YouTube transcript (1wxIfK3ocXo) — Transcript

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1. 0:00On the left side of your screen there is a disc of  diameter 600 mm and on the right side of a screen
2. 0:07there is a cylinder of diameter 200 mm and both  have equal Mass 20 kg now imagine these shapes
3. 0:17are mounted on set of bearings and connected to  electrical motors and try to rotate the shapes
4. 0:24around their Central axis so could you tell me  if this both shapes need to rotate at the same speed
5. 0:33let's say 100 RPM then which motor would require  more torque would or the required torque need to be
6. 0:41same since both have same mass and same speed or  would the motor for disc need to be more torque
7. 0:48than the one for cylinder what do you think which  one would require the greater torque how we can
8. 0:55calculate the required torque for Motors well as  we know the torque is the product of force F and the
9. 1:03distance R for example if we are applying a tangential  force of 5 Newton at a lever that is 2 m long the
10. 1:12torque required to turn around the Pivot Point  would be 5 Newton into 2 m 10 Newton meter but this
11. 1:21case is different we are not applying a force  at circumference or at a distance from the axis
12. 1:30here the motor is directly applying a turning  moment to the center of the disc or cylinder there
13. 1:39is a no lever there is no R and no tangential  Force the torque is applying directly to the
14. 1:47central axis so how we do calculate the torque  required torque for these motors moreover this
15. 1:55is not a surprising or unique case as a mechanical  design engineer we encounter such cases very often
16. 2:04we often design indexing table for assembly line  where we rotate a disc plate around their axis
17. 2:11very precisely also conveyor rollers where  cylindrical rollers rotate around their
18. 2:17Central axis also in ball screw or lead screw  we rotate the ball screw around their axis and
19. 2:25in all these cases we must calculate the required  torque to sizing the motor correctly however if your
20. 2:34physics is still good then you must be familiar  with the term moment of inertia and probably you
21. 2:42know the answer to this question as well that  the disc has a larger diameter that means the
22. 2:48mass is distributed further from the axis so  its moment of inertia is higher as compared to
23. 2:56cylinder in fact the disc has moment of inertia  of 0.9 kg meter square while the cylinder has
24. 3:05a much lower moment of inertia of 0.1 kg meter  Square so does this means the motor for the disc
25. 3:14will always require more torque not necessary the  torque required to rotate an object doesn't depend
26. 3:23solely on its moment of inertia it also depends  on angular acceleration that is how quickly we
27. 3:31need the object to reach its Target speed if  you're okay with a slower start then required
28. 3:38torque would be lower but if you need a quick  start and stop like an induction table then the
29. 3:45torque requirement increases significantly so to  choose the right motor for any rotary application
30. 3:52like indexing table or conveyor, we need to  master two things one is what does actually
31. 4:00means of moment of inertia how to calculate it  and how to optimize it for a rotary mechanism
32. 4:07and the second is how to estimate the acceleration  time for particular rotary mechanism which again
33. 4:15depends on two things one is application itself  for example, application like indexing table quick
34. 4:23start and stop are critical but for conveyors  blowers fans acceleration can be slower and the
35. 4:32second is type of motor for example Servo motor  can take much more time to accelerate whereas for
36. 4:40an induction motor too long startup time can be  lead to stalling or overheating so hi there this
37. 4:48is Ayush from Master mechanical design and today  we are going to master the motor sizing for rotary
38. 4:55applications we will go through the two examples  one for induction motor sizing for a belt conveyor
39. 5:01and another is Servo motor sizing for indexing  application and by the end of this video you
40. 5:07will have a clear understanding of how to sizing  the motor for almost any rotating application but
41. 5:14this is just an outcome of this video the main  thing the core thing we are going to must is
42. 5:21inertia and movement of inertia not just physics behind  this but we will also understand what moment of inertia
43. 5:30really is how it works and why it's so important  in machine design I will show you where we use
44. 5:38torque = F X r and where we use torque =  I X Alpha so after watching this video the
45. 5:47motor sizing won't intimidate you anymore and  you will become a master of motor sizing for
46. 5:54almost all the application before we start let  me quickly mention our today video sponsor our
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56. 7:12big thanks to JLCMC for supporting the channel so  what is inertia and moment of inertia so first of
57. 7:21the inertia is related to linear motion the motion  in a straight line and moment of inertia or we can
58. 7:29say mass moment of inertia or second movement of mass or  most accurately rotational inertia is related to
59. 7:38rotational motion but before understanding moment  of inertia let's start with inertia so you have
60. 7:46probably heard this word many times back in school  high school in college and even in job but what
61. 7:54does it actually mean is inertia a force or is  its Mass well inertia is neither a force nor just
62. 8:05Mass it is actually a property of mass a natural  tendency of mass of an object to keep doing what
63. 8:15they are already doing if something is at rest it  will naturally want to be at rest if something is
64. 8:23moving it's naturally wanting to be keep moving in  the same direction at same speed in simple terms
65. 8:32inertia is a resistance to change in its motion  in their current state and Isaac Newton refine and
66. 8:41codify the same concept of inertia in the first  law of motion also known as principle of inertia
67. 8:51that everybody Preserve in the state of rest or  uniform motion in right line unless it's compiled
68. 8:59to change the state by the force impressed  their own what this simply means is a body
69. 9:06in its rest means the velocity and acceleration  is zero the natural tendency of the body to be
70. 9:15at rest cannot automatically go into motion and  if an object is in uniform motion on a straight
71. 9:23line uniform motion means constant velocity can  be anything but must be constant which again
72. 9:31means zero acceleration then the object will  remains in uniform motion on a straight line
73. 9:38but unfortunately, we cannot experience this on  Earth surface because of gravity friction and
74. 9:44air resistance but you must have seen in videos  when an astronaut float around an object the object
75. 9:53continue to move at constant velocity because in  a space where there's no gravity it and air resistance
76. 10:02so, once the object gets accelerate it will remain  in uniform motion unless we apply a force on it so
77. 10:10here is a big question how much force is needed  to change the motion how much force is needed
78. 10:18to change the state means if something is at rest  then how much force would take it to get in moving
79. 10:27or if something is already moving in a straight  line then how much force would require changing
80. 10:32its speed or its direction so let's understand it  step by step so first if you just want to start
81. 10:41the motion regardless of how fast and slow then  any amount of unbalanced force can do the job even
82. 10:49a tiny push will work for example in deep space  far from the gravity and air resistance any
83. 10:57object no matter how heavy will start the moving  if you apply any Force at all even something as
84. 11:06small as 0.001 Newton and once it starts moving it  will keep moving in the same direction forever
85. 11:16that's the Newton first Law a body at rest stay at  rest and a body in motion keep moving in Motion in
86. 11:24a straight line unless acted by an unbalanced  force even on the Earth surface any amount of
87. 11:32unbalanced force can start the motion but here is  a catch you also need to overcome the frictional
88. 11:39resistance if the applied force is less or equal  to the resistance force nothing will happen and
89. 11:47the object will stay will not start moving  but applying even a little extra Force then
90. 11:54the frictional force the object will start moving  but if you want to achieve a specific velocity
91. 12:03in a specific time then the required Force would  depends on two things the first is mass of object
92. 12:11and second is how quickly you want to accelerate  its Newton Second Law of Motion force equal to mass * acceleration F = m.a confusing isn't it so  let's understand this both things with an example
93. 12:30imagine we have 20 kg fixture mounted on a  slider and the slider have a coefficient of
94. 12:37friction 0.5 and we want to use a Pneumatic  cylinder to slide this fixture so how do we
95. 12:45calculate the required force to move it so first  we need to calculate the force need to overcome
96. 12:51the friction the frictional force equal to Mui mg  and for our case the frictional force would be
97. 12:5898.1 Newton now if pneumatic cylinder applies  exactly 98.1 Newton or less than 98.1 Newton the
98. 13:09fixture won't move because the frictional force  and the applied force cancel each other out so
99. 13:17to move the fixture we need to apply an unbalanced  force for example if we apply 100 Newton which
100. 13:25is just 1.9 Newton more than the frictional Force  the fixture will move but here is the interesting
101. 13:33physics the amount of this unbalanced force  determines the how quickly fixture will move for
102. 13:42example let's say the slider length is 1 Meter and  you want the fixture to cover this distance in
103. 13:48Just 2 second so the fixture initial velocity is 0  m/s because our fixture is at rest and the distance
104. 13:59is 1 Meter and the time is 2 second so to calculate  the required acceleration we can use the linear
105. 14:07motion equation s = ut + 1/2 a² put the values  so, acceleration is 0.5 m/s square and now
106. 14:19we can use Newton Second Law formula f = ma the  mass of fixture is 20 kg and the acceleration is
107. 14:270.5 m/s Square so F = 10 Newton so this is a  unbalanced force required to achieve the desired
108. 14:40acceleration of 0.5 m/s squared so that fixture  could cover 1 Meter within 2 second and remember in
109. 14:50this calculation there is no friction no gravity  it's a purely based on the mass and acceleration
110. 15:00the greater the mass or the faster you want to  accelerate the more Force you will need so the
111. 15:07total Force required to move this fixture equal  to the force to overcome the friction 98.1 Newton
112. 15:16plus, the unbalanced force to achieve the desired  acceleration 10 Newton so in total we will need
113. 15:23108.1 Newton to move this fixture 1 m in 2 second  now whether the pneumatic cylinder can actually
114. 15:33provide this Force quickly enough that's depends  on other factors like air flow rate but that's the
115. 15:40different problem so in summary the inertia is a  resistance to change in motion and it depends on
116. 15:49the mass of the object greater the mass greater  the inertia and the required Force to change in
117. 15:56inertial state rest to motion or motion to rest  depends on how quickly you want to accelerate the
118. 16:04mass now what about movement of inertia the  rotational inertia so as the inertial mass
119. 16:14is related to linear motion moment of inertia is  related to rotational motion when an object rotate
120. 16:23around their own internal axis like a spinning of  something like this Rod it's rotate around their
121. 16:31own axis internal axis can be in this way also  it's also an internal axis within the body so this
122. 16:41rotation has also moment of inertia but the moment  of inertia around this axis is more as compared to
123. 16:50this axis which we will understand later also the  axis doesn't have to be necessary fixed the axis
124. 16:59can move itself when the object is rotating  like this also if you're rotating something
125. 17:06only to a specific angle there will be also  moment of inertia and the moment of inertia
126. 17:12plays the same role in rotational motion as mass  touch in linear motion just like Mass resist the
127. 17:20change in linear motion moment of inertia resist  the change in rotational motion but here is the
128. 17:28key difference the moment of inertia doesn't just  depends on amount of mass it also depends on the
129. 17:36how the mass is distributed related to the axis  of rotation if the mass is further from the axis
130. 17:44the moment of inertia would be higher means will  more resistance the change in rotation and if the
131. 17:51same amount of mass is somehow closer to the axis  of rotation then moment of inertia would be lower
132. 17:59means we'll resist less to change in rotation  now in linear motion if we want to change the
133. 18:08inertial state of a point Mass we apply a force  in the direction we want the motion to occur in
134. 18:16a straight line and the equation we use is f =  ma but what if we connect this point Mass to a
135. 18:27pivot point with a zero Mass rigid connection and  if we want to rotate this Mass around this pivot
136. 18:35point in the same plane then how much torque will  need to apply around the pivot point well we know
137. 18:44the relation between force and torque equal to  force into radius R now what's force will still
138. 18:53ma right but in this case as we are rotating  we need to replace the linear acceleration with
139. 19:02angular acceleration and we also know the linear  acceleration equal to R angular acceleration
140. 19:09Alpha if you forget linear velocity equal to R  angular velocity linear displacement equal to
141. 19:17R angular displacement so the torque T equal to m  R Alpha into R which simplified to mr² alpha now
142. 19:29if we compare this torque equation to the linear  equation force F = ma so for Force we have torque for
143. 19:38linear acceleration we have angular acceleration  and for Mass we have M r² this M r Square is
144. 19:47actually moment of inertia it's denoted as I in  most classical physics equation but in engineering
145. 19:55we often use J to represent the moment of inertia  but this is exactly the same the rotational
146. 20:02inertia moment of inertia so this was all about  the moment of inertia and their relation with torque
147. 20:08torque equal to I moment of inertia into Alpha  angular acceleration the more moment of inertia
148. 20:17an object has the more torque will require to  accelerate it and the faster you want to rotate it
149. 20:25the more torque will be required if you are wondering  if I = to Mr² then why different shapes have
150. 20:36different moment of inertia formula why not just  mr² so let me break down this a bit so first of
151. 20:46the moment of inertia is an extensive property an additive property so if there is too Mass instead
152. 20:53of one then the moment of inertia will be simple to  2 mr² if there is one more other mass at half of
153. 21:02the radius then it will become Mr² + m r / by 2  square now if suppose this is a rigid body like
154. 21:12a dis or cylinder then there is an infinite Mass  infinite Point Mass so instead of adding them we
155. 21:23integrate to find the total moment of inertia and  here I don't want to dive into math of integration
156. 21:31but the point I want you to understand is the  I = Mr² represents the moment of inertia of
157. 21:39a point mass at a distance R from the rotational  axis but in real world we often work with rigid
158. 21:47bodies like cylinder disc or any other shape so  to find the moment of inertia of these shapes we
159. 21:55perform Integrations and the results depends on  the shape of the object and the position of the
160. 22:01axis of rotation so I will leave these formula  in video description for your reference so this
161. 22:09was all about the theory of moment of inertia and  now let's understand how this theory applies in
162. 22:16motor sizing calculation so we will do the two  calculation of an index motor sizing for belt
163. 22:24conveyor and the servo motor sizing for indexing  table so let's first understand the calculation of
164. 22:33an induction motor for a belt conveyor and then  we will understand the sizing of Servo motor for
165. 22:40indexing table so let's say we have a 3 m long  belt conveyor with ruler of 200 mm diameter and
166. 22:50the load of our belt is 80 kg and the mass of  the belt itself is 20 kg now what would be the
167. 22:59required torque of motor to drive this belt  conveyor so we need to calculate the required
168. 23:06torque for two components one is pulling Force  required to overcome the friction between the Belt
169. 23:14and the Belt support and second is required torque  to overcome the moment of inertia of the whole
170. 23:22system so first let's calculate the pulling  force so the total load is 80 kg of material
171. 23:33plus 20 kg of Belt so 100 kg in total now let's  assume a solid support below the belt and the
172. 23:42coefficient of friction between the belt and  support is let's take it at 0.2 so the required
173. 23:49pulling force would be equal to frictional  resistance which is Mu mg put the values
174. 23:582 Newton so what now so think of this pulling  Force as an attached to the roller with a string
175. 24:08so when roller will rotate this pull will  act like a tangential Force something like
176. 24:15this so to find the required torque we need  to multiply this pulling Force by the radius
177. 24:23of driving roller so the roller has a 200 mm  diameter so the radius would be 100 mm 0.1 M
178. 24:33so the required torque would be 1962 Newt M but  we are not done yet we also need to calculate the
179. 24:43required torque to accelerate the whole system  from rest to motion and this depends on moment
180. 24:50of inertia of the system the torque equal to  moment of inertia into angular acceleration so
181. 24:57let's first calculate the moment of inertia so  the total moment of inertia includes the moment
182. 25:03of inertia of rulers and moment of inertia of load  over conveyor including the conveyor belt yes the
183. 25:12load and the belt is also at the rest so we need  the torque to accelerate them too so for rulers
184. 25:20the moment of inertia around their rotational  axis would be half m² so moment of inertia of
185. 25:28each roller would be 0.075 k m square so for both  roller it's 0.15 k m Square now for the load and
186. 25:39the belt the total moving load is 100 kg but it's  move linearly so most of us doesn't consider this
187. 25:49as a moment of inertia but we have to so we will  have to convert this in a equivalent moment of
188. 25:57inertia so imagine this load as a single point  mass at the roller circumference because the belt
189. 26:04and the load has to be pulled over the belt  circumference so the moment of inertia of a
190. 26:11point mass at distance R is equal to Mr square  so for 100 kg at 0.1 M radius it would be 1 kg
191. 26:21M Square so the total moment of inertia equal to  0.15 + 1 1 15 km square and now we comes to the
192. 26:33tricky part deciding the acceleration P motor more  acceleration means more torque would be required
193. 26:41and if we could start slowly the less torque would  be required so to decide the acceleration first we
194. 26:49need to understand that how an induction Motors  actually works so mostly we use a synchronous
195. 26:56induction motor means there is always a slightly  difference or slip between the rotational speed
196. 27:04of the electromagnetic field in a stator and the  router of the motor under no load condition this
197. 27:13slip is typically about 1% however when we apply  a load the slip increase for a standard motor
198. 27:21this slip should not be exceed 5% as the higher  sleep can cause overheating and reduce the motor
199. 27:29lifespan some special design motor allow sleep up  to 8% but for most of a standard Motors a staying
200. 27:37between 1% to 5% range is crucial now let's  consider a four pole induction motor operating
201. 27:47at 50 HZ for this conveyor so the synchronous  speed a four pole motor would be500 RPM I have
202. 27:57already explained synch speed of motor in this  video for two pole speed would be 3,000 RPM for
203. 28:04four pole speed would be 1500 RPM so if we want  an Optimum motor sizing we can consider a slip
204. 28:13of 5% at full load so the speed of motor at full  load would be 1,500 minus 5% of 1500 1425 RPM so
205. 28:27we have the final speed of motor at full load and  we have also initial speed 0 RPM yes we sizing the
206. 28:36motor to start the system from the rest now when  we will start this induction motor it will begin
207. 28:45at 0 RPM which means the slip will be initially  100% And as the motor accelerate the sleep will
208. 28:55gradually reduce to around 5% under the full load  operating condition so during this starting phase
209. 29:03the motor drawn a very high current 5 to seven  times more than the normal operating speed so if
210. 29:10this starting time lasts too long the motor can  overheat and potentially burn out so to avoid
211. 29:18this the starting time of induction motor must be  controlled based on the method we use for Direct
212. 29:26online do starting the acceleration time should be  between 1 to 5 Second for Star Delta starting it
213. 29:34should be range from 3 to 10 second with a soft  starter it can extend to 15 second and using a
214. 29:42VFD the starting time can be programmed to As Long  As 30 second so for optimal Motor Performance our
215. 29:51goal should be to achieve an acceleration time of  around 5 Second so we have the acceleration Time 5
216. 30:00Second so now we can calculate the acceleration of  the motor but first we need to convert the final
217. 30:08speed 1425 RPM to the radian per second so 1 RPM  equal to 0.467 radian per second so 1425 RPM 1
218. 30:2249.1 5 radian per second if you are wondering how  the one are PM equal to 10467 radian per second
219. 30:31so 1 RPM means in 1 minute in 60 second one full  Revolution 360° means 2 pi radian so in 1 second
220. 30:422 pi / 60 01467 radian per second very simple so  the angular acceleration Omega divided by T = to
221. 30:58put this values in the torque equation it would  be 34.3 Newton M so the total required torque to
222. 31:06run this conveyor with full load equal to 53.92 n  m this means the motors should provide 53.92 new
223. 31:19M or more but if you're using a gearbox let's  say 1 is to 10 ratio then the motor required
224. 31:27torque would be reduced to just 5.3 Newt M so  I hope you understood that how important is
225. 31:34to calculate the moment of inertia and choosing  the right acceleration for motor sizing now let's
226. 31:43understand the servo motor sizing for an indexing  table so let's consider an indexing application
227. 31:50an index surround plate of 1 M diameter  mass 100 kg and having two fixture and it
228. 31:58mass is 20 kg the plate is supported by a cross  roller bearing so the outer diameter of the
229. 32:06bearing is 200 mm and the cycle time to rotate  this plate from one station to another by 180°
230. 32:15is 2 second and we have to use a Servo motor to  Index this whole system so how much torque does
231. 32:23servo motor need to perform this task interesting  right so the total required torque of the servo
232. 32:31motor need to be sum of two part the first is  frictional torque to overcome the resistance in
233. 32:38bearing and the second is acceleration torque to  get the moment of inertia of system to accelerate
234. 32:45so let's tackle the frictional torque first so  let's assume the coefficient of friction of the
235. 32:51bearing is 0.05 this is ideal for bearings and  the total mass of the system is 100 kg mass of
236. 32:59the plate plus 40 kg mass of the both fixes 140  kg so the frictional force F = to Mu mg 68.6 7
237. 33:11Newton and the radius of bearing is 0.1 M half  of 200 mm so the frictional torque to overcome
238. 33:21the resistance would be torque equal to F into R  68.6 7 into 0.1 6. 867 n m so the torque needed to
239. 33:33overcome the friction is 6867 n m next we have to  calculate the torque required to accelerate this
240. 33:42whole system the torque equal to moment of inertia  into angular acceleration so the moment of inertia
241. 33:49of this whole system would be moment of inertia of  plate half m² 12.5 kg m square plus the moment of
242. 34:00inertia of each fixture 5 r M square so the total  moment of inertia of system would be 22.5 k m
243. 34:10square now in this equation we need to calculate  angular acceleration but how so here we have to
244. 34:19understand a very important thing that motion  doesn't start at full speed it's always begin
245. 34:27at zero velocity increase to the desire speed  and then de accelerate you cannot instantly
246. 34:36jump to any desired speed in 0 second there is  always a some sometimes to accelerate and to De
247. 34:45accelerate even for a fraction of second so now  in this case we got 2 second to make this whole
248. 34:55thing happens so we can divide this time into  like 0.5 second for acceleration 0.5 second for
249. 35:03de acceleration and 1 second for uniform motion  in between but if you want more smooth start and
250. 35:11stop we can go for 0.75 second for acceleration  and de acceleration and 0.5 second for uniform
251. 35:20motion in between actually this division is up  to you up to the programmer and the thing is
252. 35:26you just have to give the more time or we can say  enough time for acceleration and de acceleration
253. 35:33as much it's possible I hope you got it so we have  the total angular displacement 180° by radian so
254. 35:42we can use the angular displacement equation  for each stage angular displacement equal to
255. 35:49half initial angular velocity plus final angular  velocity into time so for angular displacement at
256. 35:57acceleration since the initial angular velocity is  zero angular displacement at acceleration equal to
257. 36:06half Omega Max T 0.75 same for the D acceleration  and for uniform motion Omega Max is constant means
258. 36:16initial velocity and final velocity is equal  so angular displacement at constant velocity
259. 36:24will be Omega Max into time 0 .5 and we can  sum up this all so pi = to 1.25 Omega Max so
260. 36:36maximum angular velocity Omega Max equal to 2.51  radian per second so the angular velocity at the
261. 36:46end of the acceleration phase is 2.51 radian  per second so the angular acceleration Alpha
262. 36:54can be calculated as angular Max 2.51 divided by  time taken 0.75 for acceleration 3.35 radian per
263. 37:05second Square so the needed torque to accelerate  and de accelerate this whole indexing system would
264. 37:14be the moment of inertia of whole system 22.5 kg  M square into angular acceleration 3.35 radian per
265. 37:24squared 75.3 7 Newton M and the total torque 6.86  the frictional resistance torque plus 7537 82.2
266. 37:373 n m so this is how we sizing and servo motor  and this is quite different from sizing of an
267. 37:45induction motor with servo motor we do not really  have to worry about the acceleration time in same
268. 37:52way the reason is simple the servo motor working  principle is different than and induction motor
269. 37:59the Sero motor is designed for the precise control  however it doesn't means that we can just decrease
270. 38:08the acceleration as much we want to rotate any  heavy system with large moment of inertia there
271. 38:16is actually a limitation when it comes to that how  much moment of inertia a Sero motor can handle and
272. 38:23we can find this limitation in CMO motor catalog  from the manufacturer let me give you an example
273. 38:31for a 400 wat Sero motor the catalog specified  the allowable moment of inertia 10 times of motor
274. 38:39moment of inertia and here the motor's moment  of inertia is 0.09 into 10 upon -4 k m square
275. 38:48so the allowable moment of inertia for this motor  would be 10 times that value which equal to 0.409
276. 38:57into 10 upon -3 K m² so once we calculated  the required torque for the system don't
277. 39:05just select the motor only based on torque and  rated speed of Sero motor we will also need to
278. 39:12verify the allowable moment of inertia for the  motor and make sure the selected motor moment
279. 39:19of inertia should be greater than the system M of  inertia or else the motor wouldn't work properly
280. 39:27so this is it for this video I hope  this will help you in sizing up motor
281. 39:31for almost any rotary application and new fresh  machine design series are coming very soon so
282. 39:38please stay tuned and thank you so much for  watching and I will see you in my next one.