The Spinning Albert Project

A device is described to demonstrate conservation of angular momentum (without getting dizzy). 

The Concept of Momentum

Objects in motion continue at the same speed and direction unless they are acted upon by a force. The more massive the object or the faster it moves, the greater the force must be to alter its speed or trajectory. A quantity that captures both properties is known as momentum (p) and is defined as the product of an object’s mass m times its velocity v as p = mv. In an xyz coordinate system, velocity can be expressed as a vector, signified by $\vec v = (v_x, v_y, v_z)$. Momentum is therefore also a vector expressed as $\vec p = m \vec v$ or $\vec p = m(v_x, v_y, v_z)$. The momentum of an object of arbitrary shape can be expressed as the sum of the momentum of all its parts: $\vec p = \sum_i m_i \vec v_i$.

If a force $\vec F = (F_𝑥, F_𝑦, F_z)$ acts over a time ∆𝑡 on an object moving with a momentum $\vec p$, it will cause the momentum to change by an amount $\Delta \vec p = \vec F \Delta t$, or more precisely in derivative form: $\vec F = \frac {d \vec p} {dt}$. This is a more general form for Newton’s second law $F = 𝑚𝑎 = 𝑚 \frac {dv} {dt}$.

In the absence of an external force $\vec F = 0$ the momentum $\vec p$ is a constant, which is to say momentum is conserved. For example, if two masses $m_1$ and $m_2$ are heading toward one another at velocities $\vec v_1$ and $\vec v_2$, and recoil with velocities $\vec v'_1$ and $\vec v'_2$, conservation of momentum requires $m_1 \vec v_1 + m_2 \vec v_2 = m_1 \vec v'_1 + m_2 \vec v'_2$. This relationship along with conservation of energy can be used to calculate $\vec v'_1$ and $\vec v'_2$.

Physics of Rotation

An object may spin in addition to moving in a straight line. This can be examined by adding the rotational motion of each component mass of the object about the axis of rotation. Fortunately, that approach is simplified with the introduction of the concepts of Moment of Inertia, Angular Momentum and Torque.

The simplest case of rotation is a mass m attached to a (massless) rod of length r circling in a plane about an axis perpendicular to the plane. We can use the xy plane for the orbit and z as the axis of rotation (see the figure below). The position of the mass at any moment is the vector $\vec r = r(\cos \theta , \sin \theta, 0)$, where $\theta$ is the angle between the rod and the x axis in the xy plane (z points out of the page). The rod restricts the motion to a circle of radius r with a velocity $\vec v_{\theta}$ pointing along the orbit ( $\theta$ direction), perpendicular to the rod vector $\vec r$, given by $\vec v_{\theta} = r \frac {d \theta}{dt} (- \sin \theta , \cos \theta , 0)$. We can use $\omega = \frac {d \theta}{dt}$ as a symbol for the rotation speed, giving $\vec v_{\theta} = r \omega (- \sin \theta , \cos \theta , 0)$. 

The kinetic energy of the mass is $KE = \frac{1}{2} m{v_\theta}^2 = \frac{1}{2}m r^2 \omega^2$. Comparing with $KE = \frac{1}{2}mv^2$ for linear motion, we can treat $\omega$ as a rotational analog to $v$ and $mr^2$ as an analog to $m$. Defining $I=mr^2$ as the Moment of Inertia gives $KE=\frac{1}{2}I\omega^2$. 

The momentum of the mass is $p_\theta=mr\omega$. Keeping with the use of $I$ and $\omega$, both sides of this equation can be multiplied by r to give $rp_\theta=rmv_\theta=I\omega$. We can then define $L=rp_\theta=I\omega$ as the Angular Momentum. 

If a force vector $\vec F$ acts on the rotating mass at an angle $\gamma$ to the rod, only the vector component of the force in the theta direction $\vec F_\theta$ can change the velocity, since the rod prevents radial motion. Since r is a fixed value, we can refer to just the magnitudes $F_\theta = \frac{dp_\theta}{dt}$. Once again, we can multiply both sides of this equation by r to obtain $rF_\theta=r \frac{dp_\theta}{dt}$. The quantity $rF_\theta$ is referred to as the Torque $\tau$ acting on the rotating mass. Using the definition for L, this becomes $\tau= \frac{dL}{dt}$, which is the fundamental equation of motion for rotating objects. Since $L=I\frac{d\theta}{dt}$, the torque equation can also be written $\tau=\frac{d^2\theta}{dt^2}$. The angular acceleration $\frac{d^2\theta}{dt^2}$ is often symbolized by $\alpha$, giving $\tau=I\alpha$. The analogous linear equation of motion is $F=ma$.

A key consequence of $\tau=\frac{dL}{dt}$ is that L remains constant in the absence of any external torque ($\frac{dL}{dt}=0$). Since $L=I\omega$, this tells us that any increase in $I$ must be compensated by a proportionate decrease in $\omega$ and vise-versa. 

Any rotating object can be considered a collection of smaller masses $m_i$, each at a distance $r_i$ from a common axis. If all these masses are part of a rigid body, then $I_{total}$ can be computed as a sum over all the masses and $L_{total}$ and $\tau_{total}$ are just a product of $I_{total}$ and the angular velocity or acceleration:

$I_{total}=\sum_i m_i r^2_i$

$L_{total}=\sum_i r_i p_{\theta i}= \sum_i r_i(m_i \omega r_i) = I_{total}\omega$

$\tau_{total}=I_{total}\alpha$

For example, the Moment of Inertia for two masses $m_1$ and $m_2$ at a distance from the axis of $R_1$ and $R_2$ is $I_{total}=m_1 R^2_1 + m_2 R^2_2$.

Spinning Albert Demonstration

An apparatus (machined and built mostly by hobby machinist Keith) is used to demonstrate conservation of angular momentum when a famous physicist is spinning and draws his arms inward.

The Albert Apparatus

A challenge in creating this apparatus is to alter the arm positions with minimal change to the angular momentum. That requires minimal torque be applied, so any force needed to move the arms should be parallel to the rotation axis. The figures below show the components of the demo used to achieve this. The central character is an Albert Einstein Plush Doll from Little Thinkers by The Unemployed Philosophers Guild.

Understanding the Observation

With the arms outward, we begin with a moment of inertia $I_{up}$. Spinning it at an angular velocity $\omega_{up}$ imparts an angular momentum $L=I_{up}\omega_{up}$. Pushing down on the sliding control forces the arms inward, reducing the moment of inertia to $I_{down}$. If most of the angular momentum is conserved, the angular velocity must change so that $L=I_{up}\omega_{up} \approx I_{down}\omega_{down}$, resulting in $$\omega_{down} \approx \omega_{up}\frac{I_{up}}{I_{down}}$$  

The outstretched arms have a higher moment of inertia so $\omega_{down} > \omega_{up}$, as can be seen in the video.

To quantify the change in angular rotation speed, a slow motion video (240 fps) was taken and the time between each half-revolution was recorded. A plot of the angular velocity as a function of time is shown below. The initial spin speed decreases over the first 3 seconds due to friction. Pushing down on the top mechanism to bring the arms down causes an increase in speed that drops over time. The ratio of the angular speeds just before and after lowering the arms is $\frac {\omega_{down}}{\omega_{up}} = 1.39$.

If angular momentum were perfectly conserved (no torques acting on the spinning system), we would expect the ratio of the spins speeds $\frac{\omega_{down}}{\omega_{up}} = \frac{I_{up}}{I_{down}}$. To see if this is true, a method is needed to determine $\frac{I_{up}}{I_{down}}$.

Tortional Pendulum

The moment of inertia of the Spinning Albert demo is very complex to calculate. It would involve adding up the moments of each portion of the figure and apparatus, which is not practical. A simpler system was constructed to moment of inertia, known as a torsional pendulum, consisting of any object suspended by a thin wire. The equation of motion for this pendulum is derived using $\tau = \frac{dL}{dt} = I_{pend}\frac{d^2\phi}{dt^2}$, where $\tau$ is the torque from the suspension wire and $I_{pend}$ is the moment of inertia of the pendulum. The wire acts as a spring whose torque increases linearly with twist angle $\phi$, allowing us to write $\tau = -\kappa \phi$. This gives $-\kappa \phi = I_{pend}\frac{d^2\phi}{dt^2}$. The solution to this equation is $\phi = \phi_0 \sin(\Omega t)$, where the oscillation frequency is $\Omega = \sqrt{\frac{\kappa}{I_{pend}}}$. The period of oscillation $T = \frac{2\pi}{\Omega}$. Calibrating the apparatus requires measuring T for an object with a known value of I to give $\kappa$. 

To build the pendulum, a brass nut with a hook soldered to it was threaded onto the center of a 1/4-20 threaded rod, and brass disks were threaded onto the ends. The suspension wire was 30 gage 316L SS wire, 21cm in length. The rod length was 17.7cm. Each brass disk had a mass $m_d$ of 25.125g and a thickness $t_d$ of 0.644cm. The pendulum was carefully set in motion, filmed in time-lapse mode (2 fps) on a smart phone, and the oscillation period was measured frame by frame.

A video for three of the pendulum configurations:

In each case, N disks are placed symmetrically at the ends, and the distance d from the center of mass to the axis of rotation was measured. The moment of inertia of the pendulum is $I_{pend} = I_{rod} + 2Nm_d d^2$. Using the above expressions for the oscillation period gives $$T^2 = \frac{(2\pi)^2}{\kappa} (I_{rod} + 2Nm_d d^2)$$

In the absence of the brass disks $T^2_{rod} = \frac{(2\pi)^2}{\kappa} I_{rod}$. A plot of $T^2$ vs. $2Nm_d d^2$ is then expected to be linear, with a slope of $\frac{(2\pi)^2}{\kappa}$ and an intercept of $T^2_{rod}$. The data is plotted below, showing a linear relationship. The intercept was verified by a direct measurement of $T^2_{rod}$ by removing all the brass disks.

The slope of the plot provides the value of $\frac{\kappa}{(2\pi)^2} = 1/0.0255$ for the suspension wire. The moment of inertia can be determined for any object that can be suspended from the same wire by measuring T, including Albert. 

The base of the Albert apparatus was removed and the remainder was suspended by the same gage and length of wire ($\kappa$ scales linearly with the length of wire). A time-lapse video was made of the oscillation with the arms up and down. 

The video analysis shows that the period for arms up was 34.6s and for arms down 26.7s. The moment of inertia can be calculated from $I= \frac{\kappa}{(2\pi)^2} T^2$ to give 

$$I_{up} = 4.7 \times 10^4 g-cm^2$$   $$I_{down} = 2.8 \times 10^4 g-cm^2$$  

Their ratio is $\frac{I_{up}}{I_{down}} = 1.67$. This is significantly greater than what was determined from the spin measurement (1.39). This indicates that there is a loss of momentum in switching from the up to down position, most likely from rotational friction. Therefore conservation of angular momentum may be sufficient to explain the qualitative behavior increasing spin speed as the moment of inertia decreases, but a quantitative explanation needs to include all the sources of torque.

It should also be noted that if the goal was just to compare the ratio $\frac{I_{up}}{I_{down}}$ from the spin experiment to that from the torsional pendulum, the measurement of $\kappa$ is not necessary. Only the ratio of the square of the tortional periods needs to be compared to the ratio of the spin periods.