Chapter 7, Problem 33P

(a)

To determine

The gravitational potential energy of block-Earth system.

Explanation of Solution

Given:

The mass of two blocks are $m_1$ and $m_2$ .

The length of mass $m_1$ from pivoted end is $l_1$ and length of mass $m_2$ from pivoted end is $l_2$ .

Formula used:

The energy contained in an object due to its position when compared to other objects is called potential energy. The two object system’s potential energy is the sum of energies of individual object.

Consider the horizontal line as datum line; it means $U=0$ at $\theta =0$ . So, mass $m_1$ is below the datum line and mass $m_2$ is above the datum line.

Write the expression for potential energy of block1.

  $U_1=m_{1}g\Delta h_1$ ........(1)

Here, $m_1$ is mass of block1, $g$ is gravitational acceleration, $\Delta h_1$ is height and $U_1$ is potential energy of block1.

Write the expression for potential energy of block2.

  $U_2=m_{2}g\Delta h_2$ ........(2)

Here, $m_2$ is mass of block2, $g$ is gravitational acceleration, $\Delta h_2$ is height and $U_2$ is potential energy of block2.

Write the expression for potential energy of two blocks system.

  $U\left(\theta \right)=U_1+U_2$ ........(3)

Here, $U\left(\theta \right)$ is the potential energy of the system.

Calculation:

Calculate the height for block1.

  $\begin{array}{c}\Delta h_1=0-l_1\sin \theta \\ =-l_1\sin \theta \end{array}$

Substitute $-l_1\sin \theta$ for $\Delta h_1$ in equation (1).

  $U_1=-m_{1}g{l}_1\sin \theta$

Calculate the height for block2.

  $\begin{array}{c}\Delta h_2=l_2\sin \theta -0\\ =l_2\sin \theta \end{array}$

Substitute $l_2\sin \theta$ for $\Delta h_2$ in equation (2).

  $U_2=m_{2}g{l}_2\sin \theta$

Substitute $-m_{1}g{l}_1\sin \theta$ for $U_1$ and $m_{2}g{l}_2\sin \theta$ for $U_2$ in equation (3).

  $\begin{array}{c}U\left(\theta \right)=-m_{1}g{l}_1\sin \theta +m_{2}g{l}_2\sin \theta \\ =\left(m_2{l}_2-m_1{l}_1\right)g\sin \theta \end{array}$

Conclusion:

Thus, the gravitational potential energy of block-Earth system is $\left(m_2{l}_2-m_1{l}_1\right)g\sin \theta$ .

(b)

To determine

The angle where potential energy is minimum.

Explanation of Solution

Given:

The mass of two blocks are $m_1$ and $m_2$ .

The length of mass $m_1$ from pivoted end is $l_1$ and length of mass $m_2$ from pivoted end is $l_2$ .

Formula used:

The energy contained in an object due to its position when compared to other objects is called potential energy. The two object system’s potential energy is the sum of energies of individual object.

Consider the horizontal line as datum line; it means $U=0$ at $\theta =0$ . So, mass $m_1$ is below the datum line and mass $m_2$ is above the datum line.

Write the expression for potential energy of block1.

  $U_1=m_{1}g\Delta h_1$ ........(1)

Here, $m_1$ is mass of block1, $g$ is gravitational acceleration, $\Delta h_1$ is height and $U_1$ is potential energy of block1.

Write the expression for potential energy of block2.

  $U_2=m_{2}g\Delta h_2$ ........(2)

Here, $m_2$ is mass of block2, $g$ is gravitational acceleration, $\Delta h_2$ is height and $U_2$ is potential energy of block2.

Write the expression for potential energy of two blocks system.

  $U\left(\theta \right)=U_1+U_2$ ........(3)

Here, $U\left(\theta \right)$ is the potential energy of the system.

The equilibrium points lies where first derivative of energy is zero. The potential energy would minimum on those points where the second derivative of energy is greater than zero.

Calculation:

Calculate the height for block1.

  $\begin{array}{c}\Delta h_1=0-l_1\sin \theta \\ =-l_1\sin \theta \end{array}$

Substitute $-l_1\sin \theta$ for $\Delta h_1$ in equation (1).

  $U_1=-m_{1}g{l}_1\sin \theta$

Calculate the height for block2.

  $\begin{array}{c}\Delta h_2=l_2\sin \theta -0\\ =l_2\sin \theta \end{array}$

Substitute $l_2\sin \theta$ for $\Delta h_2$ in equation (2).

  $U_2=m_{2}g{l}_2\sin \theta$

Substitute $-m_{1}g{l}_1\sin \theta$ for $U_1$ and $m_{2}g{l}_2\sin \theta$ for $U_2$ in equation (3).

  $\begin{array}{c}U\left(\theta \right)=-m_{1}g{l}_1\sin \theta +m_{2}g{l}_2\sin \theta \\ =\left(m_2{l}_2-m_1{l}_1\right)g\sin \theta \end{array}$

Calculate the equilibrium points.

  $\begin{array}{c}\frac{dU}{d\theta }=0\\ \left(m_2{l}_2-m_1{l}_1\right)g\cos \theta =0\\ \cos \theta =0\end{array}$

The angle for equilibrium position is $\theta =\pm \pi /2$ .

Calculate the second derivative of potential energy.

  $\frac{d^{2}U}{d{\theta }^2}=-\left(m_2{l}_2-m_1{l}_1\right)g\sin \theta$

Consider the value $m_2{l}_2-m_1{l}_1>0$ , the potential-energy function is a sine function.

For $\theta =-\pi /2$ , $\frac{d^{2}U}{d{\theta }^2}>0$ ; so, potential energy is minimum at $\theta =-\pi /2$ .

For $\theta =\pi /2$ , $\frac{d^{2}U}{d{\theta }^2}<0$ ; so, potential energy is maximum at $\theta =\pi /2$ .

Conclusion:

Thus, the angle where potential energy is minimum is $\theta =-\pi /2$ .

(c)

To determine

The gravitational potential energy of block-Earth system when $m_2{l}_2=m_1{l}_1$ .

Explanation of Solution

Given:

The mass of two blocks are $m_1$ and $m_2$ .

The length of mass $m_1$ from pivoted end is $l_1$ and length of mass $m_2$ from pivoted end is $l_2$ .

Formula used:

The energy contained in an object due to its position when compared to other objects is called potential energy. The two object system’s potential energy is the sum of energies of individual object.

Consider the horizontal line as datum line; it means $U=0$ at $\theta =0$ . So, mass $m_1$ is below the datum line and mass $m_2$ is above the datum line.

Write the expression for potential energy of block1.

  $U_1=m_{1}g\Delta h_1$ ........(1)

Here, $m_1$ is mass of block1, $g$ is gravitational acceleration, $\Delta h_1$ is height and $U_1$ is potential energy of block1.

Write the expression for potential energy of block2.

  $U_2=m_{2}g\Delta h_2$ ........(2)

Here, $m_2$ is mass of block2, $g$ is gravitational acceleration, $\Delta h_2$ is height and $U_2$ is potential energy of block2.

Write the expression for potential energy of two blocks system.

  $U\left(\theta \right)=U_1+U_2$ ........(3)

Here, $U\left(\theta \right)$ is the potential energy of the system.

Calculation:

Calculate the height for block1.

  $\begin{array}{c}\Delta h_1=0-l_1\sin \theta \\ =-l_1\sin \theta \end{array}$

Substitute $-l_1\sin \theta$ for $\Delta h_1$ in equation (1).

  $U_1=-m_{1}g{l}_1\sin \theta$

Calculate the height for block2.

  $\begin{array}{c}\Delta h_2=l_2\sin \theta -0\\ =l_2\sin \theta \end{array}$

Substitute $l_2\sin \theta$ for $\Delta h_2$ in equation (2).

  $U_2=m_{2}g{l}_2\sin \theta$

Substitute $-m_{1}g{l}_1\sin \theta$ for $U_1$ and $m_{2}g{l}_2\sin \theta$ for $U_2$ in equation (3).

  $\begin{array}{c}U\left(\theta \right)=-m_{1}g{l}_1\sin \theta +m_{2}g{l}_2\sin \theta \\ =\left(m_2{l}_2-m_1{l}_1\right)g\sin \theta \end{array}$

For $m_2{l}_2=m_1{l}_1$ , $m_2{l}_2-m_1{l}_1=0$ which gives $U=0$

Conclusion:

Thus, the gravitational potential energy of block-Earth system is zero always for $m_2{l}_2=m_1{l}_1$ ; it means it is same for all values of angle.