The amount of effort you put on your videos is simply amazing. All these examples enlighten us and keep us motivated to learn more and more. Keep up the good work! Huge thanks from Brazil
Professor Van Bizen you are still the best lecturer of physics. Happy New Year. I wish you only the best for this semester and many more. I am looking to take paid credit course in second-year physics this summer.
1:54 Wouldn't the torque be a negative quantity if we assume that the clockwise rotation provided by the tension is negative? Then the answer would be a negative acceleration right?
You can work the problem that way. I have found it better to define the direction of the acceleration (in this case down being positive) and assign the direction of the forces caused by the torque accordingly.
please mention that there is no slipping between the rope and the yoyo which is why we had taken the At from the angular accel to be equal to the translational acceleration
So if you set downwards as the positive direction and upwards as the negative direction as you did in the sum of Forces. Then why is it flipped when you are doing the sum of Torques? Where suddenly the Tension is now positive as it points upwards... shouldn't the net torque be equal to -TR?
We only need to consider the "sign" of the torque when using it as a vector quantity. Here we just want to know the magnitude of the torque which will cause an acceleration in the designated direction.
Think about what energy source is causing this work: the gravitational potential energy of the falling disk. Since gravity is a conservative force, this means the answer is independent of the path we take to calculate it. Simply use GPE=m*g*z, and plug in the two z-positions of the yo-yo's center of mass, to get the work done by gravity.
A yoyo will travel upward because the rotational kinetic energy is converted to potential energy. On the way down, the weight of the yoyo accelerates it downward. Since the mechanism is different on the way up, I would surmise that the acceleration is different.
@@MichelvanBiezen thank you for your reply dear. Is the acceleration for the whole system taken from the middle of the pulley for F=ma? While the acceleration for the angular momentum is at the outside of the pulley? Hence the it should be x2!
The amount of effort you put on your videos is simply amazing. All these examples enlighten us and keep us motivated to learn more and more. Keep up the good work! Huge thanks from Brazil
Welcome to the channel! and thank you for the comment.
Thank you so much for making these videos, I struggle with physics so much, and going through your series helps me actually grasp the concept.
Omgggg, perfect timing, I've got my assessment in three days... Thanks :)))
Professor Van Bizen you are still the best lecturer of physics. Happy New Year. I wish you only the best for this semester and many more. I am looking to take paid credit course in second-year physics this summer.
Thank you and Happy New Year to you. All the best on your future endeavors in physics. Hope you find this channel a good source for help. 🙂
@@MichelvanBiezen This channel is excellent. It has always helped me.
1:54 Wouldn't the torque be a negative quantity if we assume that the clockwise rotation provided by the tension is negative? Then the answer would be a negative acceleration right?
You can work the problem that way. I have found it better to define the direction of the acceleration (in this case down being positive) and assign the direction of the forces caused by the torque accordingly.
please mention that there is no slipping between the rope and the yoyo which is why we had taken the At from the angular accel to be equal to the translational acceleration
Excelent videos!!! Thank you Professor!
So if you set downwards as the positive direction and upwards as the negative direction as you did in the sum of Forces. Then why is it flipped when you are doing the sum of Torques? Where suddenly the Tension is now positive as it points upwards... shouldn't the net torque be equal to -TR?
Since we use the acceleration as a scalar, not a vector, the direction doesn't matter, only the magnitude.
why isn't torque negative? it seems to be rotating CW and shouldn't it have negative value?
We only need to consider the "sign" of the torque when using it as a vector quantity. Here we just want to know the magnitude of the torque which will cause an acceleration in the designated direction.
thank you so much for fast reply! It's 3:34 AM here in Korea. really appreciate it
Hello Mr.Michale, one question: the final equation: a= 3/2 g , means that the mass of the Yo-yo dose not matter ?
Thanks.
Only the shape matters.
Thank you very much! That was incredibly helpful
You are welcome. Glad you found our videos. 🙂
Hi Sir. I was wondering how to determine the Work Done by the Force lets say from time t=0 to t=2.5s. Initially at t=0, the disk is at rest.
Think about what energy source is causing this work: the gravitational potential energy of the falling disk. Since gravity is a conservative force, this means the answer is independent of the path we take to calculate it. Simply use GPE=m*g*z, and plug in the two z-positions of the yo-yo's center of mass, to get the work done by gravity.
Amazing..really helpful
Isn’t mg in the negative direction and tension in the positive direction? And clockwise torque would be negative too?
The sign assignments of the direction are arbitrary. You assign them in whatever way is most convenient for the problem at hand.
You have helped me so much! Thank you!
hi, why did you take mg-T instead of assuming T is positive?
The directions are chosen relative to the yo-yo. (Relative to the ceiling the tension would be negative).
I think if the Yoyo go up a=-(2/3)g am I right?
A yoyo will travel upward because the rotational kinetic energy is converted to potential energy. On the way down, the weight of the yoyo accelerates it downward. Since the mechanism is different on the way up, I would surmise that the acceleration is different.
@@MichelvanBiezen It'll get a little more complicated. Thank you.
hi sir
why u didn t include the torque of the mg force or it doesn t effect
cuz usuaally we do
Since the mg vector goes through the point of rotation, it does not cause a torque.
Thank so much Professor :D
You're very welcome! Glad the videos are helping. 🙂
2:31
What is your question?
Is the acceleration for F=ma and the acceleration for the angular acceleration the same?
a = (angular acceleration) x (radius) = (alpha) r
@@MichelvanBiezen thank you for your reply dear. Is the acceleration for the whole system taken from the middle of the pulley for F=ma? While the acceleration for the angular momentum is at the outside of the pulley? Hence the it should be x2!
Take a careful look at the video and see if that answers your question. (It should be apparent from the video)
Thank you sir
thanks
Thanks brazil
Glad you liked it. Welcome to the channel!
מעולה