❓What is the kinetic energy of mass moving vertically?

The formula $T_2 = \frac{1}{2}m_2 \dot{r}^2$ represents the kinetic energy ($T_2$) of mass $m_2$ which is moving vertically beneath the hole.

Here is the explanation of the expression:

1. General Kinetic Energy Formula: The basic formula for kinetic energy is $T = \frac{1}{2}m v^2$ Here, $m = m_2$.

2. Velocity of $m_2$ ($v_2$):

   • Mass $m_2$ is connected to $m_1$ by a thread of fixed total length, $L$.

   • If the length of the string on the horizontal plane is $r$, the vertical position of $m_2$ (relative to the plane) is $z_2 = L - r$ (or $r - L$ depending on the setup, but the magnitude of displacement is $r$).

   • The velocity of $m_2$ is purely vertical, and since the total string length $L$ is constant, the speed of $m_2$ is entirely dependent on the rate of change of $r$:

     $v_2 = \dot{z}_2 = \frac{d}{dt}(r - L) = \dot{r}$

   • This means the velocity of $m_2$ is simply the radial velocity ($\dot{r}$) of $m_1$.

3. Final Expression: Substituting the velocity into the kinetic energy formula:

   $T_2 = \frac{1}{2}m_2 v_2^2 = \frac{1}{2}m_2 \dot{r}^2$

In summary, $m_2$ can only move up or down, and its speed is locked to the rate at which the string length $r$ changes on the horizontal plane. It has no angular motion ($\dot{\varphi}$) because it does not rotate about the $z$-axis.

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What is the kinetic energy of mass moving vertically? | FAQsviadean on Notion