In the expression at top right shouldn't the relation 2T- 'mg - m"a"' show "mg - m"a" " in () with the sign being +? If mg and m"a" are in the same direction, they should have the same sign.
The tension in the string holding the mass "m", is indeed mg - ma. We have videos in the application of Newton's laws playlists that show why. Or you can simply draw a free body diagram around the mass and determine the tension in the rope by applying F = ma).
Nice explanation sir,. But it would be helpful if you can give a little insight into why the acceleration of the block "a" is half the acceleration of the rope a. I couldn't understand it intuitively.
For the block to lower, the length of both ropes of the pulley need to extend. So the acceleration of "rope a" needs to be distributed to the 2 ropes. Meaning if you extend your rope 1 meter, the block only moves down 1/2 meter because it's split to both ropes supporting the pulley. Acceleration works the same way as length, so when one side of the rope accelerates at "a" and the other is fixed, the blocks acceleration is "a" /2. This concept is extremely important when considering multiple pulleys in series or parallel.
Ignorant question but what validates the a priori assumption that “T” on each side of the massless pulley are (and remain) equal? If m is actually accelerating down does that not mean the disk rotates, which means one side has less tension? Put another way, if we were to hold some sort of a ball or disc by a string with equal tension on both sides, it wouldn’t move .. right? ….where is my error?
😎 Great video! Just an FYI --- in the video description, for the 'next video', clicking takes you to a page that says: "Video unavailable This video is private"
In the expression at top right shouldn't the relation 2T- 'mg - m"a"' show "mg - m"a" " in () with the sign being +? If mg and m"a" are in the same direction, they should have the same sign.
The tension in the string holding the mass "m", is indeed mg - ma. We have videos in the application of Newton's laws playlists that show why. Or you can simply draw a free body diagram around the mass and determine the tension in the rope by applying F = ma).
Nice explanation sir,. But it would be helpful if you can give a little insight into why the acceleration of the block "a" is half the acceleration of the rope a. I couldn't understand it intuitively.
For the block to lower, the length of both ropes of the pulley need to extend.
So the acceleration of "rope a" needs to be distributed to the 2 ropes. Meaning if you extend your rope 1 meter, the block only moves down 1/2 meter because it's split to both ropes supporting the pulley.
Acceleration works the same way as length, so when one side of the rope accelerates at "a" and the other is fixed, the blocks acceleration is "a" /2.
This concept is extremely important when considering multiple pulleys in series or parallel.
@@bryanwinn7911 Thank you a lot Sir
Bryan Winn. Good explanation!
You saved my ass in college physics. respect and value your work sir 💯
Thanks!
Ignorant question but
what validates the a priori assumption that “T” on each side of the massless pulley are (and remain) equal?
If m is actually accelerating down does that not mean the disk rotates, which means one side has less tension?
Put another way, if we were to hold some sort of a ball or disc by a string with equal tension on both sides, it wouldn’t move .. right?
….where is my error?
Yeah I too have the same doubt.
What you have said is true, but he is calculating the acceleration of the first instant, so you can say that both Ts are equal
In this problem only the left pulley has mass. In the next video both will have mass (and therefore both pulleys will have moment of inertia).
Super . Application of physics . Nicee
Glad you are enjoying these videos.
Amazing sir!!!!!!. 👍
Thanks a lot
😎 Great video!
Just an FYI --- in the video description, for the 'next video', clicking takes you to a page that says: "Video unavailable
This video is private"
That is correct. We haven't published that one yet. (planned for tomorrow).
@@MichelvanBiezen 😄 Ohhhh, ok. Just curious is all, and I love that you have the 'next' and 'previous' links. Extremely convenient.
Yes, we started doing that with out newer videos. Glad you like it. 🙂
@@MichelvanBiezen 🏆
It should be T = 1/2 (mg + m"a") ? Right?
No, the tension is correct as shown in the video.
Thank you Teacher
You are welcome
🌻
Thanks!
Thnku
You are welcome.