Rotational Dynamics - Basic Introduction - Free Educational videos for Students in K-12

Rotational Dynamics - Basic Introduction - By The Organic Chemistry Tutor

Transcript


00:00	a 10 kg block is attached to a 20 kg
00:04	pulley Which is a solid disk of Radius two m
00:09	. The block is released from rest , causing the
00:11	disk to spin . What is the angular acceleration of
00:16	the disk ? And how fast is the block fallen
00:19	after eight seconds . So how can we calculate the
00:24	angular acceleration of the disk ? What do we need
00:27	to do ? Well , let's identify the forces that
00:31	are in play here . So let's say that capital
00:35	M . Represents the mass of the disk and lower
00:38	case M . I'm not sure how that happened below
00:42	a case . Um represents the mass of the block
00:45	. Yeah . Now the block experience a downward weight
00:49	force of MG . And there's an upward tension force
00:54	that slows down the descent of the block and that
00:58	tension force also causes the disk to spin in the
01:03	clockwise direction . So with this information , how can
01:09	we use it to calculate the angle acceleration ? And
01:14	I'm going to describe T as being ft . So
01:17	that's the tension force . So I don't confuse that
01:21	with torque . So what we can do at this
01:27	point is right an expression for the network acting on
01:31	the disk . Now there's only one force acting on
01:34	a disk which is the tension force and torque is
01:37	force times the moment arm which in this case is
01:41	the radius of the circle . So the net torque
01:43	acting on the solid disc is the tension force multiplied
01:48	by the radius of the circle . Now let's focus
01:51	on the net force on a block . By the
01:54	way the disk is held in place so the disk
01:58	is allowed to spin but it can accelerate upward or
02:01	downward . Only the block can accelerate downward . So
02:05	we have to deal with the net force of the
02:07	block . So the net force in the y direction
02:13	of the block . It's based on these two forces
02:16	. So we have an upward tension force . So
02:19	it's going to be positive because it's directed in a
02:21	positive direction plus the downward wait for us which is
02:25	negative since it's going in a negative Y direction .
02:29	Now the net force , based on Newton's second law
02:32	, is mass times acceleration . So let's solve for
02:37	the tension force . Let's move this term to that
02:39	side . So M . A . Plus MG were
02:44	raised in the Y direction . That's going to be
02:47	equal to the tension force . Now , here's a
02:50	question for you , Emma , should it be positive
02:54	or negative ? Because the block is accelerated in the
02:59	negative Y direction is going down . We need to
03:01	put a negative in front of it . So we
03:04	could say that detention force is equal to positive MG
03:10	minus Mhm . So now at this point , what
03:15	I'm going to do is replace the tension force in
03:17	this equation with M . G minus M . E
03:22	. So the net torque acting on a disk ,
03:26	it's going to be mg minus M . A times
03:30	the radius . So now since I'm running out of
03:34	space , let's erase a few things now . The
03:40	acceleration or rather the linear acceleration is equal to the
03:44	angular acceleration times the radius . So let's replace A
03:48	with alpha times are so the network is now MG
03:54	minus M . Alpha times are times are . So
03:59	what I'm gonna do now is distribute our to these
04:01	two terms . And at the same time I'm going
04:04	to replace the network with inertia Times Alpha Based on
04:09	you in 2nd law for rotational motion . So that's
04:12	equal to mgr minus mm alpha times R squared .
04:20	Now my goal is to calculate the anglican celebration .
04:24	So I need to isolate this term . So what
04:27	I'm gonna do now is take this term . Move
04:29	it to the left side . So inertia times alpha
04:34	plus I am out for R squared is equal to
04:39	MGR Now this is the network acting on the disk
04:44	. So this is the inertia of the disc .
04:49	So right now let's factor out alpha . So what
04:53	we're going to have is alpha multiplied by the inertia
04:57	of the disc plus M R squared and that's equal
05:01	to MGR . So for this particular problem , the
05:06	angular acceleration can be calculated by taking the mass of
05:09	the block , multiplied by the gravitational acceleration times the
05:12	radius of the disk divided by the inertia of the
05:15	disk plus M R squared . Where lower case M
05:20	is the mass of the block . So now let's
05:22	go ahead and calculate alpha . So that's going to
05:28	be a massive 10 Times The gravitational acceleration of 9.8
05:33	. The radius of the disk is two m .
05:38	Now , I represents the inertia And the inertia of
05:42	a solid disc is 1/2 M . R squared ,
05:46	So the mass of the disk is 20 and the
05:51	radius is too , So half of 20 is 10
05:56	And 10 times two squared is 40 . So the
05:59	inertia of the disk is 40 Plus M . R
06:03	Squared , which is going to be 10 Times are
06:06	square , which is two square . So 10 times
06:10	9.8 is 98 , multiplied by two . So that's
06:14	196 , 10 times two squares 40 plus another 40
06:20	. That's gonna be 80 . So alpha is 196
06:25	divided by 80 . And so that's going to be
06:28	2.45 radiance per second squared . And so that's how
06:35	you can calculate the angle acceleration of the disc .
06:40	So that's how fast it's accelerated in this direction .
06:46	Now , what about heartbeat ? How Fast is the
06:50	block fallen after eight seconds . So how can we
06:53	find that answer ? Well , the first thing we
06:56	need to do is calculate the linear acceleration of the
06:59	disk . If we could find that , then we
07:02	could find out the acceleration of the block as it
07:04	falls down because they have to be the same because
07:07	they're attached to a rope and so they're going to
07:10	move at the same rate . The linear acceleration of
07:14	the disk is alpha , times are So we have
07:16	alpha , which is 2.45 And the radius of the
07:20	circle or the disk is two m , So 2.45
07:25	times two , It's 4.9 . So the linear acceleration
07:30	of the system is 4.9 m/s squared . And so
07:33	that's the acceleration of the block as it descends down
07:37	. So now we could find a final speed of
07:39	the block As it falls for eight seconds . So
07:44	the final speed is going to be the initial speed
07:47	plus 80 . The initial speed is zero because the
07:52	block was released from rest And the linear acceleration is
07:57	4.9 multiplied by eight . So 4.9 times eight Is
08:08	39.2 m/s . So that's the speed of the block
08:11	after eight seconds . Now , if we wanted to
08:14	find the velocity of the block , this would have
08:18	to be negative . The acceleration technically is negative for
08:21	the block because It's accelerating in the negative y direction
08:25	. And so the velocity is negative 39.2 m/s after
08:30	eight seconds . Now , let's say if you wanted
08:34	to find the distance , I wish to block .
08:37	So for these eight seconds , what would you do
08:43	? So once you have Alpha , and once you
08:46	have the final speed , you can easily find a
08:47	distance at which the object fell . So you can
08:51	use this equation , the final squared is equal to
08:54	the initial square Plus two A . D . So
08:58	you know the final speed is 39.2 . Once you
09:01	square , the negative sign doesn't matter . The initial
09:03	speed is zero And you have the linear acceleration ,
09:06	which is 4.9 . So now you can find D
09:09	the vertical displacement of the object . So it's 39.2
09:14	square Which is 1536.64 And two times 4.9 is 9.8
09:24	. So take 1536.64 And divided by 9.8 . And
09:30	so d is 156.8 m . So that's how far
09:37	the object fell . I mean in the block ,
09:40	that's how far I fell after eight seconds . So
09:44	, consider this problem . We have a five kg
09:47	mass attached to a 20 kg pulley , which is
09:51	a solid disc and let's say the radius of the
09:54	pulley is three m , And that's attached to a
09:59	hanging mass of 10 kg . Now , once this
10:03	system is released from rest , we know the system
10:06	is going to accelerate in this direction . How can
10:09	we calculate the acceleration of the entire system ? So
10:15	, if you want to try this problem , feel
10:18	free to do so and see if you can get
10:21	the right answer positivity if you want to try it
10:24	. So , I'm going to show you two ways
10:25	in which you get the answer . A quick and
10:28	simple way . And also a step by step procedure
10:31	if you want to write out all the equations .
10:34	So , here's the quick and simple way to find
10:36	acceleration based on you . In second law , the
10:39	acceleration of the system is the net force acting on
10:43	the system , divided by the total mass of the
10:46	system . So the only force that's really driving the
10:52	system to motion is really the weight force of this
10:55	block . So that way force is the net force
11:00	acting on the entire system . So let's call this
11:03	M . two . Let's call this M . one
11:07	and this is going to be capital . M .
11:09	The mass of the disk . So the net force
11:11	of the system is M two G . Let's put
11:16	G here and then we need to divided by the
11:20	total mass of the system . So that's the mass
11:23	of the first block , plus the mass of the
11:25	hanging block . And we can't just use this mass
11:30	. We need to take into account this equation .
11:35	The inertia of a disk is 1/2 M . R
11:38	squared . Now for different shapes this number could vary
11:46	, so I'm going to put a constancy so I
11:49	like to think of it as inertial mass . You
11:52	could call it something else if you want , but
11:54	we need that see value to be in this equation
11:56	for this to work . So that's how you can
11:59	find the acceleration of the system for this type of
12:03	system . So let's go ahead and calculate it .
12:07	So M two is 10 . G is 9.8 .
12:11	M one is 5 . M two is 10 .
12:15	See for a disc is 1/2 And the mass of
12:20	the solid disc is 20 kg . So 10 times
12:24	9.8 is 98 , half of 20 is 10 Plus
12:29	another 10 , that's 20 plus five . So this
12:33	is going to be 25 98 divided by 25 Is
12:43	3.92 m/s squared . So this is the acceleration of
12:49	the entire system . So let's see if we can
12:51	get that answer . Using another process . So how
13:05	should we do this ? We need to indicate other
13:09	forces that is acting on the system . So we
13:13	got the downward way force em to G . And
13:16	there's an upward tension force that slows down the descent
13:19	of this object . Let's call that T to detention
13:23	force acting on block two . Now we have T
13:26	. one Which pulls Black 1 to the right Now
13:32	these two tension forces are acting on the Pulley .
13:38	So if T two lifts up the object , You
13:41	can be sure that T . two causes the disk
13:44	to be poured in the clockwise direction , And T
13:50	. one causes the disk to be pulled in the
13:54	counterclockwise direction opposite to that value . And so then
14:00	that torque acting on the disk is the difference between
14:05	these two hours . Now in another video I defined
14:09	a positive torque as a talk that causes counterclockwise rotation
14:15	. So T one Torque 1 is positive . Well
14:20	T1 rather detention force that creates that work is going
14:24	to be positive T . To this tension force will
14:28	create a negative torque because it will cause the disk
14:35	to rotate in a clockwise direction , which I signed
14:38	earlier as a negative torque value . So the network
14:41	of the system , It's going to be the positive
14:44	to work . T . one minus the negative torque
14:47	teach you . And the network is equal to inertia
14:51	times alpha in the same way as net force is
14:54	mass times acceleration . So the network is the inertia
14:58	times the angle acceleration Based on Newton's 2nd Law for
15:01	Rotational Motion . So that's gonna be the difference between
15:04	T . one or talk one minus talked to .
15:09	Now I'm going to take a moment to improve this
15:12	drawn so there is no confusion , so T one
15:20	is being directed opposite to this tension . They have
15:23	to be equal in the opposite direction , and t
15:26	to acting on the disk is in this direction .
15:30	Now T one is going to create a torque that
15:35	will cause the disk to spin in the counter clockwise
15:39	direction . So that's work is positive And T two
15:45	will create . Let's see if I can fit it
15:47	here , A negative torque which will cause this to
15:54	spin in the clockwise direction . And this has been
15:56	in a counterclockwise direction . Keep in mind , clockwise
16:00	rotation is negative . Counterclockwise rotation is associated for positive
16:06	to work . So as we said before , the
16:09	network Is going to be the positive toward T one
16:15	teach you . And so network is inertia times alpha
16:22	. Now the inertia of a solid disc is one
16:26	half times the mass of the disc times r squared
16:31	. Now , linear acceleration is equal to the angular
16:34	acceleration times are . So angular acceleration is linear acceleration
16:39	divided by art . So I'm going to replace Alpha
16:42	with a over our and so that's equal to this
16:54	stuff on the right side . So now let's cancel
17:00	our So what we have left over is 1/2 M
17:06	. R . A . Which is equal to Torque
17:08	one to work . One is equal to the tension
17:12	force times the moment arm , which is the radius
17:15	of the circle . So that's gonna be T one
17:18	times are And to work to it's going to be
17:21	T two times are . So we can divide both
17:25	sides by our So now we have 1/2 mm .
17:33	Which is the difference Between the two tension forces .
17:38	Now the network , is it positive or is it
17:42	negative ? What would you say ? Now ? We
17:44	know the whole system is moving in this direction ,
17:47	which means that the disk has a network that causes
17:51	it to rotate in the clockwise direction which is associated
17:56	with a negative torque . So therefore we need to
17:58	put a negative sign . So now let's save this
18:04	equation , we're going to get back to it so
18:08	I can erase some of the stuff above here because
18:11	I'm going to need a space soon . So I'm
18:27	going to rewrite the equation up here . T 1
18:30	. -52 Is -1/2 . And me now we need
18:42	to relate the tension forces to these two masses .
18:46	So now we need to write expressions for the net
18:48	force acting on each object . So let's start with
18:51	the first object mass one . The net force acting
18:55	on mass one is equal to this tension force because
18:59	there's no other forces acting on it , at least
19:02	in the X . Direction . So the net force
19:05	and that block is equal to Emma , So T
19:09	. one is equal to Emma . So that's the
19:12	first equation that we can right now let's move on
19:16	to block too . So the net force acting on
19:21	block two , which is associate with the wind direction
19:25	, That's going to be equal to the upward tension
19:27	force . T . two minus the downward wait for
19:30	M . Two G . Now keep in mind ,
19:32	MTG is negative because it's going in negative Y direction
19:35	, that's what we have the negative sign and he
19:37	is positive because it's going in a positive direction .
19:41	Now the net force acting on Block two is equal
19:44	to M . Two times A . And this is
19:46	supposed to be M . One A . Now because
19:50	this is moving in a negative Y direction , we
19:52	need to put a negative sign in front of them
19:54	to a . M . one is moving in the
19:57	positive X . Direction . So therefore it requires a
20:00	positive sign . So now I'm going to take this
20:07	term , move it to the left side . So
20:11	therefore M . two G Plus M . two A
20:15	. Is equal to T . two . So that's
20:18	the second equation that we need . So now our
20:21	next step is to replace T . one with M
20:24	. one a . And then replace T . two
20:28	with M . Two G plus mt way . Now
20:35	there's one small correction that I need to make that
20:40	should not be a plus sign since I have a
20:43	negative sign right there . So let's put that minus
20:45	sign . So now let's replace T . one with
20:52	M . One A . And then minus T .
20:59	Two . Which is this stuff here . So let's
21:03	put that in parentheses . So that's gonna be minus
21:05	M . Two G -M2 A . And all of
21:09	that is equal to negative 1/2 . Mhm . Yeah
21:24	so now let's distribute the negative sign . So this
21:27	is going to be M1 a -M two G plus
21:33	M two A . And that's equal to negative one
21:35	half M . A . So I'm going to take
21:38	this term , move it to the right side so
21:40	it becomes positive and then I'm going to take this
21:44	term , move it to the left side so that
21:47	becomes positive . So it's gonna be M . one
21:50	a . Plus , I am too eh Plus 1/2
21:55	capital Emma And that's equal to positive M two G
22:01	. So now I can factor out a from each
22:03	of these three terms . So it's gonna be A
22:06	times M one plus M two Plus 1/2 . And
22:13	that's gonna be equal to MTG . So I need
22:16	to divide both sides by this term . So this
22:19	is my final equation . The acceleration is equal to
22:23	the Darwin weight force , which is basically the net
22:25	force of the system divided by the total mass of
22:28	the system . Where you have to incorporate the c
22:35	. value which is 1/2 for disk . So now
22:40	let's plug everything in . So M2 is 10 ,
22:43	G is 9.8 . M one is 5 M .
22:47	Two is 10 plus one half Times the mass of
22:52	the disk , which is 20 . So this is
22:54	gonna be 98 Divided by 25 just like before .
22:59	And that's going to give us the same answer of
23:01	3.92 meters per second square . So that's the acceleration
23:07	of the whole system . So that's how you can
23:10	find it . So now you have two ways in
23:12	which you can come up with this equation .

Summarizer

DESCRIPTION:

This physics video tutorial provides a basic introduction into rotational dynamics. It explains how to calculate the acceleration of a hanging mass attached to a rotating pulley. Examples include pulleys with two hanging masses (atwood machine) and inclined planes with pulleys and kinetic friction.

OVERVIEW:

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