# Atwood machine when pulley has a non-zero mass

In this post, we will analyze one Atwood machine when its pulley has a non-zero mass. As a result this pulley has a nonzero moment of inertia.

## when pulley has a non-zero mass and nonzero moment of inertia

We will discuss the *operation of an Atwood machine with a pulley that has a non-zero mass and nonzero moment of inertia*.

Consider two masses hanging from a frictionless pulley with radius 𝑅 and mass 𝑀, as shown. Such a configuration is known as an Atwood machine. We have to find out the acceleration of the masses and the tensions in the string.

The first thing to realize is:

When the pulley in such a system has nonzero mass the tension in the string is no longer constant. It is easy to see why. If the pulley has mass 𝑀 it also has a nonzero moment of inertia 𝐼 = (1/2) 𝑀𝑅^{2}.

As the masses accelerate, the pulley undergoes an angular acceleration 𝛼, and so by Eq. (T = I α) the torque 𝜏 must be nonzero.

Because the string is the only thing producing the torque and acts at right angles to the axle, Eq. (T = F x r) tells us that the total torque is 𝜏 = (𝑇_{𝐿} − 𝑇_{𝑅})𝑅, where the radius 𝑅 of the pulley is the lever arm and 𝑇_{𝐿} and 𝑇_{𝑅} are the tensions on the left and right sides of the pulley, respectively.

If 𝑇_{𝐿} and 𝑇_{𝑅} were the same, the torque would be zero; since the torque cannot be zero, the tensions must be unequal.

In solving this problem, it may help to visualize the masses stretched out in a line, with gravity pulling 𝑚_{1} to the left and 𝑚_{2 }to the right. 𝑇_{𝑅} then acts to the left and 𝑇_{𝐿} acts to the right.

Call to the left positive. Newton’s 2nd law for 𝑚_{1} becomes:

𝑚_{1}𝑎_{1} = 𝑚_{1}𝑔 − 𝑇_{L}

Newton’s 2nd law for 𝑚_{2} is:

𝑚_{2}𝑎_{2} = −𝑚_{2}𝑔 + 𝑇_{R}

However, the string provides a physical constraint that requires 𝑎_{2} = 𝑎_{1}. Call the joint acceleration 𝑎. Then

adding the two equations gives

(𝑚1 + 𝑚2)𝑎 = (𝑚1 − 𝑚2)𝑔 + 𝑇_{𝑅} − 𝑇_{L}

We know from above that 𝜏 = (𝑇_{𝐿} − 𝑇_{𝑅})𝑅 = 𝐼𝛼 = (1/2) 𝑀𝑅^{2}𝛼. If the string does not slip on the pulley we can say that 𝛼 = 𝑎/𝑅

Then 𝑇_{𝐿} − 𝑇_{𝑅} = (1/2) 𝑀𝑎 and

𝑎 =(𝑚1 − 𝑚2)g/[m1 + m2 + M/2]

If 𝑚1 = 𝑚2, then 𝑎 = 0, as expected since the force of gravity is acting on the two masses equally.

Further, 𝑎 is less when 𝑀 ≠ 0 than when 𝑀 = 0; with a nonzero 𝐼, the pulley has inertia and adds an effective mass to the system, lessening any acceleration.

One can now find the tensions by merely plugging the expression for 𝑎 back into the top two equations.