Atwood machine

The free body diagrams of the two hanging masses of the Atwood machine. Our sign convention, depicted by the acceleration vectors is that m₁ accelerates downward and that m₂ accelerates upward, as would be the case if m₁ > m₂

An equation for the acceleration can be derived by analyzing forces. Assuming a massless, inextensible string and an ideal massless pulley, the only forces to consider are: tension force (T), and the weight of the two masses (W₁ and W₂). To find an acceleration, consider the forces affecting each individual mass. Using Newton's second law (with a sign convention of ${\displaystyle m_{1}>m_{2}}$) derive a system of equations for the acceleration (a).

As a sign convention, assume that a is positive when downward for ${\displaystyle m_{1}}$ and upward for ${\displaystyle m_{2}}$. Weight of ${\displaystyle m_{1}}$ and ${\displaystyle m_{2}}$ is simply ${\displaystyle W_{1}=m_{1}g}$ and ${\displaystyle W_{2}=m_{2}g}$ respectively.

Forces affecting m₁:

${\displaystyle \;m_{1}g-T=m_{1}a}$

Forces affecting m₂:

${\displaystyle \;T-m_{2}g=m_{2}a}$

and adding the two previous equations yields

${\displaystyle \;m_{1}g-m_{2}g=m_{1}a+m_{2}a}$,

and the concluding formula for acceleration

${\displaystyle a=g{m_{1}-m_{2} \over m_{1}+m_{2}}}$

The Atwood machine is sometimes used to illustrate the Lagrangian method of deriving equations of motion.[2]

It can be useful to know an equation for the tension in the string. To evaluate tension, substitute the equation for acceleration in either of the 2 force equations.

${\displaystyle a=g{m_{1}-m_{2} \over m_{1}+m_{2}}}$

For example, substituting into ${\displaystyle m_{1}a=m_{1}g-T}$, results in

${\displaystyle T={2gm_{1}m_{2} \over m_{1}+m_{2}}={2g \over 1/m_{1}+1/m_{2}}}$

For very small mass differences between m₁ and m₂, the rotational inertia I of the pulley of radius r cannot be neglected. The angular acceleration of the pulley is given by the no-slip condition:

${\displaystyle \alpha ={a \over r},}$

where ${\displaystyle \alpha }$ is the angular acceleration. The net torque is then:

${\displaystyle \tau _{\mathrm {net} }=\left(T_{1}-T_{2}\right)r-\tau _{\mathrm {friction} }=I\alpha }$

Combining with Newton's second law for the hanging masses, and solving for T₁, T₂, and a, we get:

Acceleration:

${\displaystyle a={{g(m_{1}-m_{2})-{\tau _{\mathrm {friction} } \over r}} \over {m_{1}+m_{2}+{{I} \over {r^{2}}}}}}$

Tension in string segment nearest m₁:

${\displaystyle T_{1}={{m_{1}g(2m_{2}+{{I} \over {r^{2}}}+{{\tau _{\mathrm {friction} }} \over {rg}})} \over {m_{1}+m_{2}+{{I} \over {r^{2}}}}}}$

Tension in string segment nearest m₂:

${\displaystyle T_{2}={{m_{2}g(2m_{1}+{{I} \over {r^{2}}}+{{\tau _{\mathrm {friction} }} \over {rg}})} \over {m_{1}+m_{2}+{{I} \over {r^{2}}}}}}$

Should bearing friction be negligible (but not the inertia of the pulley and not the traction of the string on the pulley rim), these equations simplify as the following results:

Acceleration:

${\displaystyle a={{g(m_{1}-m_{2})} \over {m_{1}+m_{2}+{{I} \over {r^{2}}}}}}$

Tension in string segment nearest m₁:

${\displaystyle T_{1}={{m_{1}g(2m_{2}+{{I} \over {r^{2}}})} \over {m_{1}+m_{2}+{{I} \over {r^{2}}}}}}$

Tension in string segment nearest m₂:

${\displaystyle T_{2}={{m_{2}g(2m_{1}+{{I} \over {r^{2}}})} \over {m_{1}+m_{2}+{{I} \over {r^{2}}}}}}$

Wikimedia Commons has media related to Atwood's machine.

• Frictionless plane
• Kater's pendulum
• Spherical cow
• Swinging Atwood's machine

• Professor Greenslade's account on the Atwood Machine
• Atwood's Machine by Enrique Zeleny, The Wolfram Demonstrations Project.