CHAPTER 1part 2 Rotational motion — Transcript

Full transcript

1. 0:01asalamu alikum and very good days I'm
2. 0:04mad Zu as we continue the chapter one
3. 0:07part two uh that we cover the topic of
4. 0:10talk The Continuous um relationship
5. 0:13between the uh talk and also the angular
6. 0:17expirations and the others rotational
7. 0:20motions U parameters such as the
8. 0:24momentums and also the
9. 0:27energy okay the lesson content we'll be
10. 0:30cover about the rotational motions okay
11. 0:33the definitions of moment or angular
12. 0:36momentum and each conservations the
13. 0:38relationship between the top and also
14. 0:40the angular acceleration is before this
15. 0:43I mention that isal to top i al right
16. 0:47then we go for the rotational K energy
17. 0:50okay then also the lastly the
18. 0:52conservation of energy in trans and also
19. 0:55the rotational
20. 0:58motions
21. 1:02do you have any experience of playing a
22. 1:04bowling
23. 1:05ball I'm sure that you are uh knows that
24. 1:10although okay uh their bowling ball have
25. 1:13the same size but their weight is
26. 1:16difference okay so you need to choose
27. 1:19okay one that fits to yourself is it
28. 1:26right okay before this in the bowling
29. 1:29ball there are difference in MTH all
30. 1:32right although they are the same we uh
31. 1:34same size so let's we look uh this
32. 1:37videos how the uh the other factors that
33. 1:40also influence the Motions of the
34. 1:45rotational okay hi this is r and I have
35. 1:49two girls here they're going to help me
36. 1:50with a demo on moment of inertia what do
37. 1:53you think about that Co so which of you
38. 1:56two girls is stronger me are you sure
39. 1:59sure mhm okay we're going to do a thing
40. 2:02I have I put these together
41. 2:06already okay here I have uh two sticks
42. 2:10this one's got two juice boxes on the
43. 2:13end of the stick same two juice boxes in
44. 2:15the
45. 2:16middle so can you girls tell me which
46. 2:18one do you think is heavier okay that
47. 2:21one why do you think this one's heavier
48. 2:23because they're both in the middle and
49. 2:25um both put together make um heavier way
50. 2:27you think so what about you think the
51. 2:31same they're the same why do you think
52. 2:33they're the same because they're both
53. 2:36white they're both a stick and they both
54. 2:38have two juice boxes so they're both the
55. 2:40same stuff right so they're the same yes
56. 2:44so what we're going to do is we're going
57. 2:44to play a game to see who stronger and
58. 2:46what you're going to do is you're going
59. 2:47to take the stick and hold in the middle
60. 2:49and we're going to see you can turn it
61. 2:51back and forth 10 times as quick as you
62. 2:53can with just one hand does that sound
63. 2:55like a fun game okay you hold that one
64. 2:57in the Middle with one hand one hand
65. 3:00hold that one in the middle one hand
66. 3:02okay on your and you're going to one
67. 3:03handed on your it's going to be 10 times
68. 3:06this way one two three four like that on
69. 3:09your mark get set go 1 2 3 4 5 6 7 8
70. 3:13nine 10 she won I you get 10 no you you
71. 3:17didn't do 10 you didn't do as fast I won
72. 3:19okay now
73. 3:21switch now hold in the middle guys two
74. 3:24again your mark Get Set Let's Go 1 2 3 4
75. 3:285 6 7 8 9 10 so which one's easier this
76. 3:32one that one okay so this I couldn't do
77. 3:34it my wrist hurt red but you did a
78. 3:36really good job so this is an example of
79. 3:38two things with the same two things with
80. 3:40the same mass but different Mass
81. 3:43distribution this one has masses at the
82. 3:44ends and this one has masses
83. 3:47closer right uh as you can see here okay
84. 3:51uh the rod here have the same length and
85. 3:54they are attached with two Mass okay one
86. 3:57rod is attached their Mass uh near to
87. 4:00the center what the other R the mass is
88. 4:03located at the end of the the mass is
89. 4:06located at the end of the b as you can
90. 4:09see that which one will be rotate
91. 4:13fastest okay from the videos that you
92. 4:17see before this okay the mass located at
93. 4:20the center rotate fastest compared to
94. 4:22the mass that are located at the end of
95. 4:25The Rock okay this is okay shows that
96. 4:29besides Mass there are will be another
97. 4:32factors that will be influenc the
98. 4:42rotations okay so you can see here if
99. 4:45the uh okay because of uh the mass
100. 4:51located at the center compared to the
101. 4:55uh and they have the different tendency
102. 4:58toate
103. 5:00uh so as you can see the factors all
104. 5:03right that will be uh influence the
105. 5:07rotation the first one is mass then
106. 5:11continue to the
107. 5:13distributions
108. 5:16uh of mass okay sometimes okay one
109. 5:20another factors that will be uh
110. 5:23influence the rotational is the exis of
111. 5:28rotation
112. 5:31okay it's very hard okay to rotate an
113. 5:35object that are Center of rotation at
114. 5:37the middle compare if you try to rotate
115. 5:40that object at the end of the
116. 5:44uh uh at the End of the Age of the
117. 5:49object so another factors that also
118. 5:53contribute uh the rotational is the
119. 5:55exist of
120. 5:58rotations
121. 6:00this phenomena we call as a moment of
122. 6:03initial so moment of initial can be
123. 6:06defined as the tendency of of a body to
124. 6:09resist the angular
125. 6:11escalations right so the moment of
126. 6:14initial the symbol will be here is I is
127. 6:16a summ of the mass and also times with
128. 6:20the uh uh AIS of rotations Square okay
129. 6:25so this is because the moment of initial
130. 6:28is influenced by the factors of mass
131. 6:30distribution of mass and also AIS of
132. 6:34rotation in physics the rotation initial
133. 6:38is a product of mass times radius
134. 6:41squares is the rotational initial for a
135. 6:43mass of for the point Mass where R is
136. 6:46measured from the AIS of
137. 6:49rotations so for example here okay is
138. 6:52the AIS of rotation is 1 M compare here
139. 6:55is 2 m as you can see here all right the
140. 6:59mag ude of the moment of for 1 M
141. 7:02compared to 2 m okay is less compared to
142. 7:06the 2 m okay the 2 m um of AIS rotation
143. 7:13produce a moment of measure 4 kg M squ
144. 7:17so it means that all right
145. 7:20uh the larger the of rotation okay the
146. 7:25harder to rotate the
147. 7:28object this slide show how to calculate
148. 7:31the moment of initial so we have a three
149. 7:35different conditions here the first is a
150. 7:37single particle a many particle or
151. 7:40continuous distributions of math or
152. 7:42extended object for single particle
153. 7:46right the moment of initia is equal to
154. 7:48m² but for many particles the total
155. 7:51moment of initial is a total of the
156. 7:55individual moment of initial for example
157. 7:57we have a blue dot red dot green purple
158. 8:02and another uh em gold emal colors so
159. 8:06the total of moment of initial here is
160. 8:09by the moment of initial of the blue
161. 8:12moment of plus with the moment of
162. 8:13initial of the red okay moment of of the
163. 8:16green and purple and so
164. 8:19on okay if we have the object okay that
165. 8:23have the um they have a continuous shape
166. 8:27just as the cylinders sphere and what
167. 8:30okay the moment of initial is a moment
168. 8:33of initial of their uh length right so
169. 8:37for this shap uh they are their own
170. 8:41formulas okay uh to calculate the moment
171. 8:46of
172. 8:47initial this SL shows the moment of
173. 8:50initial formulas for a certain object
174. 8:53okay for example the hoop cylinders the
175. 8:56moment of ini is equal to m r² the solid
176. 9:00cylinder okay the moment of isal to M
177. 9:04r² okay uh and then another one is a
178. 9:08sphere uh is equal to 2 5 m r² and the
179. 9:12other side of the M of the shape that
180. 9:15you that are already given their own
181. 9:18formula of the moment of
182. 9:26emissions let's we continue for example
183. 9:29number one calculate the moment of
184. 9:31initial for a solid sphere with a mass
185. 9:33of 10 kg and R 0.2 so the solid sphere
186. 9:38right the moment of initial is equal to
187. 9:402 5 m² so 10 kg is referring to the mass
188. 9:44and the radius is referring to the 0.2 M
189. 9:48so by substitute the values all right 2
190. 9:525 * 10 and time 0.2 s can you can obtain
191. 9:56the moment of initial equal to 0.16 kg m
192. 10:00s now let's we look for example number
193. 10:03two Cate nor of me of
194. 10:07aere with mass of 10 kg and radius 0.2
195. 10:11so for theere the formula for the moment
196. 10:13of meure I = 23 m² by substitute the
197. 10:18mass 10 kg and radius 0.2 M then you
198. 10:22will obtain the I 0.27 kg
199. 10:26m² okay two small object of mass 5 kg
200. 10:30and 7 kg are Mount 4 M apart on a light
201. 10:33Rod which mean that you have uh here
202. 10:36state that here NE massan that no mass
203. 10:40for the road okay Cate the moment of me
204. 10:43of the system a when the r rotate
205. 10:46rotated about an exist halfway between
206. 10:49the masses that okay from the object
207. 10:52number one and the object number two 2 m
208. 10:54from object number one and also 2 m from
209. 10:57the object number two and then
210. 11:00when the Rotator above an AIS 0.5 m to
211. 11:03the left of the
212. 11:050.5 0 mass of 5 kg okay here is a mass
213. 11:11of 5 kilg so the r is 05 M and for the
214. 11:17object number two which is 7
215. 11:19kg will be equal to 4.5
216. 11:26M since that we have two object okay
217. 11:30so we need to calate the individual of
218. 11:33moment of initial and sum it for the
219. 11:36when the r is located at the center of
220. 11:39the object so means that this is object
221. 11:42number 5 kg here is object 7 kg okay so
222. 11:47here right as you can see the distance
223. 11:49from the AIS of rotation both of the
224. 11:52masses have the same distance so that's
225. 11:54why here uh the M1 R1 squ is 5 * a 25 2
226. 12:01and plus with the M2 R2 2 which is 7 * 2
227. 12:062 and your answer will be equal to 48 kg
228. 12:09M squ okay however when there exist uh
229. 12:14rotations okay uh at the end nearly end
230. 12:18of the 5 kg object so okay you can see
231. 12:22here uh the M1 R1 s+ M2 Mr R2 s means
232. 12:27that the difference is uh uh from the
233. 12:30previous uh questions is about the
234. 12:34locations with the exis of rotations so
235. 12:36for the object one is okay the M1 R1 squ
236. 12:40becomes 5 * 0.5 squ while the object two
237. 12:457 kg is equal to 7 * 4.5 s so their
238. 12:50moment of initi is 43.3 kg M squ so as
239. 12:54you can see that all right uh as I State
240. 12:57before okay when we change their uh AIS
241. 13:01of rotations right theis if the the
242. 13:04moment of me is larger for the uh exis
243. 13:09rotation located at the end of the
244. 13:12object compared to the center means that
245. 13:14is very harder to rotate the object when
246. 13:18okay we try to rotate at the end of
247. 13:20the uh at the end of the object compar
248. 13:24at the center to the
249. 13:28object
250. 13:29right so conclude here okay how the
251. 13:33moment of initial relates to the new
252. 13:35second of motions right so before this
253. 13:38uh for uh linear motions right the F
254. 13:41will be equal to m a t but okay uh when
255. 13:47the tangential acceleration you are uh
256. 13:50uh you are you are putting okay the
257. 13:53equation e t isal r Alpha right so
258. 13:57therefore f = m al so top means that you
259. 14:02need to multip it with r and then here
260. 14:05you also multip with r so what you can
261. 14:08obtain here the top will be equal to m
262. 14:12r² alpha okay means that the m² is
263. 14:16referring to the r and Alpha referring
264. 14:19to the
265. 14:20angular
266. 14:22um accelerations so what we can conclude
267. 14:27here the r is actually is uh uh actually
268. 14:31is reflect all right uh the value of the
269. 14:36mass in a linear motions while Alpha is
270. 14:40reflex with the tangential acceleration
271. 14:43for the linear
272. 14:46motions so you can obtain that the
273. 14:48angular accelerations the relationship
274. 14:51angular acceleration with the top and
275. 14:52moment ofure alpal to to R so what we
276. 14:57can see here when you apply a to to a
277. 14:59wheel be spin in the directions of the
278. 15:03top the greater the dot the greater the
279. 15:05angular accelerations the greater the
280. 15:08moment of meure the smaller the angular
281. 15:10acceleration and so
282. 15:16on as we look for example number four
283. 15:19what is the angular acceleration
284. 15:21experien by a uniform solid of Mass 2 kg
285. 15:25and radius 0.1 M when the net of 10 new
286. 15:30is applied so 10 is the value of top
287. 15:33assume me the this spin above Center
288. 15:37okay so toal I Alp with the i = m r²
289. 15:41alpha okay so therefore the alpha will
290. 15:44Beal 2 top m² okay by
291. 15:49substitute the top = 10 the mass = 2 K
292. 15:53and R = 0.1 s then you be obtain okay
293. 15:56the alpha = 1,00 radi second
294. 16:00squ around a of a playing uh playing
295. 16:04ground with a moment of initial 100 kg M
296. 16:07squ start at rest and is accelerated by
297. 16:10a force of 150 at the radius of 1 M from
298. 16:14Center if this force is appli at angle
299. 16:17of 90° from the line of action of the
300. 16:20same time for a time of 0.5 second what
301. 16:24is the rotational velocity of the round
302. 16:26about okay as you can see here all right
303. 16:30uh okay the Omega not is zero because it
304. 16:33start from rest the Omega you need to
305. 16:35find right and then the data Theta is
306. 16:38also unit to find and Alpha is also unit
307. 16:42to find and time taken is 0.5 seconds
308. 16:46right so they are given to you the value
309. 16:48of moment of initi so the relationship
310. 16:52all right and then uh their thought okay
311. 16:56their force is 150 NT so to net okay is
312. 17:00equal to i al to
313. 17:03netal = i
314. 17:07al okay so therefore the alpha will be
315. 17:11equal to f r/ i so here you are
316. 17:15replacing the value of F =
317. 17:18150 the value of R is 1 M and there R is
318. 17:22= 100 kg per M squ so therefore you
319. 17:26obtain the value of alpha = 1.5 gradi
320. 17:29seconds so once you get the value alha
321. 17:31you can determine the value of uh
322. 17:33rotational velocity by using the
323. 17:35equation of Omega = Omega Alpha T So
324. 17:40before this Omega is 0 and Alpha is 1.5
325. 17:44and T is 0.5 then you can obtain the
326. 17:47value of the uh rotational velocity
327. 17:50equal to
328. 17:510.75 ran per seconds the angular
329. 17:55momentum of Aid object is defined as the
330. 17:57product of the moment of initial I and
331. 17:59the angular velocity Omega
332. 18:02right so before this moment inine motion
333. 18:05isal to MV right so as I stated before
334. 18:10for the rotational motions the analogies
335. 18:13of the mass okay for the rotation is
336. 18:16becomes the I and then the V is using
337. 18:20the Omega so before this the linear
338. 18:23momentum we using a symbol p right now
339. 18:25for the angular momentum we using a
340. 18:27symbol of L okay as I mentioned is
341. 18:31analogist to a linear momentum and is
342. 18:33subjected to the fundamental constraint
343. 18:35of the conservation of angular moment
344. 18:37princi if there is no external thought
345. 18:40on the
346. 18:42object let's we watch the video number
347. 18:45two
348. 18:48here this video related with the
349. 18:50conservation of angular moment we can
350. 18:51demonstrate the conservation of angular
351. 18:53momentum the same way a figure skater
352. 18:55does by going into a tight
353. 18:57spin so I'll
354. 19:00[Music]
355. 19:03use we can demonstrate the conservation
356. 19:05of angular momentum the same way a
357. 19:07figure skater does by going into a tight
358. 19:09spin I'm not a very good skater so I'll
359. 19:12use this turntable and a pair of hammers
360. 19:15I'll hold them out at arms length so I
361. 19:17get nice large radius and I'll set
362. 19:19myself in a slow spin I'm spinning
363. 19:22around at a constant speed and then I
364. 19:24draw the Hammers close to my body and I
365. 19:28spin around very very rapidly okay this
366. 19:30video uh shows to us that the Rel the
367. 19:33meaning the definitions of the
368. 19:36conservation of the angular momentum
369. 19:38okay if you pull uh uh out your hands
370. 19:43okay you can see that the motion will be
371. 19:45slow however if you close down your
372. 19:47hands all right the motion will be
373. 19:50faster okay this is because uh this
374. 19:53related with the conservations of a
375. 19:55angular
376. 19:57momentum
377. 19:59so therefore the angular momentum is
378. 20:02equal to moment of initial times the
379. 20:04angular velocities right this is the
380. 20:06formula of it which is uh l = i Omega so
381. 20:12I = m² Omega and the unit will be in
382. 20:15kilog m per
383. 20:19seconds in the absent of the external
384. 20:21top angular momentum is we it will be
385. 20:24conserve means that angular momentum
386. 20:26initial will be equal to angular
387. 20:28momentum
388. 20:34final Angelina spin on the rotating
389. 20:37pedestal with a angular velocity of 8
390. 20:40radi per seconds
391. 20:42B and exercise ball which increase her
392. 20:46momentum of initial from 2 kg M squ all
393. 20:50right to a
394. 20:512.5 kog S what is the Ang angular
395. 20:55velocity after catching the exercise
396. 20:57ball right so as mentions that the
397. 20:59momentum initial will be equal to
398. 21:01momentum final so here the momentum is
399. 21:05equal to M
400. 21:08r² Omega is a part of initial be equal
401. 21:12to Omega initial all right
402. 21:19um or we said that uh right I Omega
403. 21:25initial initial will be equal to I final
404. 21:29Omega final okay so I initial is 2 and
405. 21:33then Omega initial is uh
406. 21:368 okay equal to Omega final is 2.5 sorry
407. 21:41I fin is 2.5 and Omega final that you
408. 21:45need to find so Omega final equal to
409. 21:48right uh
410. 21:5116 uh 2.5 so that's why you get equal to
411. 21:556.4
412. 21:57seconds
413. 22:04let's we look for the rotational kinetic
414. 22:06energy what is energy as stated here the
415. 22:10energy is the ability to do work so
416. 22:13there are two types uh energy for the
417. 22:17mechanical energy which is a potential
418. 22:19energy and also ktic
419. 22:23energy before this the mechanical energy
420. 22:26is equal to U + K right the U is
421. 22:29potential energy which is the energy
422. 22:31that related with the positions while
423. 22:33the kinetic energy is the energy related
424. 22:36with the object in motions so motion of
425. 22:40an object there are three types of
426. 22:42motion of the object such as the
427. 22:45translational rotational and also
428. 22:47vibrational okay for translational or
429. 22:49linear okay the ktic energy equal to
430. 22:52half MV s but for the rotational ktic
431. 22:56energy is equal to half I Omega
432. 23:02s g r a bowling ball of mass 7 kg radius
433. 23:0710.9 CM down a lane with a velocity of 6
434. 23:11M find the rotational energy of the body
435. 23:14ball assuming it does not sleeps right
436. 23:18so here is given to you the value of the
437. 23:21mass the value of the r here the value
438. 23:24of
439. 23:25p and the question ask you to calculate
440. 23:29the rotational ktic energy so as you
441. 23:32know that the rotational ktic energy is
442. 23:35equal to half I Omega s okay so the
443. 23:42types of volume B is a sphere right so
444. 23:45we use we need to use the formula of the
445. 23:48moment of me of sphere which is equal to
446. 23:502 5 m² okay but substitute the mass = 7
447. 23:54and R = 10.9 CM you have to to that and
448. 24:00becomes
449. 24:010109 M and then can calate the value of
450. 24:05moment of initi of the volume ball equal
451. 24:07to
452. 24:080.333
453. 24:10kg how about the Omega Omega we can find
454. 24:14from the relation of v r Omega so Omega
455. 24:18Beal V R so means that V is 6 R which is
456. 24:230109 then you can get Omega = 55 gradian
457. 24:27seconds so this
458. 24:29thetic energy of the volum ballal half I
459. 24:33Omega which isal half *
460. 24:360.033 and time 55 and isal to
461. 24:4250.4 okay find the total kinetic energy
462. 24:45of the volum ball from a previous
463. 24:47problems so before this we are calculate
464. 24:50uh uh rotational kinetic energy so right
465. 24:54now we need okay to calculate the uh
466. 25:00transational ktic energy okay so the
467. 25:03total ktic energy is is a summation of
468. 25:06the transational kinetic energy plus
469. 25:09with the kinetic energy rational okay so
470. 25:12means that all right uh energy
471. 25:15transational half m² so half * 7 * 6 s
472. 25:21and then plus 50.4 then you can get the
473. 25:24energy becomes
474. 25:26176 the law of conservation of energy
475. 25:29states that the total amount of energy
476. 25:32in an isolated system remain constant
477. 25:35okay consequence of this law is that the
478. 25:39energy cannot be created or destroyed
479. 25:42okay so means that the conservation of
480. 25:44energy uh energy initial be equal to
481. 25:47energy final so the energy initial
482. 25:49consist of the potential energy initial
483. 25:52okay and also the kinetic energy so ktic
484. 25:55energy we have divided into two parts
485. 25:58which is the energy transation or plus
486. 26:01with the energy rotational for initial
487. 26:03part it become for initial part so equal
488. 26:06to potential energy final plustic energy
489. 26:10linear final plus rotational energy
490. 26:14final example 10 a 1.2 kg dis with a
491. 26:18radius of 10 cm row without sleeping the
492. 26:22line speed of the disc is = 1.41 m² okay
493. 26:27a find the trans
494. 26:29energy find the rotational energy and
495. 26:32lastly find the total energy okay for
496. 26:35the translation energy you can use the
497. 26:36formula half m² okay and isal to half *
498. 26:411.2 is mass = 1.2 and velocity 1.41 M
499. 26:47and you can obtain the translation enery
500. 26:501.19 for the rotational ktic energy okay
501. 26:54you can use equation of half I Omega squ
502. 26:59so the this okay the I is equal to half
503. 27:04m² okay the Omega is refer to Omega to
504. 27:09Omega R right so Omega = V R so here uh
505. 27:14you multiple with the V uh v² and r² so
506. 27:17you can obtain that is actually equal to
507. 27:201 4 * 1.2 *
508. 27:261.41 squ and is = to
509. 27:300.595 and then the total energy just
510. 27:33plus the energy transal plus rotation
511. 27:36energy which is
512. 27:381.19 plus with
513. 27:420.595 and is equal to
514. 27:461.79 example 11 a hook and a cylinder of
515. 27:49equal Mass roll down a ramp with height
516. 27:53H which has the greatest speed at the
517. 27:55bottom of the ramp so we have a two
518. 27:57types hoops and also the cylinder okay
519. 28:01both of these object they have equal in
520. 28:04mass and they sleep full down the
521. 28:07incline plane here so which one will be
522. 28:11uh the greatest speed at the bottom of
523. 28:12the ramp okay let's we go for the next
524. 28:15slide to look which one is have a a
525. 28:18greater speed right here okay so to
526. 28:23determine which uh object have a great
527. 28:26speed we need to uh using the
528. 28:28conservations of energy so the
529. 28:31conservation of energy is equal to
530. 28:34kinetic energy plus with the potential
531. 28:36energy for initial point is initial and
532. 28:38equal to energy final plus potential
533. 28:41energy final initially the object is
534. 28:43rest so that's why the ktic energy
535. 28:45initial will be equal to zero and at the
536. 28:48bottom one the potential energy is equal
537. 28:50to zero because the height at the bottom
538. 28:53is always equal to zeros so the left
539. 28:56here is a uh potential energy initial
540. 29:01and also equal to ktic energy final so
541. 29:04the potential energy initial isal to MGH
542. 29:07okay here is the height of the object
543. 29:10all right is equal to half m v² + half I
544. 29:14Omega squ so the hollow cylinder i = m
545. 29:20r² while we substitute the value of I to
546. 29:25the to the equations that it becomes
547. 29:28equal to m g h okay = to half m v² +
548. 29:33half M r² okay and then here Omega squ
549. 29:38okay you know that Omega V = to R Omega
550. 29:42so therefore Omega = to V / R okay so
551. 29:48here you can cancel out the
552. 29:51mass and then you can cancel out the R
553. 29:54so the equation becomes GH Plus uhal =
554. 29:58to v² + v² and the uh the V is equal to
555. 30:05root of GH this is for hollow cylinders
556. 30:09okay for the uh solid cylinders since
557. 30:13that there R = half m² that will produce
558. 30:17the V isal root of 43 g
559. 30:22h while if we change the object to a
560. 30:25solid sphere the I of the solid
561. 30:28is a 2 5 m r² okay by doing the same
562. 30:32principle okay but substitute the value
563. 30:35the equation of I okay then you solve
564. 30:38the the equation then you can get right
565. 30:41for velocity at the bottom equal toare
566. 30:45root of 10/ 7
567. 30:48G okay let's be look for the example
568. 30:51number 12 a bing ball that has an 11 cm
569. 30:55radius and a 7.2 kg mass is rolling
570. 30:58without sleeping at 2 m/ second on the
571. 31:01horizontal ball return it continues to
572. 31:03roll without sleeping up a hill to the
573. 31:06high H before momentary become rest and
574. 31:09then rolling back down the hill mod the
575. 31:12ball ball has a uniform
576. 31:15uh spere and calculate the value of H
577. 31:19right so we doing the principle of the
578. 31:22conservation of energy which mean that
579. 31:24energy initial equal to energy final so
580. 31:26in this case okay the final part as it
581. 31:30is at the up at the rolling ball here
582. 31:33below and then the the up one is a
583. 31:37initial part so UF final plus u KF final
584. 31:42equal to UI final plus K final so as I
585. 31:45mentioned before
586. 31:49um okay initially all right uh the at
587. 31:54the up one the velocity is zero so we
588. 31:57only have equal to MGH okay and then at
589. 32:01the bottom one all right at the bottom
590. 32:04the potential energy will be equal to0
591. 32:06so this equation becomes MGH H + 0 = 0 +
592. 32:11half
593. 32:12m² uh this is the ktic transational and
594. 32:17this one is half I uh Omega squ is a
595. 32:23ktic energy of
596. 32:25rotational okay substitute the value
597. 32:28okay since that we are using the sphere
598. 32:30so the I will be equal to 2 5 m r² and
599. 32:35then the Omega = v² r² so therefore we
600. 32:40can obtain all
601. 32:44right
602. 32:45uh we can obtain
603. 32:49uh the equation BEC 7 over 10 m
604. 32:55v² okay
605. 32:58me that here MGH will be equal to 7 / 10
606. 33:04m² okay so by substitute calculations
607. 33:09simplify the formulas and then H will
608. 33:12Beal to 7 v² 10 G so V is 2 and then G
609. 33:18is 9 9.8 so you can determine the value
610. 33:21of H = 0.29
611. 33:24M okay since the objects have a
612. 33:27difference of the moment of initial
613. 33:30formulas so we can see that the object
614. 33:34will be differently reach uh the bottom
615. 33:37at a difference time because okay the
616. 33:39influence of the moment of
617. 33:52this to conclude okay for rotational
618. 33:55motion the kinetic energy equal to
619. 33:58I Omega s the top is equal to I Alpha
620. 34:02which is top is a tendency to rotate the
621. 34:05object uh and then the angular momentum
622. 34:08to I Omega the work isal to top
623. 34:12time
624. 34:13angular and so on thank you