My entire class has a group chat and we all watch your videos, even during class! We all really appreciate it because our professor isn't exactly the best. So thank you so much for taking the time to make these videos!!! It's saved the entire class from failing.
I really appreciate the fact that you have multiple videos with examples that get increasingly harder, rather than starting with one hard problem. Thank you!
Thank you, Mr Michael, Yesterday I've had to ask the my physics teacher about calculating the Angular Momentum of the Running Boy then my physics teacher shouted at me and started to ask questions I didn't even know answers for. He just wrote MVR for the boy's Angular Momentum, Thanks to you even though I didn't ask you any doubt you still showed me how to find that out. You are great, You are helping students like me. BTW you have a student from India
That's a crucial concept in this problem when the child was going negative direction and the moment he landed on the solid disk, his tangential velocity changed to angular velocity which bounded by the right hand rule. Omega becomes positive. Great video thanks again.
+Michel van Biezen Hey, thanks for the videos they are great. I have a question regarding the last equation, there are no 1/2 in the I_disk ? Is this because they cancel or what is happening?? -Thanks in advance
+Michel van Biezen correct me if i'm wrong . The moment before the boy jumps we have two bodies (child and disk) . After he jumps , they both have the same angular momentum and thus can be considered as a composite body . The w(omega)f which we are finding is relative to the ground and wf of child relative to the disk is 0 .
+merhawi tecle Using the right hand rule. Curl your fingers in the direction of the motion and your thumb will point in the direction of the angular momentum. (Your thumb will point up).
2:00 Why is the tangential velocity positive? I thought it was negative? Is it because the child is jumping on the disk in the counter clockwise direction which is positive in angular momentum?
+Newbport Angular velocity is a vector quantity. That calculation was finding the magnitude of the angular velocity. (Magnitude of a vector cannot be negative). Then when you place that into the angular momentum equation, the direction of the sign must be honored.
Oh okay thanks! And I know I've said this before, but you have the greatest physics videos I've ever seen on youtube. I've learned more from you in a day than all quarter from my professor. I've been using you all quarter! (Thank you)
Sir In the previous vid , L initial of the bag was set to be zero before it touched the disk , now I'm wondering how the child is different from that scenario.
In the previous video the bag was dropped vertically with no horizontal velocity. In this case the child has horizonal velocity prior to jumping onto the disk. After jumping onto the disk the linear momentum is convered to angular velocity.
I am lost the previously we said if the disk is rotating counterclockwise the angular velocity becomes negative so how do we come to use positive 3.33 instead of negative
in a previous example we calcualted the rotational velocity before as 0 bc the bag wasnt rotating , similarly shouldnt we accept the rotational velocity initial of the child 0 for the initial condition?
good question but disk does not have angular momentum component in the direction below. But accelerating sand box has. Its effect can be negligable. If it is not so small, we assume it is very fast, then it can make the disk rotate at a new direction for example in right but this duration would be very small because gravity causes it accelerate downwards again without making a new complete turn together.
When the child jumps on the merry go round, they exert a torque to it. However, the child had angular momentum before the jump. We know The merry go round exerts a force on the child, so the child slows down linearly. But to change their ang. Momentum, there has to be a torque on the child too? How does the merry go round exert a torque on the child since they aren't rotating.
@@MichelvanBiezen I am not understanding.yes there is friction between the two, but where is there a torque on the child. Torque is force. X radius. There is a torque on merry go round but doesn't he child just feel a force?
A torque is created when the action of the force (direction of the force) acting on the object does not go through the point of rotation. (which is the case with the child)
The child speed is at the same direction of the rotation of the disk ( unless the child is thrown on the disk) which is negative also. should we reverse the sign of angular speed of the child to -ve instead ?
Note that if the child had jumped on the disk at the bottom instead of at the top (or on the left side or right side, the situation would be different. That is why it is better to just look at the problem and determine the sign of the translated angular velocity based on the geometry.
I think the momentum of the child and the disk must be combined because the movement of the child in this case will increase the speed of disck, and it is not reasonable for the speed after the movement of the disk to be 0.2 and before 1
The direction of motion of the child is in the opposite direction of motion of the disk. That is why you would expect the final angular velocity of the disk to be smaller.
To get the sign correct could we possibly think about the child running around the disk before jumping into it as to get an idea of the direction of the childs angular velocity as he enters the system and for example say at set point his Vtan is at that particular point and once transformed into angular velocity the rules apply due to the child's Vtan indicating the angular velocities direction linearly, so that if the Vtan is negative then the Vang... Is positive and if Vtan is positive the Vang is negative??
The sign is chosen by convention. Both the angular speed and the angular momentum is considered negative in the clockwise direction and positive in the counterclockwise direction.
Why is Vnot = -5m/s but when you calculate W you use a Vt (tangential velocity) of 5 m/s. Is the tangential velocity and absolute value of the initial velocity?
Note that if the runner had been running in the opposite direction (+ 5 m/sec) and jumped on the disk on this side (rather than the opposite side), the contribution to the angular momentum would have been the same (positive) value.
The direction from which the child came is not important. What is important is in what direction that motion will cause the disk to spin (and counter clockwise is chosen as the positive direction).
Yes he did create a torque, but the essential difference is that he didn't create an external torque. The torque the child applied to the platform, was also reflected back onto the child as an equal and opposite reaction torque. Both torques are about the platform's axis of rotation. If a second child were holding the platform stationary, while standing on the ground, then we would have an external torque that would make conservation of angular momentum no longer a valid way to solve this problem. Angular momentum is still conserved in general, but the entire Earth absorbs it and moves an insignificant amount as a result.
Sir, when child jumps on the edge of the disc, will not it impart torque to it and net torque will not be zero, hence momentum will not be conserved. Pls comment, if my observation is correct.
First of all, momentum is ALWAYS conserved in any collision. But when a child jumps on the disk perpendicular to the surface, then no angular momentum would be contributed by the child.
@@MichelvanBiezen [...First of all, momentum is ALWAYS conserved in any collision...] NOT NECESSARILY. ONLY IF, during the very short time of impact, NO EXTERNAL (to the system of colliding bodies!) forces and/or NO EXTERNAL torques of IMPULSIVE CHARACTER (i.e., very intense and of a very short duration), act on the system! This might be the case in collisions between CONSTRAINED bodies, where intense reactions, appearing as EXTERNAL forces or torques, may develop, at the time of impact, at the constraints, in order to maintain an initial position and/or an initial state of motion of one of the colliding bodies during the collision. An example is the previous problem where a bag of sand lands on a disk rotating about a vertical axis passing through its center. The disk is constrained to rotate about the same axis also during and after the impact. In that example, the bag approaches the disk surface with a non-zero velocity, parallel to the axis of rotation and directed downwards. As soon it lands, its angular momentum (with respect to the center of the disk) is, according to the definition and the problem geometry, NON-ZERO and PERPENDICULAR to the axis of rotation. The disk angular momentum with respect to the same point is, at this time, non-zero and parallel to the axis of rotation. The TOTAL angular momentum of the system bag + disk is the resultant of these two components and, at this time, NOT PARALLEL to the axis of rotation. It is instead so (PARALLEL) after the impact (the bag and the disk rotate about the initial axis as a single body). IT IS HENCE NOT CONSERVED DURING THE COLLISION! What remains constant (through the collision) in that example is ONLY ITS VERTICAL COMPONENT. The torque developed at the center of the disk is: 1. ZERO, before the impact - the disk is assumed to be homogeneous and the axis of rotation passing through its center of mass; 2. NON-ZERO, PERPENDICULAR TO THE AXIS OF ROTATION AND IMPULSIVE at the time of impact - It accounts for the variation (non-conservation) of the TOTAL (vector and of the system) ANGULAR MOMENTUM through the collision. Its value is in the order of the magnitude of that variation divided by the impact duration time. 3. NON-ZERO, PERPENDICULAR TO THE AXIS OF ROTATION AND NON-IMPULSIVE after the collision. It is constantly equal and opposite to the torque of the weight of the bag with respect to the center of the disk. [...But when a child jumps on the disk perpendicular to the surface, then no angular momentum would be contributed by the child...] In this problem, the child (the equivalent of the bag) approaches the disk with a non-zero velocity, like in the previous case, but, this time, with a direction parallel to its surface. His angular momentum with respect to the center of the disk is, as soon as he lands, NON-ZERO and PARALLEL to the axis of rotation. The disk angular momentum with respect to the same point is, at this time, also non-zero and parallel to the axis of rotation. The TOTAL angular momentum of the system child + disk has at this time only a component, the one parallel to the axis of rotation. NO EXTERNAL TORQUES OF IMPULSIVE CHARACTER ACT ON THE SYSTEM. The torque developed at the center of the disk is: 1. ZERO, before the impact - the disk is assumed to be homogeneous and the axis of rotation passing through its center of mass; 2. NON-ZERO, PERPENDICULAR TO THE AXIS OF ROTATION AND NON-IMPULSIVE during the collision. It is constantly equal and opposite to the torque of the weight of the child with respect to the center of the disk. 3. NON-ZERO, PERPENDICULAR TO THE AXIS OF ROTATION AND NON-IMPULSIVE after the collision, again constantly equal and opposite to the torque of the weight of the child with respect to the center of the disk.
Professor, I'm surprised at the answer, I thought the final angular velocity would increase when the child jumped on, but the answer -0.2rad/sec is less than the initial angular velocity of -1.0rad/sec. Especially as the child jumped onto the outer edge of the disk for maximum effect and the running speed is quite fast.
We can convert the linear momentum of the child (p = mv) to ancular momentum of the child ( L = Iw = mr^2 (v/r)) knowing that the child will be traveling in circles after getting onto the rotating disk.
Work done can be defined as the change in energy. Thus you can calculate the work done by the child on the Merry Go Around by calculating the change in rotational kinetic energy.
Sir, does direction of child running (before jump) has relevance on direction of initial angular velocity of the child for use in the momentum conservation equation. i:e if -5m/s is given as +5m/s.
so is angular velocity direction assigned as cw=negative and ccw=positive, but for angular momentum we use right hand rule to determine direction, am I correct?
+Jinnie Tinianow It is the same for both. Positive direction for angular velocity is CCW. The direction for angular momentum can be determined by: 1) L = r x p where L, r, and p are all vectors. 2) L = I w where both L and w are vectors (w = omega here) So you can see that the sign for L and w are the same.
Hi! I think you forgot that the boy was running at a velocity of Negative 5. so the boy's angular speed should have been -3.33 and the final answer is -1.43
***** The video is correct as is. When the boy jumps on the disk his linear momentum is translated into angular momentum and since his angular momentum is counter-clockwise by definition that is positive. At any rate, his direction is opposite to the rotation of the disk, so they cannot have the same sign
Technically the linear momentum is equal to the cross product of linear momentum and radius, I don't think your method would work if you are given vectors
Man, you are the Master of physics! No doubt.
My entire class has a group chat and we all watch your videos, even during class! We all really appreciate it because our professor isn't exactly the best. So thank you so much for taking the time to make these videos!!! It's saved the entire class from failing.
Thank you for sharing. My wife and I are glad that we are able to help.
I really appreciate the fact that you have multiple videos with examples that get increasingly harder, rather than starting with one hard problem. Thank you!
I've gotta say, you're helping me enjoy learning physics after multiple bad experiences. Thank you.
I'm so glad I found this on youtube, now Angular Momentum(and others topics) feels extremely clear in my mind! Thank you so much!
Thank you, Mr Michael, Yesterday I've had to ask the my physics teacher about calculating the Angular Momentum of the Running Boy then my physics teacher shouted at me and started to ask questions I didn't even know answers for. He just wrote MVR for the boy's Angular Momentum, Thanks to you even though I didn't ask you any doubt you still showed me how to find that out.
You are great, You are helping students like me. BTW you have a student from India
Glad we could help and welcome to the channel!
That's a crucial concept in this problem when the child was going negative direction and the moment he landed on the solid disk, his tangential velocity changed to angular velocity which bounded by the right hand rule. Omega becomes positive. Great video thanks again.
could you elaborate im really confused about why the child's angular velocity is positive
You are a life savior, I learn a lot from your video and now I can say with confidence that I can pass my mechanics final. THANK YOU
You are saving my project for a second time, literally!
You are having the best education channel in the world full education resources
Fred,
That is good. Learning physics should be fun and interesting.
+Michel van Biezen
Hey, thanks for the videos they are great. I have a question regarding the last equation, there are no 1/2 in the I_disk ? Is this because they cancel or what is happening??
-Thanks in advance
+Michel van Biezen correct me if i'm wrong . The moment before the boy jumps we have two bodies (child and disk) . After he jumps , they both have the same angular momentum and thus can be considered as a composite body . The w(omega)f which we are finding is relative to the ground and wf of child relative to the disk is 0 .
After the boy jumps, they will not have the same angular momentum, but they will have the same angular velocity.
oh , yeah you are right .
The best physics videos on youtube!
You're the bestest ever ....thanks man
Thank you. Glad you find our videos helpful.
I CANT THANK YOU ENOUGH FOR THIS VIDEOS
Glad it was helpful. 🙂
Thanks again!! but a little concerned about the 3.33 rad/sec...isn't it negative..i did the negative way and got -1.4285.
+merhawi tecle
Using the right hand rule. Curl your fingers in the direction of the motion and your thumb will point in the direction of the angular momentum. (Your thumb will point up).
@@MichelvanBiezen that means the answer is false ? should we take negative 3,33?
yeah me too i got the answer -1.42
i has been 5 years you must be a physics professor now
2:00 Why is the tangential velocity positive? I thought it was negative?
Is it because the child is jumping on the disk in the counter clockwise direction which is positive in angular momentum?
+Newbport
Angular velocity is a vector quantity. That calculation was finding the magnitude of the angular velocity.
(Magnitude of a vector cannot be negative).
Then when you place that into the angular momentum equation, the direction of the sign must be honored.
Oh okay thanks!
And I know I've said this before, but you have the greatest physics videos I've ever seen on youtube. I've learned more from you in a day than all quarter from my professor.
I've been using you all quarter! (Thank you)
The kid got a rude awakening after landing on the disk
Yes, I would not recommend trying this. 🙂
@@MichelvanBiezen well, it’s really not much different than the merry-go-round located in the playground at my old elementary school. 🤔
Sir In the previous vid , L initial of the bag was set to be zero before it touched the disk , now I'm wondering how the child is different from that scenario.
In the previous video the bag was dropped vertically with no horizontal velocity. In this case the child has horizonal velocity prior to jumping onto the disk. After jumping onto the disk the linear momentum is convered to angular velocity.
I am lost the previously we said if the disk is rotating counterclockwise the angular velocity becomes negative so how do we come to use positive 3.33 instead of negative
As a vector quantity, counter clockwise is considered the positive direction.
u save my final. The last question of the exam is almost the same!! thank!
Good for you. Glad we were able to help
in a previous example we calcualted the rotational velocity before as 0 bc the bag wasnt rotating , similarly shouldnt we accept the rotational velocity initial of the child 0 for the initial condition?
If the child is moving (as is in this case), it cannot be zero
but in the previous example the mass 50 kg is accelerating downward shouldnt we have calculated the rotational velocity of that one too then?
good question but disk does not have angular momentum component in the direction below. But accelerating sand box has. Its effect can be negligable. If it is not so small, we assume it is very fast, then it can make the disk rotate at a new direction for example in right but this duration would be very small because gravity causes it accelerate downwards again without making a new complete turn together.
When the child jumps on the merry go round, they exert a torque to it. However, the child had angular momentum before the jump. We know The merry go round exerts a force on the child, so the child slows down linearly. But to change their ang. Momentum, there has to be a torque on the child too? How does the merry go round exert a torque on the child since they aren't rotating.
Friction between the child and the merry go round.
@@MichelvanBiezen I am not understanding.yes there is friction between the two, but where is there a torque on the child. Torque is force. X radius. There is a torque on merry go round but doesn't he child just feel a force?
A torque is created when the action of the force (direction of the force) acting on the object does not go through the point of rotation. (which is the case with the child)
Thank you so much!!! Great explanations !!!
The child speed is at the same direction of the rotation of the disk ( unless the child is thrown on the disk) which is negative also.
should we reverse the sign of angular speed of the child to -ve instead ?
Note that if the child had jumped on the disk at the bottom instead of at the top (or on the left side or right side, the situation would be different. That is why it is better to just look at the problem and determine the sign of the translated angular velocity based on the geometry.
I think the momentum of the child and the disk must be combined because the movement of the child in this case will increase the speed of disck, and it is not reasonable for the speed after the movement of the disk to be 0.2 and before 1
The direction of motion of the child is in the opposite direction of motion of the disk. That is why you would expect the final angular velocity of the disk to be smaller.
To get the sign correct could we possibly think about the child running around the disk before jumping into it as to get an idea of the direction of the childs angular velocity as he enters the system and for example say at set point his Vtan is at that particular point and once transformed into angular velocity the rules apply due to the child's Vtan indicating the angular velocities direction linearly, so that if the Vtan is negative then the Vang... Is positive and if Vtan is positive the Vang is negative??
Also apologies for pestering you with questions and being patient, thank you very much for the help so far and any in the future hahaha :)
The sign is chosen by convention. Both the angular speed and the angular momentum is considered negative in the clockwise direction and positive in the counterclockwise direction.
Why is Vnot = -5m/s but when you calculate W you use a Vt (tangential velocity) of 5 m/s. Is the tangential velocity and absolute value of the initial velocity?
Note that if the runner had been running in the opposite direction (+ 5 m/sec) and jumped on the disk on this side (rather than the opposite side), the contribution to the angular momentum would have been the same (positive) value.
Why didn't you use that negative when computing the initial angular velocity?
The direction from which the child came is not important. What is important is in what direction that motion will cause the disk to spin (and counter clockwise is chosen as the positive direction).
@@MichelvanBiezen thank you
when the boy run and jump on the disc isn't he creat a torque? and by doing it make the angular momentum not to be conserved?
thank you for your help!
Momentum (including angular momentum) is ALWAYS conserved through ANY collision.
Yes he did create a torque, but the essential difference is that he didn't create an external torque. The torque the child applied to the platform, was also reflected back onto the child as an equal and opposite reaction torque. Both torques are about the platform's axis of rotation.
If a second child were holding the platform stationary, while standing on the ground, then we would have an external torque that would make conservation of angular momentum no longer a valid way to solve this problem. Angular momentum is still conserved in general, but the entire Earth absorbs it and moves an insignificant amount as a result.
Sir, when child jumps on the edge of the disc, will not it impart torque to it and net torque will not be zero, hence momentum will not be conserved. Pls comment, if my observation is correct.
First of all, momentum is ALWAYS conserved in any collision. But when a child jumps on the disk perpendicular to the surface, then no angular momentum would be contributed by the child.
@@MichelvanBiezen Yes, that was my point. Thanks
@@MichelvanBiezen
[...First of all, momentum is ALWAYS conserved in any collision...]
NOT NECESSARILY.
ONLY IF, during the very short time of impact, NO EXTERNAL (to the system of colliding bodies!) forces and/or NO EXTERNAL torques of IMPULSIVE CHARACTER (i.e., very intense and of a very short duration), act on the system!
This might be the case in collisions between CONSTRAINED bodies, where intense reactions, appearing as EXTERNAL forces or torques, may develop, at the time of impact, at the constraints, in order to maintain an initial position and/or an initial state of motion of one of the colliding bodies during the collision.
An example is the previous problem where a bag of sand lands on a disk rotating about a vertical axis passing through its center. The disk is constrained to rotate about the same axis also during and after the impact.
In that example, the bag approaches the disk surface with a non-zero velocity, parallel to the axis of rotation and directed downwards.
As soon it lands, its angular momentum (with respect to the center of the disk) is, according to the definition and the problem geometry, NON-ZERO and PERPENDICULAR to the axis of rotation. The disk angular momentum with respect to the same point is, at this time, non-zero and parallel to the axis of rotation.
The TOTAL angular momentum of the system bag + disk is the resultant of these two components and, at this time, NOT PARALLEL to the axis of rotation. It is instead so (PARALLEL) after the impact (the bag and the disk rotate about the initial axis as a single body). IT IS HENCE NOT CONSERVED DURING THE COLLISION! What remains constant (through the collision) in that example is ONLY ITS VERTICAL COMPONENT.
The torque developed at the center of the disk is:
1. ZERO, before the impact - the disk is assumed to be homogeneous and the axis of rotation passing through its center of mass;
2. NON-ZERO, PERPENDICULAR TO THE AXIS OF ROTATION AND IMPULSIVE at the time of impact - It accounts for the variation (non-conservation) of the TOTAL (vector and of the system) ANGULAR MOMENTUM through the collision. Its value is in the order of the magnitude of that variation divided by the impact duration time.
3. NON-ZERO, PERPENDICULAR TO THE AXIS OF ROTATION AND NON-IMPULSIVE after the collision. It is constantly equal and opposite to the torque of the weight of the bag with respect to the center of the disk.
[...But when a child jumps on the disk perpendicular to the surface, then no angular momentum would be contributed by the child...]
In this problem, the child (the equivalent of the bag) approaches the disk with a non-zero velocity, like in the previous case, but, this time, with a direction parallel to its surface. His angular momentum with respect to the center of the disk is, as soon as he lands, NON-ZERO and PARALLEL to the axis of rotation. The disk angular momentum with respect to the same point is, at this time, also non-zero and parallel to the axis of rotation. The TOTAL angular momentum of the system child + disk has at this time only a component, the one parallel to the axis of rotation. NO EXTERNAL TORQUES OF IMPULSIVE CHARACTER ACT ON THE SYSTEM.
The torque developed at the center of the disk is:
1. ZERO, before the impact - the disk is assumed to be homogeneous and the axis of rotation passing through its center of mass;
2. NON-ZERO, PERPENDICULAR TO THE AXIS OF ROTATION AND NON-IMPULSIVE during the collision. It is constantly equal and opposite to the torque of the weight of the child with respect to the center of the disk.
3. NON-ZERO, PERPENDICULAR TO THE AXIS OF ROTATION AND NON-IMPULSIVE after the collision, again constantly equal and opposite to the torque of the weight of the child with respect to the center of the disk.
Love the numerous examples!!!
Professor, I'm surprised at the answer, I thought the final angular velocity would increase when the child jumped on, but the answer -0.2rad/sec is less than the initial angular velocity of -1.0rad/sec. Especially as the child jumped onto the outer edge of the disk for maximum effect and the running speed is quite fast.
Note that the disk war rotating in the clockwise direction initially and the child is jumping onto the disk in a counter clockwise direction
Sir did you use the parallel axis theorem for the total moment of inertial or was it simply an addition?
In this case it is simply addition.
hello sir,
shouldn't the inertia of the disk be 1/2mr²
That is correct, but we don't need the equation, since we were given the value of the moment of inertial of the disk.
can we have momentum of child as mvr?
We can convert the linear momentum of the child (p = mv) to ancular momentum of the child ( L = Iw = mr^2 (v/r)) knowing that the child will be traveling in circles after getting onto the rotating disk.
How do we find how much work is done by the child in this example?
Work done can be defined as the change in energy. Thus you can calculate the work done by the child on the Merry Go Around by calculating the change in rotational kinetic energy.
Sir, does direction of child running (before jump) has relevance on direction of initial angular velocity of the child for use in the momentum conservation equation. i:e if -5m/s is given as +5m/s.
Yes, the resulting angular momentum will be positive or negative depending on the direction of motion of the child.
why did the angular velocity of the boy become positive?
Angular momentum is defined here as positive in the counter-clockwise direction.
you are a god
No no, just another person trying to do his best.
so is angular velocity direction assigned as cw=negative and ccw=positive, but for angular momentum we use right hand rule to determine direction, am I correct?
+Jinnie Tinianow
It is the same for both. Positive direction for angular velocity is CCW. The direction for angular momentum can be determined by:
1) L = r x p where L, r, and p are all vectors.
2) L = I w where both L and w are vectors (w = omega here)
So you can see that the sign for L and w are the same.
Oh, I get it now! Thanks! You're amazing!
Isn't it angular velocity of child is 5/1.5=-3.33m.s in 1:53?
Ngozi,
According to convention, the angular momentum is positive when rotating in a counterclockwise direction.
Oh ok thank you. Your videos are really helpful. Your helping pass physics!
Michel van Biezen ahh the conventions are confusing me. Can I switch the convention to clockwise with positive and cc with negative in this case too?
Anthony Vu
Anthony
Yes. In the end it doesn't matter which convention you use as long as you stay consistent throughout the problem.
Mr.Fosana: Thank you sir
Great! thank you very much.
You are welcome! Glad you found the videos helpful.
Is kinetic energy constant?
No, kinetic energy will be lost.
Hi! I think you forgot that the boy was running at a velocity of Negative 5. so the boy's angular speed should have been -3.33 and the final answer is -1.43
*****
The video is correct as is.
When the boy jumps on the disk his linear momentum is translated into angular momentum and since his angular momentum is counter-clockwise by definition that is positive.
At any rate, his direction is opposite to the rotation of the disk, so they cannot have the same sign
@@MichelvanBiezen but when I used the relationship L= r p sin(theta) it messed up my calculation
Why is the moment of inertia of child is equal to mR^2? I though it was 1/2 mR^2?
Once the child is on the edge of the disk, the child can be considered a "point mass".
This is true if the child jumps to the edge of the disc and remains stationary and does not move with respect to the disc
That is the assumption here indeed.
@@MichelvanBiezen yes thats what i expected
I have an important research and I want to discuss it with you, if possible by e-mail
My e-mail address is mvanbiezen@elcamino.edu, but unfortunately I am very busy with my own research at the moment.
thank you :)
Thank u!!!!!!!!!!!!!!
You're welcome!
Thanks
Shouldn't that be -3.3
since the child is moving at a counterclockwise direction the angular velocity is positive.
Ok professor now i understtod thank you sir
Missed the welcome intro :(
Oh darn. 🙂
Technically the linear momentum is equal to the cross product of linear momentum and radius, I don't think your method would work if you are given vectors
The problem can be worked both with vector notation or by calculating the magnitude.