noc19-ee36-lec58

Namaskar, in last class we discussed about the mathematical modeling of translational motion of mechanical system. Today we will discuss about the mathematical modeling of rotational system. So let us start the discussion of today's lecture. So the rotational motion of a body can be defined as motion about a fixed axis. The extension of Newton's law of motion for rotational motion states that the algebraic sum of moments or torque about a fixed axis is equal to the product of the inertia and the angular acceleration about the axis. So we define that total sum of torques which is applicable on a body which is rotating is equal to J into alpha. The J will be inertia of the body and alpha is called as angular acceleration. So other variables which we generally use to describe the motion of rotation are angular velocity omega and angular displacement theta. So if you take d theta by dt you will get angular velocity omega and d 2 theta by dt 2 you get the angular acceleration. Now the elements involved in the rotational motions are first inertia. So inertia is considered as the property of an element that stores the kinetic energy of the rotational motion. So you can correlate it with respect to the mass m which we discussed in case of translational motions. So here also the inertia is the property of the element which stores the kinetic energy of rotational motion. Now the inertia of a given body depends on the geometric composition about the axis of the rotation and its density. So for instance if the inertia of a circular disc or staff or shaft of radius r and mass m the value of inertia about its geometric axis is given as J is equal to half of m into r square. Now when the torque is applied to a body with the inertia J if you see the figure the body which is rotating having inertia J and it is having theta as the displacement when the torque is applied the governing equation which you can write is torque T is equal to J alpha that is equal to J d omega by dt or you can write J d 2 theta by dt 2. So here theta is the angular displacement which is because of the applied torque omega is the angular velocity and alpha is the angular acceleration. Now similar to the spring linear spring which we saw in case of translational motion we can also define torsional spring. Torsional spring let us say we have a torsional spring constant K that is given in torque per unit angular displacement. So we can device to represent the compliance of a rod or a shaft when it is subject to applied torque. So torsional spring is the property of the shaft or the rod which is connected to your rotating body. So when you give torque it twists slightly which give the property of torsional spring. So what you will write simple torque spring system can be represented by this equation that is T capital T that is the torque at any time T can be given as K theta. Theta is also at that time T. Now if the torsional spring is preloaded by a torque say T p the above equation can be modified or we can write torque T minus T p that is the preloaded torque equal to K theta. Now similar to what we saw in case of translational motion in this case also we will see various frictions. What are those frictions? First is viscous friction that is torque T equal to b d theta by dt. This b is called a viscous friction coefficient. Then a static friction in that case the friction torque T can be defined as plus minus f s that is the value of a static friction for theta dot equal to 0 that is the object is just about to move. In that case the value of friction is given by f s plus minus is basically used to show the direction of the friction which will be applicable in the body. Then Coulomb friction T at any time T can be given as f c d theta by dt divided by mod of d theta by dt. So this is how we define viscous friction, static and Coulomb friction in the rotating body. Now let us take one example so that we can understand how we will create the mathematical model of a mechanical system. So in this mechanical system we have one motor connected to a load through shaft which has the friction coefficient as we will rather say stiffness coefficient as k which defines the torsional spring coefficient. Now T m is the mechanical torque applied by the motor and theta l is the displacement at any time T for the load which is having inertia equal to j l. Motor inertia is j m and viscous friction coefficient is b m. Now in this case the motor is coupled to a inertial load through shaft with the spring constant k. The non rigid coupling between two mechanical components which we see here often causes torsional resonance that can be transmitted to all parts of the system. Now let us see how we can create the free body diagram. The variable which we will use to define the free body diagram of this mechanical system the variables are T m, T m is nothing but motor torque, b m is viscous friction of the motor. So this is viscous friction coefficient related to motor, k is a spring constant of the shaft, theta m is motor displacement, omega m is given as motor velocity, j m is motor inertia, theta l we consider it as load displacement, omega l is load velocity and j l is the load inertia. So we will use these variables to define the free body diagram of the system. So from the motor side if you see the free body diagram you will see that the torque T m when it is applied you will have the viscous friction part plus spring torque part applicable in the opposite direction of the motor torque applied. So if motor torque is T m you can write that the torque which is reflecting at the output of the motor that is j m d 2 theta m by d t 2 equal to T m minus the viscous friction coefficient and multiplied by the angular velocity. So this is your viscous friction torque that is b m d theta m by d Similarly for torsional spring you will have torque as k theta m minus theta l. So when you rearrange you take T m at the right side and other components you take on the left side. So the equation from motor side you can write j m d 2 theta m by d t 2 plus b m d theta m by d t plus k theta m minus theta l equal to the torque which is applied by the motor. Now if you rearrange this equation by taking higher order highest order derivative on the left side you can write d 2 theta m by d t 2 equal to minus b m by j m d theta m by d t minus k by j m theta m minus theta l plus 1 by j m T m. Now on the right side the equation which you can write is because here we are assuming that viscous friction coefficient is 0. So what we can write we can write k theta m minus theta l equal to the torque which is applied in the load that is j l d 2 theta l by d t 2. Now in this case the system contains three energy storage elements that is j m j l and the k. So what we will do we will define three state variables in this case the equations which we just saw in case of the motor side and the load side free body diagram we came with these two equations that is d 2 theta m by d t 2 equal to minus b m by j m d theta by d theta m by d t minus k by j m theta m minus theta l plus 1 by j m T m. That is the first equation we got when we considered free body diagram from the motor side similarly from the load side we got d 2 theta l by d t 2 equal to k by j l theta m minus theta l. Now let us assume that we have three state variables say x 1 is defined as theta m minus theta l x 2 is defined as theta l d theta l by d t and x 3 is defined as d theta m by d t. So the state equations which you can write if you differentiate this equation that is x 1 is equal to theta m minus theta l what you can write d x 1 by d t is equal to d theta m by d t minus d l by d t. So when you differentiate you will write d x 1 by d t equal to d theta m by d t minus d theta l by d t now this d theta m by d t we have considered as x 3. So we write x 3 in place of d theta m by d t and similarly d theta l by d t is written as x 2. So we got our first equation for a state variable that is d x 1 by d t is equal to x 3 minus x 2. Now from the previous equations that is d 2 theta l by d t 2 is equal to k by j l theta m minus theta l. So theta m minus theta l we have considered as x 1 and this is nothing but the differentiation of x 2. So what we will write we will write d x 2 by d t equal to k by j l x 1 t. Similarly for the first equation what you can write d 2 theta m by d t 2 is nothing but x 3 dot or you can write d x 3 by d t and then it is minus b m by j m d theta m by d t we will write as x 3. So minus b m by j m x 3 we have got then minus k by j m d theta m minus theta l. So this will become again x 1 and finally we write 1 upon j m t m. So what we have got minus k by j m x 1 minus b m by j m x 3 plus 1 upon j m t m. So we got the third one this is your second equation. Now you can compile it in state space format. So what you can write x 1 dot x 2 dot x 3 dot equal to the matrix A. So matrix A you can compile easily from here by putting x 1 x 2 x 3 here. So what will be the value in case of first equation for x 1 it is 0 x 2 is minus 1 x 3 is plus 1. For second you have k by j l rest are 0. For third one you write minus k by j m 0 minus b m by j m. So this is what you have got is matrix A. We can represent the state equations which we have just got using the signal flow graph. So how will represent we have defined the state variables as x 1 is equal to theta m minus theta l x 2 is d theta l by d t and x 3 is d theta m by d t. First we will create the node related to these variables. So x 1 we have considered that is theta m minus theta l here. Then you get x 2 that is nothing but theta l dot then x 3 which is nothing but theta m dot. Now for the first equation what you are getting d x 1 by d t is equal to x 3 minus x 2. So here you have x 3 you have x 2 you have x 3 minus x 2. So x 3 will be having gain of 1 x 2 will be having gain of minus 1 when you sum up you get theta m dot minus theta l dot that is nothing but equal to d x 1 by d t. So for from first equation what you have got is these two links. Now from d x 2 by d t d x 2 by d t is nothing but theta l double dot. So d x 2 by d t we replace we place here that is theta l double dot and that is nothing but equal to k by j l x 1. So we will take the loop from x 1 with gain k by j l and connect it to the theta l double dot. Now theta l double dot if you integrate you will get theta l dot that is nothing but x 2. So you have one integrator link with the s to the power minus 1 that is nothing but 1 by s. So means you integrate theta l double dot to get theta l dot. So this section is complete with respect to these two equations. Then when you again integrate theta m dot minus theta l dot you will get x 1. So again you create one integrator block and you get the x 1. Now next is the equation that is d x t by d t is equal to minus k by j m x 1 minus b m by j m x 3 plus 1 upon j m t m. So we put t m as a input node and then we put theta m double dot. Theta m double dot is nothing but d x 3 by d t. So we use this equation and find out the links. So first is minus k by j m into x 1. So from x 1 we get minus k by j m. Then from x 3 we get minus b m by j m. So from x 3 we connect the link again with gain of minus b m by j m and then you have 1 upon j m t m. So t m is connected to the theta m double dot that is with the gain 1 upon j m. So with this you complete the summation at theta m double dot. Now theta m double dot is connected to x 3 with integrator integration. If you take the integral of theta m double dot you will get theta m dot which is nothing but x 3. So using these three equations you have now created the signal flow graph presentation of the system. Now let us talk about the gear train. So gear train or lever or the timing belt over a pulley is a mechanical device that transmits energy from one part of the system to another in such a way that force, torque, speed and displacement may be altered. So these devices can also be regarded as matching devices used to attain maximum power transfer. Now we see these in the figure we see two gears are coupled together. The inertia and friction of the gears are neglected when you consider the ideal case. So in this case t 1 through these gears is transferred to the second gear that is having torque t 2. Now how we will create the relationship between t 1 and t 2? So the relationship between torque t 1 and t 2 and then angular displacement theta 1 and theta 2 and the teeth numbers in the gear that is n 1 and n 2 can be correlated with the following facts. First is that number of teeth on the surface of the gears is proportional to the radius r 1 and r 2 of the gears. So in that case we can define r 1 n 2 is equal to r 2 n 1 because r 1 is proportional to n 1 and r 2 is proportional to n 2. So using this you can correlate that r 1 n 2 is equal to r 2 n 1. Now the distance traveled along the surface of each gear is same. Therefore you can write theta 1 r 1 that is the distance traveled equal to theta 2 r 2. Now third one is that the work done by one gear is equal to that of other since there are assumed to be no losses. In that case t 1 theta 1 will be equal to t 2 theta 2. So using these three correlations we can write that t 1 upon t 2 is equal to theta 2 by theta 1 and equal to n 1 by n 2 and equal to omega 2 by omega 1 and equal to r 1 by r 2. So this is the most important correlation which we use in case of gear system to correlate the various parameters from one side to other side. Now in practice the gears also have inertia and friction between the coupled gear teeth and that often cannot be neglected. So in that case the equivalent representation of the gear train with viscous friction, coulomb friction and inertia as lump parameter is considered which is shown in the figure. So here you have the coulomb frictions, then b 1 and b 2 are the viscous frictions, t is the torque applied from the gear 1. So the opposite torque which is t 1 applied on the gear n 1 and translated to gear 2 is t 2 and the energy transferred from gear 2 the object which is having inertia equal to j 2. Now since we have two gears so we need two torque equations. So for gear 2 what we can write that is t 2 t that is the torque applied because of the movement of gear 1 we have torque t 2 in the gear 2. So we can write equal to j 2 d 2 theta 2 by dt 2 plus the coefficient related to viscous friction and coefficient related to coulomb friction. We can write b 2 d theta 2 by dt plus f c 2 omega 2 by omega 2 basically the magnitude of omega 2. Now the torque equation on the first side what you can write the applied torque t is equal to j 1 d 2 theta 1 by dt 2 plus b 1 d theta 1 by dt plus f c 1 into omega 1 divided by the magnitude that is mod of omega 1 plus t 1. Now we have got these two equations related to gear 1 and gear 2 using the previous correlation which we just saw that is t 1 by t 2 is equal to n 1 by n 2 t 1 is nothing but t 2 into n 1 by n 2. So we will use t 1 is equal to n 1 by n 2 into t 2 we write the equation of t 2 here and simplify with the help of the these two equations. So we can write these theta in terms of theta 1 with the help of again the correlation which we have theta 1 by theta 2 by theta 1 is nothing but n 1 by n 2. So we use these correlation and finally when we simplify we get t 1 as equal to n 1 upon n 2 square j 2 d 2 theta 1 by dt 2 plus n 1 upon n 2 square b 2 d theta 1 by dt plus n 1 upon n 2 f c 2 omega 2 upon mod of omega 2. So this indicates that it is possible to reflect inertia friction compliance torque speed and displacement from one side of the gear train to other. So when you see this equation you can correlate these equations with respect to the transformer equations which we derived in earlier discussions. So you can also see that as we saw in case of transformer we transfer the power from one side to other side. Similarly the gear is the mechanical arrangement which transfers power from one side to other side. So in this way you can see that how closely the mechanical system of gears is analogous to the electrical system of the transformers. Now the following quantities we obtained when reflecting from gear 2 to gear 1 we define them as the inertia that is n 1 upon n 2 square j 2 viscous friction coefficient we say as n 1 upon n 2 square b 2 torque n 1 upon n 2 into t 2 angular displacement we say with respect to the gear 1 that is n 1 upon n 2 into theta 2. Similarly angular velocity we convert into the gear 1 side is n 1 upon n 2 omega 2 coulomb friction coefficient is n 1 upon n 2 f c 2 omega 2 divided by mod omega 2. So if you consider the equivalent of all those components what we can write total torque t you can just put the value of t 1 which we have got into this equation because that will be total torque on the gears on the gear 1 is nothing but j 1 d 2 theta 1 by d t 2 plus b 1 d theta 1 by d t plus f c 1 omega 1 upon mod omega 1 plus t 1. This t 1 we replace with the equation which we have just got with the help of transformation. So when you further simplify you can write total torque t is nothing but j 1 equivalent d 2 theta 1 by d t 2 plus b 1 equivalent d theta 1 by d t plus t f that is equivalent coulomb torque. So what is j 1 equivalent that is j 1 plus n 1 upon n 2 square j 2 b 1 equivalent is b 1 plus n 1 upon n 2 square b 2 and t f is nothing but f c 1 omega 1 divided by omega 1 mod plus n 1 upon n 2 f c 2 omega 2 upon omega 2 mod. So if you see these equations you can easily correlate in case of transformer also when you transfer the resistance and reactance value from secondary side to primary side you will see similar kind of relationship as we see in this case of gear the reflection of the various properties from gear 2 to gear 1. So in this way you can say that the gear arrangement is closely analogous to the electrical system in case of the transformer. So with this we can close our today's discussion in this discussion we mainly focused about the rotational motion. So from next lecture onwards we will discuss about the correlation between electrical and the mechanical or other than the mechanical system. So that you can understand how the mechanical system can be equivalently represented with the help of electrical circuit thank you.