Chapter 6.5, Problem 17E

Interpretation Introduction

Interpretation:

To show explicitly that bead’s kinetic and potential energy is not conserved when it is in constant motion. Find the physical interpretation of the conserved quantity.

Concept Introduction:

Express the total energy by considering rotational and translation motion of the hoop and bead.

Find the derivative of the energy to determine the energy is conserved or not.

Answer to Problem 17E

Solution:

It is shown that the bead’s kinetic and potential energy is not conserved when it is in motion. The physical interpretation of the conserved quantity is stated.

Explanation of Solution

Given information:

The system is with bead and hoop in motion.

The total energy is expressed as

$\text{E = }\frac{\text{1}}{\text{2}}{\text{m}\dot{\text{x}}}^{\text{2}}\text{+ V(x) }$

Here, $\text{V(x) }$ is the potential energy and $\frac{\text{1}}{\text{2}}{\text{m}\dot{\text{x}}}^{\text{2}}$ is the kinetic energy.

This energy $\text{E }$ is called conserved energy because it is constant with time.

For the bead’s motion, the energy equation is given as,

$\text{mr}\ddot{\varphi }\text{ = - b}\ddot{\varphi }\text{- mg sin(}\varphi \text{) + mr}{\omega }^2\text{sin(}\varphi \text{)cos(}\varphi \text{)}$

Here $\text{m}$ is the mass of the bead, $\text{r}$ is the radius of the hoop. Hoop is rotated at an angular velocity $\omega$ about its vertical axis, $\varphi$ is the angle between the bead and the downward vertical direction. The parameter $\text{b}$ is constant.

The conserved energy when $\text{b = 0}$,

$\text{mr}\ddot{\varphi }\text{ = - mg sin(}\varphi \text{) + mr}{\omega }^2\text{sin(}\varphi \text{)cos(}\varphi \text{)}$

By dividing both sides by $\text{mr}$ as shown below

$\ddot{\varphi }\text{ = - }\frac{\text{g}}{r}\text{ sin(}\varphi \text{) + }{\omega }^2\text{sin(}\varphi \text{)cos(}\varphi \text{)}$

$\ddot{\varphi }\text{ = }{\omega }^2\text{sin(}\varphi \text{) }\left(\text{cos(}\varphi \text{) - }\frac{\text{g}}{{\omega }^{2}r}\right)$

The energy equation for the bead system is as,

$E=\frac{1}{2}{\dot{\varphi }}^2\text{ - }{\displaystyle \int yd\varphi }$

By substituting $\text{y = }{\omega }^2\text{sin(}\varphi \text{) }\left(\text{cos(}\varphi \text{) - }\frac{\text{g}}{{\omega }^{2}r}\right)$

$E=\frac{1}{2}{\dot{\varphi }}^2\text{ - }{\displaystyle \int {\omega }^2\text{sin(}\varphi \text{) }\left(\text{cos(}\varphi \text{) - }\frac{\text{g}}{{\omega }^{2}r}\right)d\varphi }$

$E=\frac{1}{2}{\dot{\varphi }}^2\text{ + }\frac{{\omega }^2}{2}{\text{cos}}^2\text{(}\varphi )\text{ - }\frac{\text{g}}{{\omega }^{2}r}\text{cos(}\varphi \text{)}$

Hence the conserved energy when $\text{b = 0}$ is $\frac{1}{2}{\dot{\varphi }}^2\text{ + }\frac{{\omega }^2}{2}{\text{cos}}^2\text{(}\varphi )\text{ - }\frac{\text{g}}{{\omega }^{2}r}\text{cos(}\varphi \text{)}$

We have to show that the bead’s kinetic energy and potential energy is not conserved. Consider the system with hoop and bead.

$E_{Total}=PE+K{E}_{translational}+K{E}_{rotational}$

When the hoop has zero potential energy, if the bead is at the vertical height of the hoop then e $PE$ is the potential energy of bead is

$PE=mgr\left(1-\text{cos(}\varphi \text{)}\right)$

Here are the two kinetic energies, translational and rotational energy.

$KE=K{E}_{translational}+K{E}_{rotational}$

$K{E}_{translational}=\frac{1}{2}m{v}^2$, Here $\text{ν}=r\dot{\varphi }$, By plugging the value

$K{E}_{translational}=\frac{1}{2}m{\left(r\dot{\varphi }\right)}^2$

Considering the kinetic energy due to the rotation of bead around the vertical rotational axis of the hoop. This rotational kinetic energy depends on the horizontal distance from the rotational axis and the rotation rate of the hoop $\dot{\varphi }$ .

It is given by,

$K{E}_{rotational}=\frac{1}{2}m{\left(r\sin (\varphi )\omega \right)}^2$

Total kinetic energy is given by,

$KE=K{E}_{translational}+K{E}_{rotational}$

$KE=\frac{1}{2}m{\left(r\dot{\varphi }\right)}^2+\frac{1}{2}m{\left(r\sin (\varphi )\omega \right)}^2$

Hence total energy is given as,

$E_{Total}=PE+K{E}_{translational}+K{E}_{rotational}$

$E_{Total}=mgr\left(1-\text{cos(}\varphi \text{)}\right)+\frac{1}{2}m{\left(r\dot{\varphi }\right)}^2+\frac{1}{2}m{\left(r\sin (\varphi )\omega \right)}^2$

To determine the conserved energy we will take the derivative with respect to $\varphi$,

${\dot{E}}_{Total}=mgr\left(\text{sin(}\varphi \text{)}\right)\dot{\varphi }+m{r}^2\ddot{\varphi }\dot{\varphi }+m{r}^2{\omega }^2\sin (\varphi )\cos (\varphi )\dot{\varphi }$

${\dot{E}}_{Total}\text{= mr}\dot{\varphi }\left(\text{gsin(}\varphi \text{) + r}\ddot{\varphi }\text{ + r}{\omega }^2\sin (\varphi )\cos (\varphi )\right)$

By substituting $\ddot{\varphi }\text{ = }{\omega }^2\text{sin(}\varphi \text{) }\left(\text{cos(}\varphi \text{) - }\frac{\text{g}}{{\omega }^{2}r}\right)$

${\dot{E}}_{Total}\text{= mr}\dot{\varphi }\left(\text{gsin(}\varphi \text{) + r}{\omega }^2\text{sin(}\varphi \text{)}\left(\cos (\varphi )-\frac{g}{\text{r}{\omega }^2}\right)\text{ + r}{\omega }^2\sin (\varphi )\cos (\varphi )\right)$

${\dot{E}}_{Total}\text{= mr}\dot{\varphi }\text{gsin(}\varphi {\text{) + mr}}^2\dot{\varphi }{\omega }^2\text{sin(}\varphi \text{)}\left(\cos (\varphi )-\frac{g}{\text{r}{\omega }^2}\right){\text{ + mr}}^2\dot{\varphi }{\omega }^2\sin (\varphi )\cos (\varphi )$

${\dot{E}}_{Total}\text{= mr}\dot{\varphi }\text{gsin(}\varphi {\text{) + mr}}^2\dot{\varphi }{\omega }^2\text{sin(}\varphi \text{)}\cos (\varphi )-{\text{mr}}^2\dot{\varphi }{\omega }^2\text{sin(}\varphi \text{)}\frac{g}{\text{r}{\omega }^2}{\text{ + mr}}^2\dot{\varphi }{\omega }^2\sin (\varphi )\cos (\varphi )$

${\dot{E}}_{Total}\text{= mrg}\dot{\varphi }\text{sin(}\varphi {\text{) + mr}}^2{\omega }^2\dot{\varphi }\text{sin(}\varphi \text{)}\cos (\varphi )-\text{mr}g\dot{\varphi }\text{sin(}\varphi {\text{) + mr}}^2{\omega }^2\dot{\varphi }\sin (\varphi )\cos (\varphi )$

${\dot{E}}_{Total}{\text{= 2mr}}^2{\omega }^2\text{sin(}\varphi \text{)}\cos (\varphi )\dot{\varphi }$

${\dot{E}}_{Total}\ne \text{ 0}$

Since ${\dot{E}}_{Total}\ne \text{ 0}$, this implies that the total energy is not constant and hence it is not conserved.

As the hoop transfer the energy to the bead when it goes away from the bottom. This hoop has a constant rotation rate and it is denoted by $\omega$. Thus, the hoop needs to put in more energy to maintain the constant rotation rate when the bead goes farther away from the rotation axis. When the bead gets closer to the rotation axis it takes away energy.

For a system, the conserved quantity is the Hamiltonian denoted by $H$, $H$ is said to be the total energy, when the potential energy $V$ is a function only of the co-ordinates for a stationary coordinate system. But in this problem moving constraints are there, this $H$ is not the total energy. The work is done by the force moving the hoop. This work is that energy needs to be included in the accounting of energy.

Conclusion

It is shown that the bead’s kinetic and potential energy is not conserved when it is in constant motion. The physical interpretation of the conserved quantity is stated.