An Atwood machine consists of two hanging masses connected by a light string over a pulley.
The purpose of the model is to find the acceleration of the system and the tension in the string. In this calculator,
the positive direction is chosen so that \(m_2\) moves downward and \(m_1\) moves upward. Therefore, a positive value of
\(a\) means the right mass descends, while a negative value means the left mass descends.
1. Ideal pulley model
In the simplest model, the pulley is massless and frictionless. The string tension is the same on both sides of the pulley.
Newton’s second law is applied to each mass along the vertical direction.
Equations of motion for an ideal pulley.
\[
\begin{aligned}
m_2g - T &= m_2a \\
T - m_1g &= m_1a
\end{aligned}
\]
Adding the two equations eliminates the tension and gives the acceleration:
Ideal-pulley acceleration.
\[
\begin{aligned}
m_2g - m_1g &= (m_1+m_2)a \\
(m_2-m_1)g &= (m_1+m_2)a \\
a &= \frac{(m_2-m_1)g}{m_1+m_2}
\end{aligned}
\]
Once the acceleration is known, the common string tension can be found from either mass:
Ideal-pulley tension.
\[
\begin{aligned}
T &= m_1(g+a) \\
T &= m_2(g-a)
\end{aligned}
\]
2. Massive solid-disk pulley
If the pulley has mass, the two tensions are no longer equal. The difference between the tensions produces a net torque
that rotates the pulley. For a solid disk, the moment of inertia is \(I=\tfrac12 M_pR^2\).
Force equations for the two masses.
\[
\begin{aligned}
m_2g - T_2 &= m_2a \\
T_1 - m_1g &= m_1a
\end{aligned}
\]
The pulley rotation adds the torque equation:
Rotational equation for the pulley.
\[
\begin{aligned}
(T_2-T_1)R &= I\alpha \\
\alpha &= \frac{a}{R} \\
T_2-T_1 &= \frac{I}{R^2}a
\end{aligned}
\]
For a solid disk, the rotational inertia term becomes:
Solid-disk effective inertia term.
\[
\begin{aligned}
\frac{I}{R^2} &= \frac{\tfrac12 M_pR^2}{R^2} \\
&= \tfrac12 M_p
\end{aligned}
\]
This acts like an additional effective mass in the denominator of the acceleration formula:
Massive-pulley acceleration.
\[
\begin{aligned}
a &= \frac{(m_2-m_1)g}{m_1+m_2+\tfrac12 M_p}
\end{aligned}
\]
The side tensions are then:
Massive-pulley tensions.
\[
\begin{aligned}
T_1 &= m_1(g+a) \\
T_2 &= m_2(g-a)
\end{aligned}
\]
3. Direction and interpretation
The sign of \(a\) is just as important as its magnitude. If \(m_2>m_1\), the numerator \((m_2-m_1)g\) is positive,
so the system accelerates with \(m_2\) downward. If \(m_1>m_2\), the value is negative, meaning the actual acceleration
is opposite the chosen sign convention. If \(m_1=m_2\), the net driving force is zero and the acceleration is zero.
Balanced case.
\[
\begin{aligned}
m_1 &= m_2 \\
a &= 0
\end{aligned}
\]
A massive pulley does not change the direction of motion, but it reduces the magnitude of the acceleration because part
of the gravitational driving effect must rotate the pulley. That is why the denominator contains the extra term
\(\tfrac12M_p\). A heavier pulley makes the system accelerate more slowly.
| Model |
Acceleration |
Tension result |
Main assumption |
| Ideal pulley |
\(a=\dfrac{(m_2-m_1)g}{m_1+m_2}\) |
\(T=m_1(g+a)=m_2(g-a)\) |
Massless pulley, same tension on both sides |
| Solid-disk pulley |
\(a=\dfrac{(m_2-m_1)g}{m_1+m_2+\tfrac12M_p}\) |
\(T_1=m_1(g+a),\quad T_2=m_2(g-a)\) |
Pulley inertia included as \(I/R^2=\tfrac12M_p\) |
If the pulley is not a solid disk, replace \(\tfrac12M_p\) with the correct value of \(I/R^2\). If bearing friction is important,
an additional friction torque must be included in the rotational equation.
