📂Classical Mechanics

Mass Dependent on Position: The Motion of Balls Connected by Chains

Overview¹

Consider the case where a ball with a chain moves in the vertical direction. If the chain is long enough that the ball moves up and down, causing the length of the chain that is floating in the air to continuously change, then the total mass of the object changes according to the position of the ball. In other words, the mass of the object becomes dependent on its position. This motion is analyzed.

Motion of a Ball Connected to a Chain

Suppose a ball of mass $m$ is connected to a chain as shown in the figure above. Let’s assume the length of one chain is $1$ and its mass is $\rho$. To simplify the problem, imagine the chains are volumeless and not overlapping, with ends connected. Then the mass of this object, “ball connected to a chain”, becomes a function dependent on position. Let $x$ be the distance from the ground to the bottom of the ball (the connecting point of the ball and the chain), then the mass of the ball connected to a chain is $M(x) = m + \rho x$. Then, the motion equation of this object is as follows.

$$ \dfrac{d p}{d t} = F \implies \dfrac{d}{dt}\left[ (m + \rho x)\dot{x} \right] = -mg $$

By integrating the above equation, we obtain the following.

$$ \begin{equation} m\dot{x} + \rho x \dot{x} = -mgt + C_{1} \end{equation} $$

Meanwhile, by changing the left side of the equation and integrating, we obtain the following.

$$ \dfrac{d}{dt} \left[ mx + \dfrac{\rho x^{2}}{2} \right] = -mgt + C_{1} $$

$$ \begin{equation} \implies mx + \dfrac{\rho x^{2}}{2} = -\dfrac{mg}{2}t^{2} + C_{1}t + C_{2} \end{equation} $$

If the initial conditions are $x(0) = x_{0}$ and $\dot{x}(0) = v_{0}$, then from $(1)$ and $(2)$, the two constants are as follows.

$$ m v_{0} + \rho x_{0} v_{0} = C_{1} \implies C_{1} = (m + \rho x_{0}) v_{0} $$

$$ m x_{0} + \dfrac{\rho x_{0}^{2}}{2} = C_{2} \implies C_{2} = \left( m + \dfrac{\rho x_{0}}{2} \right) x_{0} $$

Let’s say the time to reach the maximum height is $t_{\ast}$. Then, because of $\dot{x}(t_{\ast}) = 0$, according to $(2)$,

$$ t_{\ast} = \dfrac{C_{1}}{mg} = \dfrac{m + \rho x_{0} v_{0}}{mg} $$

Also, let the maximum height be $x(t_{\ast}) = x_{\ast}$. Inserting $t_{\ast}$ into $(3)$,

$$ m x_{\ast} + \dfrac{\rho x_{\ast}^{2}}{2} = -\dfrac{(m + \rho x_{0})^{2} v_{0}^{2}}{2mg} + \dfrac{(m + \rho x_{0})^{2}v_{0}^{2}}{mg} + \left( m + \dfrac{\rho x_{0}}{2} \right)x_{0} $$

Organizing this as a quadratic equation for $x_{\ast}$,

$$ x_{\ast}^{2} + \dfrac{2m}{\rho}x_{\ast} - \left[ \dfrac{(m + \rho x_{0})^{2} v_{0}^{2}}{mg\rho} + \left(\dfrac{2m}{\rho} + x_{0}\right)x_{0} \right] = 0 $$

If the initial position is considered as the ground $x_{0} = 0$,

$$ x_{\ast}^{2} + \dfrac{2m}{\rho}x_{\ast} - \dfrac{m v_{0}^{2}}{g \rho} = 0 $$

Using the quadratic formula to calculate the value of $x_{\ast}$, except for the case of $x_{\ast} \lt 0$, we get the following. $$ x_{\ast} = -\dfrac{m}{\rho} + \dfrac{1}{2} \sqrt{\dfrac{4m}{\rho}\left( \dfrac{m}{\rho} + \dfrac{v_{0}^{2}}{g}\right)} $$

Compared to the case of Vertical Motion of a Constant Mass Ball, where the maximum height is $x_{\ast} = \dfrac{v_{0}^{2}}{2g}$, it is represented quite complexly.

See Also

• Equations of Motion for Variable Mass Systems