Systems of Two Shafts: Rotational Mechanics

Shafts are used everywhere mechanical power needs to be transmitted from one location to another. Shaft rotational mechanics appear in machines, vehicles, tools, appliances, and industrial systems. Below is a clear breakdown.

Shafts are found in cars – Crankshaft converts piston motion into rotation.
Almost all factory machines use shafts – gear pump shafts, conveyors.
Shafts are used in electric motors and generators.
They are found in turbines – steam/gas/wind turbines, jet engines.
Household appliances – washing machines, vacuum cleaner.

It is important to understand how different forces affect shaft rotational mechanics. This article primarily focuses on shafts used in turbines and briefly examines a few other simple cases.

Table of Contents

Rotation Caused by Stream of Water (Two Shafts are in Equilibrium)

Figure 1. Motion caused by a fast stream of water

Imagine two shafts pictured in Figure 1 are at stable equilibrium. This means that two shafts rotate at angular velocities w1 and w2 such that their tangential velocities are the same:

\[ \textbf{(1) } v_{tg} = \omega_1 r_1 = \omega_2 r_2 \]

When two shafts rotate in this way, there is no friction between them. Equation (1) is also called the geometric constraint.

Two shafts experience the following torques:

\[ \textbf{(2) } \] \[ \tau_1 = F_{app} * r_1 \] \[ \tau_2 = F_{app} * r_2 \]

Fapp is the applied force caused by flowing stream of water and is the same in amount for both shafts.

You might be tempted to differentiate the geometric constraint (1). What you get is:

\[ \textbf{(3)} \frac{d v_{tg}}{d t} = \frac{d \omega_1}{d t} r_1 = \frac{d \omega_2}{d t} r_2 \] \[ (\frac{d \omega_1}{d t} r_1 = \frac{d \omega_2}{d t} r_2) \textbf{ means } (\alpha_1 r_1 = \alpha_2 r_2) \] \[ \frac{d v_{tg}}{d t} = \alpha_1 r_1 = \alpha_2 r_2 = a_t \] \[ \textbf{OR} \] \[ \textbf{(4)} \alpha_1 r_1 = \frac{\tau_1 r1}{I_1} = \frac{\tau_2 r_2}{I_2} = \alpha_2 r_2 \]

Note that a1 and a2 are angular accelerations. What we get is the constraint that tangential acceleration at is the same in both shafts.

Key Point 1. However, it is still okay for shaft 1 and shaft 2 to have different tangential accelerations a1r1 and a2r2 (or at1 and at2) at the instant of time when shafts are at equilibrium – equation (1). When they rotate out of equilibrium, the restoring forces drive them back into equilibrium (what is known as stable equilibrium).

Key Point 2. Differentiation of the geometric constraint yields a1r1 and a2r2 that are equal only if you assume equality in the tangential accelerations for two shafts:

\[ \frac{d v_{tg1}}{d t} = \frac{d v_{tg2}}{d t} \]

To find relationship between force, torques, angular and tangential accelerations between two touching shafts you need only Key Point 1 and equation (2):

\[ (\tau_1 = F_{app} r_1) \textbf{ gives } (\alpha_1 = \frac{\tau_1}{I_1}) \] \[ (\tau_2 = F_{app} r_2) \textbf{ gives } (\alpha_2 = \frac{\tau_2}{I_2}) \]

Then substitute angular accelerations a1 and a2 into equations for tangential acceleration:

\[ a_{t1} = \alpha_1 r_1 \] \[ a_{t2} = \alpha_2 r_2 \]

Tangential accelerations do not have to be equal.

However, if you desire to keep tangential accelerations for both shafts equal (Key Point 2), you can use equation (4):

\[ \frac{\tau_1}{\tau_2} = \frac{m_1}{m_2} \frac{r_1}{r_2} \]

Note that the derivation of equation (4) holds for any difference in tangential velocities vtg1 and vtg2:

\[ v_{tg1} – v_{tg2} = \omega_1 r_2 – \omega_2 r_2 = C \]

Where C is a constant.

Differentiating with respect to time yields equation (4).

So, equation (4) is a generalization for any difference w1r1 – w2r2. Equation (1) is a special case when difference in tangential speeds is zero:

\[ v_{tg1} – v_{tg2} = \omega_1 r_1 = \omega_2 r_2 = 0 \]

So, we can use torques T1 and T2 that we get from (2) in place of T1 and T2 in (4). This way you can find the relationship between masses m1 and m2:

\[ \frac{\tau_1}{\tau_2} = \frac{m_1}{m_2} \frac{r_1}{r_2} \] \[ \frac{F_{app} r_1}{F_{app} r_2} = \frac{m_1}{m_2} \frac{r_1}{r_2} \] \[ 1 = \frac{m_1}{m_2} \] \[ m_1 = m_2 \]

Masses should be equal for tangential accelerations to be the same in both shafts.

Stream of Water (Two Shafts are out of Equilibrium)

Figure 2. Same stream of water but shaft 2 (right) now has greater tangential velocity than shaft 1 (left). Assume no heat is produced by kinetic friction Ffr.

Two shafts rotate out of equilibrium when one shaft accelerates tangentially more than the other shaft. In other words, when shaft 2 has smaller mass than shaft 1.

\[ \alpha_1 r_1 < \alpha_2 r_2 \] \[ \textbf{OR} \] \[ \frac{\tau_1}{I_1} r_1 < \frac{\tau_2}{I_2} r_2 \] \[ \frac{\tau_1}{\tau_2} < \frac{m_1}{m_2} \frac{r_1}{r_2} \] \[ 1 < \frac{m_1}{m_2} \] \[ m_2 < m_1 \]

When shaft 2 starts rotating with tangential speed vtg2 that is greater than tangential speed vtg1 in shaft 1, there is a friction force that slows down shaft 2 and accelerates shaft 1. This reduces vtg2 and increases vtg1.

The following relationship between forces and torques holds:

\[ \tau_1 = (F_{app} + F_{fr}) r_1 \] \[ \tau_2 = (F_{app} – F_{fr}) r_2 \]

There must be normal force N between two shafts for kinetic friction Ffr to exist:

\[ F_{kinetic fr} = (\mu_{kinetic} N) \]

Geometric constraint no longer holds:

\[ \omega_1 r_1 \neq \omega_2 r_2 \] \[ \omega_1 r_1 < \omega_2 r_2 \]

The shafts are free to rotate with different tangential accelerations. However, to keep tangential accelerations equal for both shafts, we get:

\[ a_t = r_1 \alpha_1 = r_2 \alpha_2 \\ \alpha_1 = \frac{a_t}{r_1} \\ \alpha_2 = \frac{a_t}{r_2} \\ \textbf{(1)} F_{app} + F_{fr} = \frac{\tau_1}{r_1} = \frac{I_1 \alpha_1}{r_1} = \frac{I_1 a_t}{r_1^2} \\ \textbf{(2)} F_{app} -F{fr} = \frac{\tau_2}{r_2} = \frac{I_2 \alpha_2}{r_2} = \frac{I_2 a_t}{r_2^2} \\ \textbf{Adding (1) and (2):} \\F_{app} = \frac{a_t (m_1 + m_2)}{2} \]

Here we can find the applied force that is necessary to keep both shafts accelerate tangentially by the same amount at. Or, similarly, given applied force Fapp and m1 and m2, we can find the amount of tangential acceleration on both shafts.

To see an example of dynamic and static frictions as used in cycling please read The Physics of Cycling: Understanding Bicycle Motion and Stability – SciTechFocus

No Water Stream, Contact Force applied To Shaft 2

This case happens when instead of water stream; we apply a contact force Fapp to rotate only one of the shafts:

Figure 3. Two shafts are out of equilibrium. Tangential velocity vtg1 < vtg2. Constant contact force Fapp is applied to Shaft 2. Force of friction accelerates Shaft 1 and subtracts from applied force on Shaft 2. Assume no heat is produced by kinetic friction Ffr.

For there to be kinetic friction, the shafts must rotate out of equilibrium. In other words, Shaft 2 might have greater tangential velocity than shaft 1. The following equations hold:

\[ \tau_1 = F_{kinetic fr} r_1 \] \[ \tau_2 = (F_{app} – F_{kinetic fr}) r_2 \] \[ F_{kinetic fr} = (\mu_{kinetic} N) \]

Keeping tangential accelerations equal for both shafts, it can be shown that:

\[ F_{app}= a_t (m_1 + m_2) \]

Two Shafts are out of Equilibrium, No External Forces

Figure 4. No External Force. Shaft 2 (right) rotates with faster tangential velocity vtg2 than shaft 1. Shaft rotational mechanics enables force of friction to transfer kinetic rotational energy from shaft 2 to shaft 1. Assuming no energy loss due to kinetic friction.

Here the initial tangential velocities differ for both shafts:

\[ v_{tg1 i} = \omega_1 r_1 \] \[ v_{tg2 i} = \omega_2 r_2 \]

w1 and w2 are initial angular velocities for shafts 1 and 2. When friction brings two shafts to equilibrium (assuming no loss in form of heat energy due to kinetic friction), it transfers rotational kinetic energy from shaft 2 to shaft 1:

\[ KE_{net final} = KE_{net initial} \] \[ \textbf{(1)} \frac{I_1 \omega_1^2}{2} + \frac{I_2 \omega_2^2}{2} = \frac{I_1 \omega_{f1}^2}{2} + \frac{I_2 \omega_{f2}^2}{2} \]

Now we have function to find final angular velocities wf1 and wf2 from initial angular velocities w1 and w2. We then use the geometric constraint to express final angular velocity of shaft 1 in terms of final angular velocity of shaft 2 (Friction goes away only when two shafts are in equilibrium, i.e. satisfy the geometric constraint):

\[ v_{ftg1} = v_{ftg2} \] \[ \omega_{f1} r_1 = \omega_{f2} r_2 \] \[ \textbf{(2)} \omega_{f1} = \frac{\omega_{f2} r_2}{r_1} \]

Substituting this into equation for final and initial rotational energy (1):

\[ \frac{I_1}{2} (\frac{\omega_{f2} r_2}{r_1})^2 – \frac{I_1 \omega_1^2}{2} = -\frac{I_2 \omega_{f2}^2}{2} + \frac{I_2 \omega_2^2}{2} \] \[ \textbf{Which simplifies into} \] \[ \omega_{f2} = \frac{r_1^2 (I_2 \omega_2^2 + I_1 \omega_1^2)}{I_1 r_2^2 + I_2 r_1^2} \]

It is easy now to find final angular velocity for shaft 1. Final tangential velocities can be found from w1*r1 and w2*r2. Final tangential acceleration is zero since there will be no net external/internal force once the two shafts are at equilibrium.

Effect of Static Friction on Two Shaft Rotations

For this section it is important to understand two concepts: static friction and kinetic friction.

When two touching shafts rotate together without slipping at their point of contact, the friction between them is static friction. Static friction is the frictional force that acts without relative motion between the surfaces at the contact point. Because there is no sliding, kinetic friction does not occur, and no heat is produced from the friction itself. Instead, static friction transmits the tangential force/torque between the shafts to keep them rolling without slipping.

If slipping begins, i.e. one shaft’s surface speed differs from the other’s, friction becomes kinetic friction, and heat generation begins.

Consider Figure 5. The two shafts are motionless initially, but there is a contact force applied to shaft 2 that is less than maximum static friction. In the next instant that force is shared by both shaft 1 and shaft 2.

Figure 5. Two shafts touching, no initial motion. Contact force Fapp applied to shaft 2 is equal to static friction but less than maximum static friction. Fapp is shared by two shafts simultaneously.

Static friction Fstatic increases in proportion of applied force Fapp:

\[ F_{static} = F_{app} \]

Static friction Fstatic stays equal the applied force Fapp until Fapp exceeds maximum static friction:

\[ F_{app} > \mu_{static} N = F_{static fr} \]

If applied force is less than maximum static friction, there can be no kinetic friction meaning the two shafts are at equilibrium. No kinetic friction implies equal tangential velocities and equal tangential accelerations:

\[ v_{tg1 initial} = v_{tg2 initial} = 0 \] \[ a_t = \alpha_1 r_1 = \alpha_2 r_2 \]

Tangential velocities are zero initially. To ensure that tangential velocities stay equal with time, the tangential acceleration must be the same for two shafts.

The applied force Fapp is shared between the two shafts:

\[ \tau = I \alpha = F r = I \frac{a_t}{r} \] \[ F = I \frac{a_t}{r^2} \] \[ F_{net app} = F_1 + F_2 = \frac{I_1 a_t}{r_1^2} + \frac{I_2 a_t}{r_2^2} = \frac{1}{2} a_t (m_1 + m_2) \]

The following relationship is key:

\[ F_{net app} = \frac{1}{2} a_t (m_1 + m_2) \]

As a side note, in case when applied force Fapp exceeds maximum static friction, static friction Fstatic turns into kinetic friction Fkinetic:

\[ F_{app} > \mu_{static} N = F_{static fr} \] \[ F_{kinetic fr} = \mu_{kinetic} N \] \[ F_1 = F_{kinetic fr} \] \[ F_2 = F_{app} – F_{kinetic fr} \]

Kinetic friction accelerates shaft 1 – Net force on shaft 1 F1 equals kinetic friction. It also slows down shaft 2 – Net force F2 applied to shaft 2 is applied force Fapp minus kinetic friction. Force Fapp is no longer shared in this case.

Additional Considerations and Practical Applications of Shaft Rotational Mechanics

In practical engineering systems, shaft rotational mechanics rarely operates in isolation. They are typically part of larger assemblies involving gears, pulleys, couplings, bearings, and control systems. The interactions among these components introduce additional forces, constraints, and energy transfer mechanisms that influence the behaviour of the shafts.

One key factor is the role of damping. Real shafts experience energy losses due to bearing friction, air resistance, and internal material hysteresis. Incorporating damping into rotational models results in more realistic predictions of shaft dynamics.

Conclusion

This article explored certain scenarios where an external force affects the rotations of two touching shafts. The cases include moving stream of water and its force effects on two shafts. Both cases of rotational equilibrium and out-of-equilibrium motions are considered.

This article also gives an intuition how forces apply and when there is internal force in form of friction. The final section examines situation of out of equilibrium rotations with frictional force as the only force driving/slowing down rotations.