Chapter 8, Problem 84AP

A light rod of length 2L is free to rotate in a vertical plane about a frictionless pivot through its center. A particle of mass m₁ is attached at one end of the rod, and a mass m₂ is at the opposite end, where m₁ > m₂. The system is released from rest in the vertical position shown in Figure P8.84a, and at some later time, the system is rotating in the Position shown in Figure P8.84b. Take the reference point of the gravitational potential energy to be at the pivot, (a) Find an expression for the system's total mechanical energy in the vertical position. (b) Find an expression for the total mechanical energy in the rotated position shown in Figure P8.84b. (c) Using the fact that the mechanical energy of the system is conserved, how would you determine the angular speed co of the system in the rotated position? (d) Find the magnitude of the torque on the system in the vertical position and in the routed position. Is the torque constant? Explain what these results imply regarding the angular momentum of the system, (c) Find an expression for the magnitude of the angular acceleration of the system in the rotated position. Does your result make sense when the rod is horizontal? When it is vertical? Explain.

Figure P8.84

To determine

The total mechanical energy of the system about vertical position.

Answer to Problem 84AP

The total mechanical energy of the system about vertical position is $m_{1}gL-m_{2}gL$

Explanation of Solution

Given Info:

The mass of the particles are $m_1$ and $m_2$ and light rod of length 2L . Mass $m_1$ is attached at one end and $m_2$ attached at other end.

In Figure P8.84 (a) the system has only potential energy and at centre the potential energy is zero since the reference point of the gravitational potential energy is at the pivot that is at the midpoint.

The formula to calculate total mechanical energy of the system about vertical position is

$E=m_{1}gL-m_{2}gL$

• g is the acceleration due to gravity
• L is the distance of the masses from the pivot

Thus, the total mechanical energy of the system about vertical position is $m_{1}gL-m_{2}gL$

Conclusion:

Therefore the total mechanical energy of the system about vertical position is $m_{1}gL-m_{2}gL$

To determine

Total mechanical energy in the rotated position.

Answer to Problem 84AP

The total mechanical energy in the rotated position is $\frac{1}{2}\left(m_1+m_2\right)L^2{\omega }^2+\left(m_1-m_2\right)gL\sin \theta$

Explanation of Solution

Given Info: The mass of the particles are $m_1$ and $m_2$ and light rod of length 2L . Mass $m_1$ is attached at one end and $m_2$ attached at other end.

In Figure P8.84 (b) the system has potential energy and rotational kinetic energy.

The expression to calculate total mechanical energy of the rotating system is,

$E^{\prime }=K{E}^{\prime }+P{E}^{\prime }$

• $K{E}^{\prime }$ is the rotational kinetic energy
• $P{E}^{\prime }$ is the new potential energy

The potential energy of the system is given by

$P{E}^{\prime }=m_{1}gL\sin \theta -m_{2}gL\sin \theta$ (I)

The formula to calculate the rotational kinetic energy is given by

$K{E}^{\prime }=\frac{1}{2}\left(I_1+I_2\right){\omega }^2$

• $I_1$ is moment of inertia of mass $m_1$
• $I_2$ is the moment of inertia of mass $m_2$
• $\omega$ is the angular speed of the rotated system

The formula to calculate moment of inertia of mass $m_1$ is

$I_1=m_1{L}^2$

The formula to calculate moment of inertia of mass $m_2$ is

$I_2=m_2{L}^2$

Rewrite the above equation for $K{E}^{\prime }$ using the expression for $I_1$ and $I_2$

$\begin{array}{c}K{E}^{\prime }=\frac{1}{2}\left(m_1{L}^2+m_2{L}^2\right){\omega }^2\\ =\frac{1}{2}\left(m_1+m_2\right)L^2{\omega }^2\end{array}$ (II)

Formula to calculate the total mechanical energy is,

$E^{\prime }=P{E}^{\prime }+K{E}^{\prime }$

• $E^{\prime }$ is the total mechanical energy at the rotated position

Substitute equation (I) and (II) in above equation to calculate $E^{\prime }$.

$\begin{array}{c}{E}^{\prime }=\frac{1}{2}\left(m_1+m_2\right)L^2{\omega }^2+m_{1}gL\sin \theta -m_{2}gL\sin \theta \\ =\frac{1}{2}\left(m_1+m_2\right)L^2{\omega }^2+\left(m_1-m_2\right)gL\sin \theta \end{array}$

Conclusion: The total mechanical energy in the rotated position is $\frac{1}{2}\left(m_1+m_2\right)L^2{\omega }^2+\left(m_1-m_2\right)gL\sin \theta$

To determine

The angular speed of the rotated system

Answer to Problem 84AP

The angular speed of the rotated system is  ${\left(\frac{2g\left(m_1-m_2\right)\left(1-\sin \theta \right)}{\left(m_1+m_2\right)L}\right)}^{\frac{1}{2}}$

Explanation of Solution

Given Info:

According to conservation of total mechanical energy of the system, the total mechanical energy of the vertical system is same as total mechanical energy of the rotated system.

The formula to calculate total mechanical energy of the system about vertical position is

$E=m_{1}gL-m_{2}gL$

Formula to calculate the mechanical energy of the rotated system is,

$E^{\prime }=\frac{1}{2}\left(m_1+m_2\right)L^2{\omega }^2+\left(m_1-m_2\right)gL\sin \theta$

Equate the above expressions for $E$ and $E^{\prime }$ to calculate $\omega$

$\begin{array}{c}\frac{1}{2}\left(m_1+m_2\right)L^2{\omega }^2+\left(m_1-m_2\right)gL\sin \theta =m_{1}gL-m_{2}gL\\ \frac{1}{2}\left(m_1+m_2\right)L^2{\omega }^2=\left(m_1-m_2\right)gL\left(1-\sin \theta \right)\\ \omega ={\left(\frac{2g\left(m_1-m_2\right)\left(1-\sin \theta \right)}{\left(m_1+m_2\right)L}\right)}^{\frac{1}{2}}\end{array}$

Conclusion:

Therefore the angular speed of the rotated system is  ${\left(\frac{2g\left(m_1-m_2\right)\left(1-\sin \theta \right)}{\left(m_1+m_2\right)L}\right)}^{\frac{1}{2}}$

To determine

The magnitude of torque in the vertical and rotated position

Answer to Problem 84AP

Therefore the net torque on the vertical system is zero and rotated system is $\left(m_1-m_2\right)gL\cos \theta$ and torque is not a constant therefore angular momentum will change with a non-uniform rate

Explanation of Solution

Given Info:

In the case of vertical position the gravitational force and the position of the masses from the axis of rotation lies in the same line so the net torque on the system is zero.

Formula to calculate the torque on the mass $m_1$ is,

$\begin{array}{c}{\tau }_1=m_{1}gL\sin \left(\frac{\pi }{2}-\theta \right)\\ =m_{1}gL\cos \theta \end{array}$

Formula to calculate the torque on the mass $m_2$ is,

$\begin{array}{c}{\tau }_2=-m_{2}gL\sin \left(\frac{\pi }{2}-\theta \right)\\ =-m_{2}gL\cos \theta \end{array}$

Formula to calculate net torque on the system is,

$\tau ={\tau }_1+{\tau }_2$

Substitute $m_{1}gL\cos \theta$ for ${\tau }_1$ and $-m_{2}gL\cos \theta$ for ${\tau }_2$ to calculate $\tau$

$\begin{array}{c}\tau =m_{1}gL\cos \theta -m_{2}gL\cos \theta \\ =\left(m_1-m_2\right)gL\cos \theta \end{array}$

Conclusion:

Therefore the net torque on the vertical system is zero and rotated system is $\left(m_1-m_2\right)gL\cos \theta$

Torque is not a constant therefore angular momentum will change with a non-uniform rate.

To determine

The magnitude of angular acceleration of the system in the rotated position.

Answer to Problem 84AP

The angular acceleration of the rotated system is $\left(\frac{\left(m_1-m_2\right)g\cos \theta }{\left(m_1+m_2\right)L}\right)$

Angular acceleration has maximum value when rod is horizontal and has minimum value when rod is vertical.

Explanation of Solution

Explanation:

Given Info:

Formula to calculate the angular acceleration of the rotated system is given by

$\alpha =\frac{\tau }{\left(I_1+I_2\right)}$

Substitute $\left(m_1-m_2\right)gL\cos \theta$ for $\tau$ and $\left(m_1{L}^2+m_2{L}^2\right)$ for $I_1+I_2$ in the above equation to calculate $\alpha$

$\begin{array}{c}\alpha =\frac{\left(m_1-m_2\right)gL\cos \theta }{\left(m_1{L}^2+m_2{L}^2\right)}\\ =\frac{\left(m_1-m_2\right)g\cos \theta }{\left(m_1+m_2\right)L}\end{array}$

Thus the angular acceleration of the rotated system is $\left(\frac{\left(m_1-m_2\right)g\cos \theta }{\left(m_1+m_2\right)L}\right)$

Conclusion:

When rod is horizontal that is $\theta$ is zero

Substitute 0 for $\theta$ in the above equation to calculate angular acceleration to calculate angular acceleration

$\begin{array}{c}\alpha =\left(\frac{\left(m_1-m_2\right)g\cos 0}{\left(m_1+m_2\right)L}\right)\\ =\left(\frac{\left(m_1-m_2\right)g}{\left(m_1+m_2\right)L}\right)\end{array}$

Angular acceleration of the rod when the rod is horizontal is $\left(\frac{\left(m_1-m_2\right)g}{\left(m_1+m_2\right)L}\right)$

When rod is vertical that is $\theta$ is 90.

Substitute 0 for $\theta$ in the above equation to calculate angular acceleration to calculate angular acceleration

$\begin{array}{c}\alpha =\left(\frac{\left(m_1-m_2\right)g\cos 90}{\left(m_1+m_2\right)L}\right)\\ =0\end{array}$

Angular acceleration of the rod when the rod is vertical is 0. Therefore the rod has maximum angular acceleration when rod is horizontal. This is position corresponds to unstable equilibrium position.