# What is positive rotation direction of a pulley in the Atwood machine?

In the Atwood machine the mass $$m_1$$ hangs on the left and $$m_2$$ hangs on the right, with $$m_2 > m_1$$. When released from rest the system accelerates clockwise which we define to be the positive direction.

The pulley has non-negligible mass and also rotates clockwise. The tension $$T_1$$ on $$m_1$$ is in the positive direction, and the tension $$T_2$$ of $$m_2$$ is in the negative direction. Since $$m_2 > m_1$$, it follows $$|T_2| > |T_1|$$. The tension on the sides of the pulley generate torque. Why is there a positive net torque when simply combining $$T_1$$ and $$T_2$$ gives a negative result?

• There is no preferred direction for torque. Clockwise rotation is negative according to the standard convention for torque. Mar 2, 2022 at 22:27

If the tension on the sides of the pulley is what causes the torque, why is there a positive net torque when simply combining T1 and T2 gives a negative result?

Positive and negative rotation directions are not strictly defined, which means you are free to define positive rotation direction. What is important here is that you write all the equations with respect to the defined positive directions.

Figure below shows free-body diagram for the Atwood machine. Note that positive directions are not indicated on purpose because we will discuss both cases. Regardless of the positive direction, the tension $$T_1$$ contributes to counterclockwise and $$T_2$$ to clockwise rotation of the pulley.

Figure: Free-body diagram for the Atwood machine

#### Translational motion

If the positive direction for linear acceleration is downwards then equations of (linear) motion are

$$m_1 a_1 = m_1 g - T_1 \qquad \text{and} \qquad m_2 a_2 = m_2 g - T_2$$

Since string is assumed to be non-elastic, both hanging masses have the same acceleration magnitude but opposite direction: $$a_2 = -a_1 \equiv a$$. The above equations now become

$$-m_1 a = m_1 g - T_1 \qquad \text{and} \qquad +m_2 a = m_2 g -T_2 \tag {1a}$$

If $$a$$ is positive then $$a_2$$ is positive (mass $$m_2$$ is going downwards) and $$a_1$$ is negative (mass $$m_1$$ is going upwards). This means that for positive $$a$$ the pulley is rotating clockwise, and for negative $$a$$ the pulley is rotating counterclockwise.

Note that we could have defined the accelerations as $$a_1 = -a_2 \equiv a$$, in which case

$$-m_1 a = m_1 g - T_1 \qquad \text{and} \qquad +m_2 a = m_2 g -T_2 \tag {1b}$$

and positive $$a$$ means that pulley is rotating counterclockwise. You can define positive direction to be whatever you want, as long as you use the same direction for all other variables.

#### Rotational motion

Let's assume that the acceleration is defined as $$a_2 = -a_1 \equiv a$$ in which case Eq. (1a) applies for the translational motion. If the string does not slip over the pulley, the angular and linear accelerations are related as follows

$$a = -\alpha r \qquad \text{or} \qquad a = +\alpha r$$

where $$r$$ is radius of the pulley. Which sign is to be used (positive or negative) depends on the positive rotation direction.

If the positive rotation direction is defined to be clockwise then $$a = +\alpha r$$ and

$$-T_1 r + T_2 r = I \alpha = I \cdot (+a/r) \tag {2a}$$

If the positive rotation direction is defined to be counterclockwise then $$a = -\alpha r$$ and

$$+T_1 r - T_2 r = I \alpha = I \cdot (-a/r) \tag {2b}$$

Note that Eqs. (2a) and (2b) are the same with respect to acceleration $$a$$.

#### Final equations of motion

The pulley can be modelled as a solid cylinder with moment of inertia $$I = \frac{1}{2} M r^2$$, where $$M$$ is mass and $$r$$ is radius of the pulley.

Final equations of motion for the Atwood's machine with $$a_2 = -a_1 \equiv a$$ are

\begin{aligned} m_1 a &= T_1 - m_1 g \\ m_2 a &= m_2 g - T_2 \\ 0.5 M a &= T_2 - T_1 \end{aligned}

From this set of equations it is relatively simple to find expressions for $$a$$, $$T_1$$, and $$T_2$$ as a function of $$m_1$$, $$m_2$$, $$M$$, and $$g$$.

Consider the following diagram of the pulley with some axes added.

Torque, $$\tau = \vec r \times \vec F$$

The torque about the centre of the pulley due to force $$\vec T_1$$ is $$\vec \tau_1=(-r\hat i)\times (-T_1 \hat j) = +r\,T_1\,\hat z$$ which with the right hand convention corresponds to an anticlockwise rotation.

The torque about the centre of the pulley due to force $$\vec T_2$$ is $$\vec \tau_1=(+r\hat i)\times (-T_2 \hat j) = -r\,T_2\,\hat z$$ which with the right hand convention corresponds to clockwise rotation.

So the net torque is $$\vec \tau_1 +\vec \tau_2 = r(T_1-T_2)\,\hat z$$ which corresponds to a clockwise rotation if $$T_2 >T_1$$.

There is no positive direction in Atwood machine problems, unlike most kinematical problems you might encounter where you assume right and up is positive while left and right is negative,in Atwood machines, you simple assume a flow, be it to the right or to the left, and then you assume any force acting in the direction of the flow is positive and vice versa.

Then you solve for the unknown force or acceleration according to that flow, if your answer was negative, don’t panic, it merely means that the correct flow proved to be opposite to the one you assumed.

The tension on m2 (T2) is in the negative direction

Tension is a scalar. Assuming a "light" string, the tension is uniform and becomes a force (of equal magnitude) at both ends.

The force from the segment attached to m2 on the pulley is clockwise. The force from the segment attached to m1 on the pulley is counterclockwise.

• tension is force. So tension is vector. Mar 3, 2022 at 9:44
• What direction is tension? Mar 3, 2022 at 16:32
• upward and downward, which means along the z-axis(for simple Atwood machine). Mar 3, 2022 at 17:09
• Mar 3, 2022 at 17:19