Chapter 10, Problem 84P

(a)

To determine

The speed of the skateboarder at the bottom of the half pipe.

Answer to Problem 84P

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

Explanation of Solution

The total energy of the system is conserved. Since the skateboarder is starting from rest the initial kinetic energy will be zero, and when he reaches at the bottom of the half pipe the potential energy will be zero. Thus the potential energy is being converted to kinetic energy.

Write the conservation equation for energy.

  ${\left(K+U_g\right)}_A={\left(K+U_g\right)}_B$        (I)

Here, $K_A$ is the kinetic energy at point A, $U_{gA}$ is the potential energy at A, $K_B$ is the kinetic energy at B, and $U_{gB}$ is the potential energy at B.

Substitute, $0$ for $K_A$, and $U_{gB}$, $mg{y}_A$ for $U_{gA}$, and $\frac{1}{2}m{\upsilon }_B^2$ for $K_B$ in equation (I).

  $0+mg{y}_A=\frac{1}{2}m{\upsilon }_B^2+0$        (II)

Here, $m$ is the mass of the skateboarder, $g$ is the acceleration due to gravity, $y_A$ is the distance from point A to B, ${\upsilon }_B$ is the speed when reaching at B.

Rearrange to obtain an expression for ${\upsilon }_B$.

  $\begin{array}{c}0+mg{y}_A=\frac{1}{2}m{\upsilon }_B^2+0\\ {\upsilon }_B=\sqrt{2g{y}_A}\end{array}$        (III)

Conclusion

Substitute, $9.80\text{m}/{\text{s}}^2$ for $g$, $6.30\text{m}$ for $y_A$ in equation (III) to find the speed of the skateboarder at the bottom of the half pipe.

  $\begin{array}{c}{\upsilon }_B=\sqrt{2\left(9.80\text{m}/{\text{s}}^2\right)\left(6.30\text{m}\right)}\\ =11.1\text{m}/\text{s}\end{array}$

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

(b)

To determine

The angular momentum about the center of curvature at the bottom of the half pipe.

Answer to Problem 84P

The angular momentum about the center of curvature at the bottom of the half pipe is $\underset{\_}{5.32\times {10}^3\text{kg}\cdot {\text{m}}^2/\text{s}}$.

Explanation of Solution

Write the expression for the angular momentum.

  $L=m\upsilon r$        (IV)

Here, $r$ is the radius of the half pipe.

Conclusion:

Substitute, $76.0\text{kg}$ for $m$, $11.1\text{m}/\text{s}$ for $\upsilon$, $\left(6.80\text{m}-\text{0}\text{.500}\text{m}\right)$ for $r$ in equation (IV) to obtain the angular momentum about the center of curvature.

  $\begin{array}{c}L=\left(76.0\text{kg}\right)\left(11.1\text{m}/\text{s}\right)\left(6.80\text{m}-\text{0}\text{.500}\text{m}\right)\\ =5.32\times {10}^3\text{kg}\cdot {\text{m}}^2/\text{s}\end{array}$

Therefore, the angular momentum about the center of curvature at the bottom of the half pipe is $\underset{\_}{5.32\times {10}^3\text{kg}\cdot {\text{m}}^2/\text{s}}$.

(c)

To determine

To explain why the angular momentum is a constant and kinetic energy changes when the skateboarder stand up and lift his arms after passing through the bottom point.

Answer to Problem 84P

Since there is no external torque, the angular momentum is conserved, but at the same time the chemical energy of the skateboarder is being converts to mechanical energy, hence the kinetic energy is not conserved.

Explanation of Solution

As the person stand up and lift his arms after passing through the bottom point there is no torque acting about the axis of the channel on him, and also the wheels of the skateboard prevent the tangential force acting on him. Since there is no torque and no external force the angular momentum will be kept conserved.

But as the person stand up and lift his arms after passing through the bottom point, his legs convert the chemical energy to mechanical energy required to moving forward. According to work energy theorem the work done by the skateboarder results in increasing of kinetic energy. Hence the energy of the system is not conserved.

Conclusion:

Therefore, since there is no external torque, the angular momentum is conserved, but at the same time the chemical energy of the skateboarder is being converts to mechanical energy, hence the kinetic energy is not conserved.

(d)

To determine

The speed of the skateboarder immediately after he stood up.

Answer to Problem 84P

The speed of the skateboarder immediately after he stood up is $\underset{\_}{12.0\text{m}/\text{s}}$.

Explanation of Solution

Consider equation (II) to obtain the answer.

Conclusion:

Substitute, $5.32\times {10}^3\text{kg}\cdot {\text{m}}^2/\text{s}$ for $L$, $76.0\text{kg}$ for $m$, and $\left(6.80\text{m}-\text{0}\text{.950}\text{m}\right)$ for $r$ in equation (II) to obtain the value of $\upsilon$.

$\begin{array}{c}5.32\times {10}^3\text{kg}\cdot {\text{m}}^2/\text{s}=76.0\text{kg}\times \left(6.80\text{m}-\text{0}\text{.950}\text{m}\right)\times \upsilon \\ \upsilon =\frac{5.32\times {10}^3\text{kg}\cdot {\text{m}}^2/\text{s}}{76.0\text{kg}\times \left(6.80\text{m}-\text{0}\text{.950}\text{m}\right)}\\ =12.0\text{m}/\text{s}\end{array}$

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

(e)

To determine

The amount of chemical energy converted to mechanical energy.

Answer to Problem 84P

The amount of chemical energy converted to mechanical energy is $\underset{\_}{1.08\text{kJ}}$.

Explanation of Solution

From point B to C chemical energy is being converted to mechanical energy. the total energy in this case is also a constant.

Write the equation for the conservation of energy from point B to C.

  ${\left(K+U_g\right)}_B+U_{\text{chemical,}B}={\left(K+U_g\right)}_C$        (V)

Here, $U_{\text{chemical}}$ is the chemical energy at point B.

Substitute, $\frac{1}{2}m{\upsilon }_B^2$ for $K_B$, $0$ for $U_{g,B}$, $\frac{1}{2}m{\upsilon }_C^2$ for $K_C$, and $mg{y}_C$ for $U_{gC}$ in equation (V) to find $U_{\text{chemical}}$.

  $\frac{1}{2}m{\upsilon }_B^2+0+U_{\text{chemical,}B}=\frac{1}{2}m{\upsilon }_C^2+mg{y}_C$        (VI)

Here, $ylC$ is the distance between B and C.

Conclusion:

Substitute, $76.0\text{kg}$ for $m$, $11.1\text{m}/\text{s}$ for ${\upsilon }_B$, $12.0\text{m}/\text{s}$ for ${\upsilon }_C$, $9.80\text{m}/{\text{s}}^2$ for $g$, and $\left(0.950\text{m}-\text{0}\text{.500}\text{m}\right)$ for $y_C$ in equation (VI) to find $U_{\text{chemical}}$.

  $\begin{array}{c}\frac{1}{2}\left(76.0\text{kg}\right){\left(11.1\text{m}/\text{s}\right)}^2+0+U_{\text{chemical,}B}=\frac{1}{2}\left(76.0\text{kg}\right){\left(12.0\text{m}/\text{s}\right)}^2+\left(76.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\left(0.950\text{m}-\text{0}\text{.500}\text{m}\right)\\ U_{\text{chemical,}B}=5.44\text{kJ}-\text{4}\text{.69}\text{kJ+335}\text{J}\\ \text{=1}\text{.08}\text{kJ}\end{array}$

Therefore, the amount of chemical energy converted to mechanical energy is $\underset{\_}{1.08\text{kJ}}$.