Conservation of Energy

Conservation of Energy

Often, the most efficient way to solve a problem in physics or chemistry is to consider the system in terms of starting energy and final energy. We then examine the dissipation of energy from the system. The law of conservation of energy states that energy is neither created nor destroyed. Practically, we usually write this as $\Sigma E_{initial} = \Sigma E_{final}$.

Practice Problems

1. A yo-yo is made of two solid cylindrical disks, each of mass 0.050 kg and diameter 0.075 m joined by a concentric thin solid cylindrical hub of mass 0.0050 kg and diameter 0.010 m. Use the law of conservation of energy to calculate the linear speed of the yo-yo just before it reaches the end of its 1.0 m length of string if it is released from rest. What fraction of its kinetic energy is rotational?

The intention of this problem is that, at rest, we have only potential energy and, in this case, gravitational potential energy.

$\Sigma E_{initial} = \Sigma E_{final}$

$E_{gravitational} = \Sigma E_{kinetic}$

$E_{gravitational} = E_{kinetic-linear} + E_{kinetic-rotational}$

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

For a cylinder, we could derive or look up $I = \frac{1}{2}mr^2$ and recall $\omega = \frac{v}{r}$

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}\frac{1}{2}mr^2\frac{v^2}{r^2}$

$gh = \frac{1}{2}v^2 + \frac{1}{4}v^2$

$gh = \frac{3}{4}v^2$

$v = \sqrt{\frac{4}{3}gh} =\sqrt{\frac{4}{3}\cdot9.8\cdot1} = 3.6 m/s$

We compare the two terms above and see that $\frac{\frac{1}{4}v^2}{\frac{3}{4}v^2} = \frac{1}{3}$ of its kinetic energy is rotational.


2. An Atwood’s machine consists of two masses $m_a$ and $m_b$, which are connected by a massless inelastic cord that passes over a pulley. If the pulley has radius R and moment of inertia I about its axle then determine the acceleration of the masses $m_a$ and $m_b$ and compare this to the situation in which the moment of inertia of the pull is ignored.

We may make one nice, simplifying assumption. It is certainly reasonable to assume that the acceleration of $m_A$ upwards is equal to the acceleration of $m_B$ downwards.

We need to consider the net forces of $m_B$ and $m_A$. Here, we have:

$F_{\text{net on }m_B} = T_B – m_B \cdot g$

$m_B \cdot a = T_B – m_B \cdot g$

$T_B = m_B \cdot a + m_B \cdot g$


$F_{\text{net on }m_A} = T_A – m_A \cdot g$

$m_A \cdot a = T_A – m_A \cdot g$

$T_A = m_A \cdot a + m_A \cdot g$

We expect the pulley to rotate in the negative direction and here we will appeal to the equation for torque. We assume given the drawing that $\theta$ is $90^0$.

$\tau = F\cdot R\cdot sin(\theta)$

$\tau = F\cdot R\cdot sin(90)$

$\tau = F\cdot R$


$\tau= -T_B \cdot R + T_A \cdot R$

$I\alpha= -T_B \cdot R + T_A \cdot R$

$\frac{1}{2}mR^2\frac{a}{R}= -T_B \cdot R + T_A \cdot R$

$\frac{1}{2}m\cancel{R}a = -T_B \cdot \cancel{R} + T_A \cdot \cancel{R}$

$ma = 2T_A – 2T_B $

$ma = 2(m_A \cdot a + m_A \cdot g) – 2(m_B \cdot a + m_B \cdot g)$

$ma = 2m_A \cdot a + 2m_A \cdot g – 2m_B \cdot a + 2m_B \cdot g$

$ma – 2m_A \cdot a + 2m_B \cdot a = 2m_A \cdot g + 2m_B \cdot g$

$a(m – 2m_A + 2m_B) = 2m_A \cdot g + 2m_B \cdot g$

$a = \frac{2g(m_A + m_B)}{m – 2m_A + 2m_B}$

The major difference between consideration of a mass pulley and a pulley with nonneglible mass is that it requires force to turn the pulley. That force is exactly equal to the DIFFERENCE in tension between $m_A$ and $m_B$. In the case of a massless pulley, we could have assumed the tensions were equal.

This shows up in the derivation in the following way:
$a_{massless-pulley} = \frac{2g(m_A + m_B)}{\cancel{m} – 2m_A + 2m_B} =\frac{g(m_A + m_B)}{-m_A + m_B}$

3. A merry-go-round with a moment of inertia equal to 1260 $\frac{kg}{m^2}$ and a radius of 2.5 m rotates with negligible friction at 1.70 $\frac{rad}{s}$. A child initially standing still next to the merry-go-round jumps onto the edge of the platform straight toward the axis of rotation causing the platform to slow to 1.25 $\frac{rad}{s}$. What is her mass?

We have several choices on how to do this problem. The simplest idea would do to pursue a solution through the law of conservation of momentum.

$\Sigma L_{initial} = \Sigma L_{final}$

$L_{child initial} + L_{merry-go-round-initial} = L_{child + merry-go-round-final}$

$I_{child-initial}\omega_{child-initial} + I_{merry-go-round-initial}\omega_{merry-go-round-initial}= I_{child + merry-go-round-final}\omega_{child + merry-go-round-final}$

$\cancel{I_{child-initial}\omega_{child-initial}} + I_{merry-go-round-initial}\omega_{merry-go-round-initial}= (I_{child} + I_{merry-go-round-final})\omega_{child + merry-go-round-final}$

$1260(1.7)= (m_{child}r_{disk}^2 +1260)1.25$

$\frac{1260\cdot1.7}{1.25} -1260 = m_{child}2.5^2$

$m_{child} = 72.5 \text{ kg}$

4. Let $m_1$ be the mass of the bullet and $m_2$ be the mass of the block. Suppose the block starts from rest and the bullet has horizontal velocity $v_1$. Suppose we have two possible experiments. In the first experiment the bullet lodges in the block and they fly foward together with velocity $v_3$. In the other experiment, the bullet bounces off the block elastically leaving the block with new velocity $v_4$ while the block flies forward with velocity $v_5$. Determine the ratio $\frac{v_3}{v_5}$.

Here, we start with the conservation of momentum and write:

$P_i = P_f$

$m_1v_1 + m_2v_2 = (m_1 + m_2)v_3$

Since the block is at rest $v_2 = 0$ and the second term will be 0.

$m_1v_1 + \cancel{m_2v_2} = (m_1 + m_2)v_3$

$m_1v_1 = (m_1 + m_2)v_3$

$v_3 = \frac{m_1v_1}{m_1 + m_2}$

Now, we consider combining the result from the right column to obtain

$\frac{v_3}{v_5} = \frac{ \frac{m_1v_1}{m_1 + m_2}}{\frac{2m_1v_1}{m_2 + m_1}}$

$\frac{v_3}{v_5} = \frac{m_1v_1}{m_1 + m_2}\cdot\frac{m_2 + m_1}{2m_1v_1}$

$\frac{v_3}{v_5} = \frac{m_1v_1}{\cancel{m_1 + m_2}}\cdot\frac{\cancel{m_2 + m_1}}{2m_1v_1}$

$\frac{v_3}{v_5} = \cancel{m_1v_1}\cdot\frac{1}{2\cancel{m_1v_1}} = \frac{1}{2}$

Again, we start with the conservation of momentum and write:

$P_i = P_f$

$m_1v_1 + m_2v_2 = m_1v_4 + m_2v_5$

Since the block is at rest $v_2 = 0$ and the second term will be 0.

$m_1v_1 + \cancel{m_2v_2} = m_1v_4 + m_2v_5$

$m_1v_1 = m_1v_4 + m_2v_5$

$m_1v_1 – m_2v_5 = m_1v_4 $

$\frac{m_1v_1 – m_2v_5}{m_1} = v_4$

Now, we notice that i’ve introduced a $v_4$ that I would like to substitute for.

$E_i = E_f$

$\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1v_4^2 + \frac{1}{2}m_2v_5^2$

Since the block is at rest $v_2 = 0$ and the second term will be 0.

$\frac{1}{2}m_1v_1^2 + \cancel{\frac{1}{2}m_2v_2^2} = \frac{1}{2}m_1v_4^2 + \frac{1}{2}m_2v_5^2$

$m_1v_1^2 = m_1v_4^2 + m_2v_5^2$

$m_1v_1^2 – m_2v_5^2 = m_1v_4^2$

$m_1v_1^2 – m_2v_5^2 = m_1\left(\frac{m_1v_1 – m_2v_5}{m_1}\right)^2$

$m_1v_1^2 – m_2v_5^2 = \frac{(m_1v_1 – m_2v_5)^2}{m_1}$

$m_1^2v_1^2 – m_1m_2v_5^2 = (m_1v_1 – m_2v_5)^2$

$m_1^2v_1^2 – m_1m_2v_5^2 = m_1^2v_1^2 – 2m_1m_2v_1v_5 + m_2^2v_5^2$

$- m_1m_2v_5^2 = – 2m_1m_2v_1v_5 + m_2^2v_5^2$

$- m_1v_5^2 = – 2m_1v_1v_5 + m_2v_5^2$

$0 = – 2m_1v_1v_5 + m_2v_5^2 + m_1v_5^2$

$0 = (- 2m_1v_1 + m_2v_5 + m_1v_5)v_5 \rightarrow$

$0 = – 2m_1v_1 + m_2v_5 + m_1v_5$

$2m_1v_1 = (m_2 + m_1)v_5$

$\frac{2m_1v_1}{m_2 + m_1} = v_5$