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Volume
Equation
The volume ($V$) out of ($$) that does not vary along the height ($h$) is equal to
 $ V = S h $ |
The expression holds even if the shape, but not the value, of section the section ($S$) varies along the height, as long as its total area remains constant.
ID:(3792, 0)
Translational Kinetic Energy
Equation
In the case of studying translational motion, the definition of energy
| $ \Delta W = \vec{F} \cdot \Delta\vec{s} $ |
is applied to Newton's second law
| $ F = m_i a $ |
resulting in the expression
| $ K_t =\displaystyle\frac{1}{2} m_i v ^2$ |
The energy required for an object to transition from velocity $v_1$ to velocity $v_2$ can be calculated using the definition with
| $ \Delta W = \vec{F} \cdot \Delta\vec{s} $ |
Using the second law of Newton, this expression can be rewritten as
$\Delta W = m a \Delta s = m\displaystyle\frac{\Delta v}{\Delta t}\Delta s$
Employing the definition of velocity with
| $ \bar{v} \equiv\displaystyle\frac{ \Delta s }{ \Delta t }$ |
we obtain
$\Delta W = m\displaystyle\frac{\Delta v}{\Delta t}\Delta s = m v \Delta v$
where the difference in velocities is
$\Delta v = v_2 - v_1$
Furthermore, the velocity itself can be approximated by the average velocity
$v = \displaystyle\frac{v_1 + v_2}{2}$
Using both expressions, we arrive at
$\Delta W = m v \Delta v = m(v_2 - v_1)\displaystyle\frac{(v_1 + v_2)}{2} = \displaystyle\frac{m}{2}(v_2^2 - v_1^2)$
Thus, the change in energy is given by
$\Delta W = \displaystyle\frac{m}{2}v_2^2 - \displaystyle\frac{m}{2}v_1^2$
In this way, we can define kinetic energy
| $ K_t =\displaystyle\frac{1}{2} m_i v ^2$ |
ID:(3244, 0)
Kinetic Energy of Rotation
Equation
In the case being studied of translational motion, the definition of energy
| $ \Delta W = T \Delta\theta $ |
is applied to Newton's second law
| $ T = I \alpha $ |
resulting in the expression
| $ K_r =\displaystyle\frac{1}{2} I \omega ^2$ |
The energy required for an object to change its angular velocity from $\omega_1$ to $\omega_2$ can be calculated using the definition
| $ \Delta W = T \Delta\theta $ |
Applying Newton's second law, this expression can be rewritten as
$\Delta W=I \alpha \Delta\theta=I\displaystyle\frac{\Delta\omega}{\Delta t}\Delta\theta$
Using the definition of angular velocity
| $ \bar{\omega} \equiv\displaystyle\frac{ \Delta\theta }{ \Delta t }$ |
we get
$\Delta W=I\displaystyle\frac{\Delta\omega}{\Delta t}\Delta\theta=I \omega \Delta\omega$
The difference in angular velocities is
$\Delta\omega=\omega_2-\omega_1$
On the other hand, angular velocity itself can be approximated with the average angular velocity
$\omega=\displaystyle\frac{\omega_1+\omega_2}{2}$
Using both expressions, we obtain the equation
$\Delta W=I \omega \Delta \omega=I(\omega_2-\omega_1)\displaystyle\frac{(\omega_1+\omega_2)}{2}=\displaystyle\frac{I}{2}(\omega_2^2-\omega_1^2)$
Thus, the change in energy is given by
$\Delta W=\displaystyle\frac{I}{2}\omega_2^2-\displaystyle\frac{I}{2}\omega_1^2$
This allows us to define kinetic energy as
| $ K_r =\displaystyle\frac{1}{2} I \omega ^2$ |
ID:(3255, 0)
Total Kinetic Energy
Equation
Kinetic energy can be of translation and/or rotation. Therefore, the total kinetic energy is the sum of both:
| $ K = K_t + K_r $ |
ID:(3686, 0)
Gravitational potential energy at the planet's surface
Equation
At the surface of the planet, the gravitational force is
| $ F_g = m_g g $ |
and the energy
| $ \Delta W = \vec{F} \cdot \Delta\vec{s} $ |
can be shown to be
| $ V = m g z $ |
As the gravitational force is
| $ F_g = m_g g $ |
with $m$ representing the mass. To move this mass from a height $h_1$ to a height $h_2$, a distance of
| $ V = m g ( h_2 - h_1 )$ |
is covered. Therefore, the energy
| $ \Delta W = \vec{F} \cdot \Delta\vec{s} $ |
with $\Delta s=\Delta h$ gives us the variation in potential energy:
$\Delta W = F\Delta s=mg\Delta h=mg(h_2-h_1)=U_2-U_1=\Delta V$
thus, the gravitational potential energy is
| $ V = m g z $ |
ID:(3245, 0)