Chapter 8, Problem 53P

A giant swing at an amusement park consists of a 365-kg uniform arm 10.0 m long, with two seats of negligible mass connected at the lower end of the arm (Fig. P8.53). (a) How far from the upper end is the center of mass of the arm? (b) The gravitational potential energy of the arm is the same as if all its mass were concentrated at the center of mass. If the arm is raised through a 45.0‘ angle, find the gravitational potential energy, where the zero level is taken to be 10.0 m below the axis, (c) The arm drops from rest from the position described in part (b). Find the gravitational potential energy of the system when it reaches the vertical orientation. (d) Find the speed of the seats at the bottom of the swing.

To determine

The distance from the upper end to the centre of mass of the arm of the giant swing.

Answer to Problem 53P

Solution: The distance from the upper end to the centre of mass of the arm of the giant swing is $y=10.0\text{m}-\left(5.00\text{m}\right)\cos \theta$.

Explanation of Solution

Given info: The mass of the uniform arm is $365\text{kg}$. Length of the arm is $10.0\text{m}$.

The representative diagram of swing is given below.

• $c$ is the centre of gravity

Since the arm is a uniform rod, its centre of gravity will be exactly at the centre of the rod. The height of the centre of the rod from ground will be,

$y_c=10.0\text{m}-\left(5.00\text{m}\right)\cos \theta$

• $y_c$ is the y-component of centre of gravity

Conclusion:

The distance from the upper end to the centre of mass of the arm of the giant swing is $y_c=10.0\text{m}-\left(5.00\text{m}\right)\cos \theta$.

To determine

The gravitational potential energy of the arm.

Answer to Problem 53P

Solution: The distance from the upper end to the centre of mass of the arm of the giant swing is $y_c=10.0\text{m}-\left(5.00\text{m}\right)\cos \theta$.

Explanation of Solution

Given info: The mass of the uniform arm is $365\text{kg}$. Length of the arm is $10.0\text{m}$. The arm is raised through an angle of $45.0°$. The zero level of the potential is taken $10.0\text{m}$ below the axis.

The representative diagram of swing is given below.

Since the arm is a uniform rod, its centre of gravity will be exactly at the centre of the rod. The height of the centre of the rod from ground will be,

$y_c=10.0\text{m}-\left(5.00\text{m}\right)\cos \theta$ (1)

The potential energy will be given by,

$P{E}_g=mg{y}_c$ (2)

• $P{E}_g$ is the gravitational potential energy
• $m$ is the mass
• $g$ is the free fall acceleration

Combining equation (1) and (2),

$P{E}_g=mg\left(10.0\text{m}-\left(5.00\text{m}\right)\cos \theta \right)$

Substitute $365\text{kg}$ for $m$, $9.8\text{m}\cdot {\text{s}}^{\text{-2}}$ for $g$, $45.0°$ for $\theta$ to determine the gravitational potential energy.

$\begin{array}{c}P{E}_g=\left(365\text{kg}\right)\left(9.8\text{m}\cdot {\text{s}}^{\text{-2}}\right)\left(10.0\text{m}-\left(5.00\text{m}\right)\cos 45.0°\right)\\ =2.31\times {10}^4\text{J}\end{array}$

Conclusion:

The potential energy is found to be $2.31\times {10}^4\text{J}$.

To determine

The gravitational potential energy of the system when it reaches the vertical orientation.

Answer to Problem 53P

Solution: The potential energy at vertical positon is found to be $1.79\times {10}^4\text{J}$.

Explanation of Solution

Given info: The mass of the uniform arm is $365\text{kg}$. Length of the arm is $10.0\text{m}$. The zero level of the potential is taken $10.0\text{m}$ below the axis.

The representative diagram of swing is given below.

Since the arm is a uniform rod, its centre of gravity will be exactly at the centre of the rod. The height of the centre of the rod from ground will be,

$y_c=10.0\text{m}-\left(5.00\text{m}\right)\cos \theta$ (1)

• $\theta$ is the angle of the arm with vertical

The potential energy will be given by,

$P{E}_g=mgy$ (2)

• $m$ is the mass
• $g$ is the free fall acceleration

Combining equation (1) and (2),

$P{E}_g=mg\left(10.0\text{m}-\left(5.00\text{m}\right)\cos \theta \right)$

In vertical position $\theta$ will be $0°$.

Substitute $365\text{kg}$ for $m$, $9.8\text{m}\cdot {\text{s}}^{\text{-2}}$ for $g$, $0°$ for $\theta$ to determine the gravitational potential energy.

$\begin{array}{c}P{E}_g=\left(365\text{kg}\right)\left(9.8\text{m}\cdot {\text{s}}^{\text{-2}}\right)\left(10.0\text{m}-\left(5.00\text{m}\right)\cos 0°\right)\\ =1.79\times {10}^4\text{J}\end{array}$

Conclusion:

The potential energy at vertical positon is found to be $1.79\times {10}^4\text{J}$.

To determine

The speed of the seats at the bottom of the swing.

Answer to Problem 53P

Solution: The linear speed of the system at the bottom of the swing is $9.28\text{m}\cdot {\text{s}}^{\text{-1}}$.

Explanation of Solution

Given info: The mass of the uniform arm is $365\text{kg}$. Length of the arm is $10.0\text{m}$. The zero level of the potential is taken $10.0\text{m}$ below the axis.

The representative diagram of swing is given below.

Using the conservation of mechanical energy of the system it’s obtained,

$\frac{1}{2}{I}_e{\omega }_f^2+mg{y}_f=0+mg{y}_i$

• $I_e$ is moment of inertia about one end of the rod
• ${\omega }_f$ is the angular speed
• $m$ is the mass
• $g$ is the acceleration due to gravity
• $y_f$ is the final height of the system
• $y_i$ is the initial height of the system

Re-write the above equation to determine an expression for ${\omega }_f$,

${\omega }_f=\sqrt{\frac{2mg\left(y_i-y_f\right)}{I_e}}$ (1)

For a rod the moment of inertia about one end is,

$I_e=\frac{m{L}^2}{3}$ (2)

• $L$ is the length of the rod

Combining (1) and (2),

${\omega }_f=\sqrt{\frac{6g\left(y_i-y_f\right)}{L^2}}$

The relation between the angular speed and linear speed is given by,

$v=r\omega$

• $v$ is the linear speed
• $\omega$ is the angular speed
• $r$ is the radius

In the given case $r=L$ and $\omega ={\omega }_f$.

Thus,

$\begin{array}{c}v=L\sqrt{\frac{6g\left(y_i-y_f\right)}{L^2}}\\ =\sqrt{6g\left(y_i-y_f\right)}\end{array}$

The initial height of the system is,

$y_i=10.0\text{m}-\left(5.00\text{m}\right)\cos 45.0°$

The final height of the system is,

$y_f=10.0\text{m}-\left(5.00\text{m}\right)\cos 0°$

Substitute $\text{9}\text{.8}\text{m}\cdot {\text{s}}^{\text{-2}}$ for $g$ to determine the speed,

$\begin{array}{c}v=\sqrt{6g\left(10.0\text{m}-\left(5.00\text{m}\right)\cos 45.0°-10.0\text{m}-\left(5.00\text{m}\right)\cos 0°\right)}\\ =9.28\text{m}\cdot {\text{s}}^{\text{-1}}\end{array}$

Conclusion:

The linear speed of the system at the bottom of the swing is $9.28\text{m}\cdot {\text{s}}^{\text{-1}}$