}\text{m}\right)}\\ =11.95\text{m}/\text{s}\\ \approx 12.0\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of skateboarder after he stands up is $12.0\text{m}/\text{s}$.

(e)

To determine

The amount of chemical energy in the skateboarder’s leg was converted into mechanical energy in skateboarder-Earth system when he stood up.

Answer to Problem 11.48AP

The amount of chemical energy in the skateboarder’s leg was converted into mechanical energy in skateboarder-Earth system when he stood up is $\text{1}\text{.125}\text{kJ}$.

Explanation of Solution

Given info: The mass of particle is $76.0\text{kg}$, the distance from the ground of its center of mass is $0.500\text{m}$. The radius of the half of the pipe is $6.80\text{m}$.

At point B, the skate boarder has kinetic and chemical energy is,

$E_{\text{b}}=K{E}_{\text{B}}+W$

Here,

$E_{\text{b}}$ is the total energy at point B.

$W$ is the chemical; energy.

$K{E}_{\text{B}}$ is the kinetic energy at point B.

At point C, he has kinetic energy due and the potential energy is,

$E_{\text{c}}=K{E}_{\text{c}}+U_{\text{c}}$

Here,

$E_{\text{c}}$ is the total energy at point C.

$K{E}_{\text{c}}$ is the kinetic energy at point C.

$U_{\text{c}}$ is the potential energy at point C.

Write the expression of the conservation of energy.

$E_{\text{b}}=E_{\text{c}}$

Substitute $K{E}_{\text{B}}+W$ for $E_{\text{b}}$ and $K{E}_{\text{c}}+U_{\text{c}}$ for $E_{\text{c}}$ in above equation.

$K{E}_{\text{B}}+W=K{E}_{\text{c}}+U_{\text{c}}$ . (2)

Write the expression for the kinetic energy at point B.

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

Substitute $76.0\text{kg}$ for $m$ and $11.1\text{m}/\text{s}$ for $v$ in above equation.

$\begin{array}{c}K{E}_{\text{B}}=\frac{1}{2}\left(76.0\text{kg}\right){\left(11.1\text{m}/\text{s}\right)}^2\\ =4681.98\text{kg}\cdot {\left(\text{m}/\text{s}\right)}^2\end{array}$

Thus, the kinetic energy at point B is $4681.98\text{kg}\cdot {\left(\text{m}/\text{s}\right)}^2$.

Write the expression for the kinetic energy at point C.

$K{E}_{\text{c}}=\frac{1}{2}m{v^{\prime }}^2$

Substitute $76.0\text{kg}$ for $m$ and $12.0\text{m}/\text{s}$ for $v^{\prime }$ in above equation.

$\begin{array}{c}K{E}_{\text{c}}=\frac{1}{2}\left(76.0\text{kg}\right){\left(12.0\text{m}/\text{s}\right)}^2\\ =5472\text{kg}\cdot {\left(\text{m}/\text{s}\right)}^2\end{array}$

Thus, the kinetic energy at point C is $5472\text{kg}\cdot {\left(\text{m}/\text{s}\right)}^2$.

Write the expression for potential energy at point C.

$U_{\text{c}}=mg{x}^{″}$ . (3)

Here,

$x^{″}$ is the radius of the pipe at point C.

The radius of the pipe at point C,

$\begin{array}{c}{x}^{″}=0.950\text{m}-0.500\text{m}\\ =\text{0}\text{.450}\text{m}\end{array}$

Substitute $\text{0}\text{.450}\text{m}$ for $x^{″}$, $76.0\text{kg}$ for $m$ and $9.8\text{m}/{\text{s}}^2$ for $g$ in equation (3).

$\begin{array}{c}{U}_{\text{c}}=\left(76.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^2\right)\left(\text{0}\text{.450}\text{m}\right)\\ =335.16\text{kg}\cdot {\left(\text{m}/\text{s}\right)}^2\end{array}$

Thus, the potential energy at point C is $335.16\text{kg}\cdot {\left(\text{m}/\text{s}\right)}^2$.

Substitute $335.16\text{kg}\cdot {\left(\text{m}/\text{s}\right)}^2$ for $U_{\text{c}}$, $5472\text{kg}\cdot {\left(\text{m}/\text{s}\right)}^2$ for $K{E}_{\text{c}}$ and $4681.98\text{kg}\cdot {\left(\text{m}/\text{s}\right)}^2$ for $K{E}_{\text{B}}$ in equation (2).

$\begin{array}{c}4681.98\text{kg}\cdot {\left(\text{m}/\text{s}\right)}^2+W=5472\text{kg}\cdot {\left(\text{m}/\text{s}\right)}^2+335.16\text{kg}\cdot {\left(\text{m}/\text{s}\right)}^2\\ W=5807.16\text{J}-4681.98\text{J}\\ =1125.18\text{J}\\ \text{=1}\text{.125}\text{kJ}\end{array}$

Conclusion:

Therefore, the amount of chemical energy in the skateboarder’s leg was converted into mechanical energy in skateboarder-Earth system when he stood up is $\text{1}\text{.125}\text{kJ}$.