The equations of motion for a rotating mass are shown in Figure 5.1 Basic properties of rotation. Given the angular position, the angular velocity can be found by differentiating once, the angular acceleration can be found by differentiating again. The angular acceleration can be integrated to find the angular velocity, the angular velocity can be integrated to find the angular position. The angular acceleration is proportional to an applied torque, but inversely proportional to the mass moment of inertia.
Figure 5.1 Basic properties of rotation
Figure 5.2 Drill problem: Find the position with the given conditions
Free Body Diagrams (FBDs) are required when analyzing rotational systems, as they were for translating systems. The force components normally considered in a rotational system include,
· inertia - opposes acceleration and deceleration
· springs - resist deflection
· dampers - oppose velocity
· levers - rotate small angles
· gears and belts - change rotational speeds and torques
When unbalanced torques are applied to a mass it will begin to accelerate, in rotation. The sum of applied torques is equal to the inertia forces shown in Figure 5.3 Summing moments and angular inertia.
Figure 5.3 Summing moments and angular inertia
The mass moment of inertia determines the resistance to acceleration. This can be calculated using integration, or found in tables. When dealing with rotational acceleration it is important to use the mass moment of inertia, not the area moment of inertia.
The center of rotation for free body rotation will be the centroid. Moment of inertia values are typically calculated about the centroid. If the object is constrained to rotate about some point, other than the centroid, the moment of inertia value must be recalculated. The parallel axis theorem provides the method to shift a moment of inertia from a centroid to an arbitrary center of rotation, as shown in Figure 5.4 Parallel axis theorem for shifting a mass moment of inertia.
Figure 5.4 Parallel axis theorem for shifting a mass moment of inertia
Figure 5.5 Parallel axis theorem for shifting a area moment of inertia
An example of calculating a mass moment of inertia is shown in Figure 5.6 Example: Mass moment of inertia. In this problem the density of the material is calculated for use in the integrals. The integrals are then developed using slices for the integration element dM. The integrals for the moments about the x and y axes, are then added to give the polar moment of inertia. This is then shifted from the centroid to the new axis using the parallel axis theorem.
Figure 5.6 Example: Mass moment of inertia
Figure 5.7 Drill problem: Mass moment of inertia calculation
Figure 5.8 Drill problem: Find the velocity of the rotating shaft
Twisting a rotational spring will produce an opposing torque. This torque increases as the deformation increases. A simple example of a solid rod torsional spring is shown in Figure 5.9 A solid torsional spring. The angle of rotation is determined by the applied torque, T, the shear modulus, G, the area moment of inertia, JA, and the length, L, of the rod. The constant parameters can be lumped into a single spring coefficient similar to that used for translational springs.
Figure 5.9 A solid torsional spring
The spring constant for a torsional spring will be relatively constant, unless the material is deformed outside the linear elastic range, or the geometry of the spring changes significantly.
When dealing with strength of material properties the area moment of inertia is required. The calculation for the area moment of inertia is similar to that for the mass moment of inertia. An example of calculating the area moment of inertia is shown in Figure 5.10 Example: Area moment of inertia, and based on the previous example in Figure 5.6 Example: Mass moment of inertia. The calculations are similar to those for the mass moments of inertia, except for the formulation of the integration elements. Note the difference between the mass moment of inertia and area moment of inertia for the part. The area moment of inertia can be converted to a mass moment of inertia simply by multiplying by the density. Also note the units.
Figure 5.10 Example: Area moment of inertia
Figure 5.11 Drill problem: Find the torsional spring coefficient
An example problem with torsional springs is shown in Figure 5.12 Example: A rotational spring. There are three torsional springs between two rotating masses. The right hand spring is anchored solidly in a wall, and will not move. A torque is applied to the left hand spring. Because the torsional spring is considered massless the torque will be the same at the other end of the spring, at mass J1. FBDs are drawn for both of the masses, and forces are summed. (Note: the similarity in the methods used for torsional, and for translational springs.) These equations are then rearranged into state variable equations, and finally put in matrix form.
Figure 5.12 Example: A rotational spring
Rotational damping is normally caused by viscous fluids, such as oils, used for lubrication. It opposes angular velocity with the relationships shown in Figure 5.13 The rotational damping equation. The first equation is used for a system with one rotating and one stationary part. The second equation is used for damping between two rotating parts.
Figure 5.13 The rotational damping equation
Figure 5.14 Drill problem: Find the deceleration
The example in Figure 5.12 Example: A rotational spring is extended to include damping in Figure 5.15 Exampl: A torsional mass spring damper ystem. The primary addition from the previous example is the addition of the damping forces to the FBDs. In this case the damper coefficients are indicated with 'B', but 'Kd' could have also been used. The state equations were developed in matrix form. Visual comparison of the final matrices in this and the previous example reveal that the damping terms are the only addition.
Figure 5.15 Exampl: A torsional mass spring damper ystem
The lever shown in Figure 5.16 Force transmission with a level can be used to amplify forces or motion. Although theoretically a lever arm could rotate fully, it typically has a limited range of motion. The amplification is determined by the ratio of arm lengths to the left and right of the center.
Figure 5.16 Force transmission with a level
Figure 5.17 Drill problem: Find the required force
While levers amplify forces and motions over limited ranges of motion, gears can rotate indefinitely. Some of the basic gear forms are listed below.
Spur - Round gears with teeth parallel to the rotational axis.
Rack - A straight gear (used with a small round gear called a pinion).
Helical - The teeth follow a helix around the rotational axis.
Bevel - The gear has a conical shape, allowing forces to be transmitted at angles.
Gear teeth are carefully designed so that they will mesh smoothly as the gears rotate. The forces on gears acts at a tangential distance from the center of rotation called the pitch diameter. The ratio of motions and forces through a pair of gears is proportional to their radii, as shown in Figure 5.18 Basic Gear Relationships. The number of teeth on a gear is proportional to the diameter. The gear ratio is used to measure the relative rotations of the shafts. For example a gear ratio of 20:1 would mean the input shaft of the gear box would have to rotate 20 times for the output shaft to rotate once.
Figure 5.18 Basic Gear Relationships
For lower gear ratios a simple gear box with two gears can be constructed. For higher gear ratios more gears can be added. To do this, compound gear sets are required. In a compound gear set two or more gears are connected on a single shaft, as shown in Figure 5.19 Example: A compound gear set. In this example the gear ratio on the left is 4:1, and the ratio for the set on the right is 4:1. Together they give a gear ratio of 16:1.
Figure 5.19 Example: A compound gear set
A manual transmission is shown in Figure 5.20 Example: A manual transmission. In the transmission the gear ratio is changed by sliding (left-right) some of the gears to change the sequence of gears transmitting the force. Notice that when in reverse an additional compound gear set is added to reverse the direction of rotation.
Figure 5.20 Example: A manual transmission
Rack and pinion gear sets are used for converting rotation to translation. A rack is a long straight gear that is driven by a small mating gear called a pinion. The basic relationships are shown in Figure 5.21 Relationships for a rack and pinion gear set.
Figure 5.21 Relationships for a rack and pinion gear set
Belt based systems can be analyzed with methods similar to gears (with the exception of teeth). A belt wound around a drum will act like a rack and pinion gear pair. A belt around two or more pulleys will act like gears.
Figure 5.22 Drill problem: Find the gear speed
Friction between rotating components is a major source of inefficiency in machines. It is the result of contact surface materials and geometries. Calculating friction values in rotating systems is more difficult than translating systems. Normally rotational friction will be given as static and kinetic friction torques.
An example problem with rotational friction is shown in Figure 5.23 Example: A system with friction. Basically these problems require that the model be analyzed as if the friction surface is fixed. If the friction force exceeds the maximum static friction the mechanism is then analyzed using the dynamic friction torque. There is friction between the shaft and the hole in the wall. The friction force is left as a variable for the derivation of the state equations. The friction value must be calculated using the appropriate state equation. The result of this calculation and the previous static or dynamic condition is then used to determine the new friction value.
Figure 5.23 Example: A system with friction
The friction example in Figure 5.23 Example: A system with friction can be analyzed using the C program in Figure 5.24 Example: A C program for the friction example in Figure 5.23. For the purposes of the example some component values are selected and the system is assumed to be at rest initially. The program loops to integrate the state equations. Each loop the friction conditions are checked and then used for a first-order solution to the state equations.
Figure 5.24 Example: A C program for the friction example in Figure 5.23 Example: A system with friction
DC motors create a torque between the rotor (inside) and stator (outside) that is related to the applied voltage or current. In a permanent magnet motor there are magnets mounted on the stator, while the rotor consists of wound coils. When a voltage is applied to the coils the motor will accelerate. The differential equation for a motor is shown in Figure 5.25 Model of a permanent magnet DC motor, and will be derived in a later chapter. The value of the constant 'K' is a function of the motor design and will remain fixed. The value of the coil resistance 'R' can be directly measured from the motor. The moment of inertia 'J' should include the motor shaft, but when a load is added this should be added to the value of 'J'.
Figure 5.25 Model of a permanent magnet DC motor
The speed response of a permanent magnet DC motor is first-order. The steady-state velocity will be a straight line function of the torque applied to the motor, as shown in Figure 5.26 Torque speed curve for a permanent magnet DC motor. In addition the line shifts outwards as the voltage applied to the motor increases.
Figure 5.26 Torque speed curve for a permanent magnet DC motor
The energy and power relationships for rotational components are given in Figure 5.27 Energy and power relations for rotation. These can be useful when designing a system that will store and release energy.
Figure 5.27 Energy and power relations for rotation
A large machine is to be driven by a permanent magnet electric motor. A 20:1 gear box is used to reduce the speed and increase the torque of the motor. The motor drives a 10000kg mass in translation through a rack and pinion gear set. The pinion has a pitch diameter of 6 inches. A 10 foot long shaft is required between the gear box and the rack and pinion set. The mass moves on rails with static and dynamic coefficients of friction of 0.2 and 0.1 respectively. We want to select a shaft diameter that will keep the system critically damped when a voltage step input of 20V is applied to the motor.
To begin the analysis the velocity curve in Figure 5.28 Example: Motor speed curve and the derived differential equation was obtained experimentally by applying a voltage of 15V to the motor with no load attached. In addition the resistance of the motor coils was measured and found to be 40 ohms. The steady-state speed and time constant were used to determine the constants for the motor.
Figure 5.28 Example: Motor speed curve and the derived differential equation
The remaining equations describing the system are developed in Figure 5.29 Example: Additional equations to model the machine. These calculations are done with the assumption that the inertial effects of the gears and other components are insignificant.
Figure 5.29 Example: Additional equations to model the machine
If the gear box is assumed to have relatively small moment of inertia, then we can say that the torque load on the motor is equal to the torque in the shaft. This then allows the equation for the motor shaft to be put into a useful form, as shown in Figure 5.30 Example: Numerical analysis of system response. Having this differential equation now allows the numerical analysis to proceed. The analysis involves iteratively solving the equations and determining the point at which the system begins to overshoot, indicating critical damping.
Figure 5.30 Example: Numerical analysis of system response
These results indicate that a spring value of XXX is required to have the system behave as if it is critically damped. (Note: Clearly this system is not second order, but in the absence of another characteristics we approximate it as second order.)
· The basic equations of motion were discussed.
· Mass and area moment of inertia are used for inertia and springs.
· Rotational dampers and springs.
· A design case was presented.
1. Draw the FBDs and write the differential equations for the mechanism below. The right most shaft is fixed in a wall.
2. For the system pictured below a) write the differential equations (assume small angular deflections) and b) put the equations in state variable form.
3. Draw the FBDs and write the differential equations for the mechanism below.
4. The system below consists of two masses hanging by a cable over mass `J'. There is a spring in the cable near M2. The cable doesn't slip on `J'.
a) Derive the differential equations for the following system.
b) Convert the differential equations to state variable equations
5. Write the state equations for the system to relate the applied force 'F' to the displacement 'x'. Note that the rotating mass also experiences a rotational damping force indicated with Kd1
6. For the system pictured below a) write the differential equations (assume small angular deflections) and b) put the equations in state variable form.
7. For the system pictured below a) write the differential equations (assume small angular deflections) and b) put the equations in state variable form.
8. For the system pictured below a) write the differential equations (assume small angular deflections) and b) put the equations in state variable form.
9. For the system pictured below a) write the differential equations (assume small angular deflections) and b) put the equations in state variable form.
10. For the system pictured below a) write the differential equations (assume small angular deflections) and b) put the equations in state variable form.
11. For the system pictured below a) write the differential equations (assume small angular deflections) and b) put the equations in state variable form.
12. Find the polar moments of inertia of area and mass for a round cross section with known radius and mass per unit area. How are they related?
13. The rotational spring is connected between a mass `J', and the wall where it is rigidly held. The mass has an applied torque `T', and also experiences damping `B'.
a) Derive the differential equation for the rotational system shown.
b) Put the equation in state variable form (using variables) and then plot the position (not velocity) as a function of time for the first 5 seconds with your calculator using the parameters below. Assume the system starts at rest.
c) A differential equation for the rotating mass with a spring and damper is given below. Solve the differential equation to get a function of time. Assume the system starts at rest.
14. Find the response as a function of time (i.e. solve the differential equation to get a function of time.). Assume the system starts undeflected and at rest.
15. For the system pictured below a) write the differential equation for the system with theta2 as the output (assume small angular deflections) and b) put the equations in state variable form.
16. Analyze the system pictured below assuming the rope remains tight.
a) Draw FBDs and write the differential equations for the individual masses.
b) Write the equations in state variable matrix form.
c) Use Runge-Kutta integration to find the system state after 1 second.
1. Write the state equation(s) for the following mechanical system of two gears. Assume that the cables always remain tight and all deflections are small.
2. Draw FBDs for the following mechanical system containing two gears.
3. Draw FBDs for the following mechanical system. Consider both friction cases.
4. Draw FBDs for the following mechanical system.
5. Develop a differential equation of motion for the system below assuming that the cable always remains tight.
6. Write the state equations for the following mechanical system. Assume that the cables always remain tight and all deflections are small.
7. Write the state equations for the following mechanical system. Assume that the cables always remain tight and all deflections are small.
8. Write the state equations for the following mechanical system. Assume that the cables always remain tight and all deflections are small.
9. Analyze the system pictured below assuming the rope remains tight and gravity acts downwards.
a) Draw FBDs and write the differential equations for the individual masses.
b) Combine the equations and simplify the equations as much as possible.
c) Write the equations in state variable matrix form.
d) Use Runge-Kutta to find the system state after 1 second.