Torque & Angular Acceleration Example
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- Опубликовано: 11 апр 2013
- Torque & Angular Acceleration Example: Dr. Eric Abraham: (Edit: There is a calculation error towards the end, and an edit is noted in the video. The angular acceleration should be 1.76 rad/sec^2 instead of 1.37. All the formulas are correct.)
Very Very Very well done Mr A
no words, just awesome, thank you
You Are Awesome :)
To whom who is not satisfied with the speed, you can speed the video up.. it's better when it's 1.25
Abdulrahman Mahdaly thanks for the kind words! FYI: Listening at x2 is always good for a laugh. Since it is computerized and my voice is a high pitch anyway, it sounds like a 12-year-old speaking into an autotuner.
AbrahamPhysics hahah that's when it gets to too high speeds but when it's 1.25 it feels good lol
Thank you so much sir for your great stuff :-)
this is awesome thanks alot man
thank god i passed mechanics and never had to solve a problem like this on the final
Most helpful video, if u guys crying for speed then change the playback speed to 1.5
Helped a lot bud.
+Tom Coombs First, sorry, I accidently deleted you comment looking to reply to it! Second, I think you have to change your settings in google+ to allow someone to reply to your comments. Third, I had already edited the video for the angular acceleration calculation. There is a calculation error there, and it should be 1.76 instead of 1.37 rad/sec^2. All the formulas are correct, and the other accelerations and tensions can be found with the correct angular acceleration as you did.
Yeah me too. I got your same answer... I have double check my calculation for 10 times.
How would it be different if m2 was great enough to cause clockwise rotation? Would the equations for each Fnet just be the opposite? I am attempting this and am getting myself very confused. Being able to understand this in both directions would be very beneficial for the class I am currently in. Thanks!
The final equation for angular acceleration wouldn't have to change at all, your value of alpha would just be negative! If you wanted to repeat the whole thing, I would switch the coordinate system directions to make the directions the objects were going positive. You should get the same equation for alpha, except the numerator would be switched, essentially multiplying this answer by -1. The only differences throughout the calculation will be minus signs.
Thanks so much for the quick reply, and great video btw. I had switched the coordinates as you say, but when I move on to Torque(1) - Torque(2) = (Moment of inertia)(Alpha) then solve for alpha, the signs distribute in a funky way which makes it feel incorrect (such as the denominator having I - m1r1^2 - m2r2^2). I then tried doing Torque(2) - Torque(1) = (Moment of inertia)(Alpha), and it creates something as you say (the negative). It just seems weird to me to have torque2 subtract torque1 as the right hand rule says it should be T1 - T2
Thanks! Well, I did not think for a second that reversing the direction had any subtlety at all, but it does! You are right, if you just switch axes for the forces like I suggested, you get I - m1r1^2 - m2r2^2, which is not correct. It must be I + m1r1^2 + m2r2^2 since that is the total resistance to the change in rotation. By choosing the rotation to be positive, I avoided this subtlety. The problem is this: In the video, I establish relationships between a1, a2, and alpha. Because of the choice of coordinate systems, these are all positive numbers (which is what we wanted). However, we now have an implicit assumption that alpha is positive, too! If the system is rotating clockwise, by right hand rule, alpha should be a negative number.
The most correct relationships for approaching it this way would be a1 = r1*|alpha| and a2 = r2*|alpha| (keeping everything positive). Then, when you substitute into the equation for torques, you need a relationship between |alpha| and alpha. This requires knowledge about which way the system is rotating, which again, we assumed we did when we chose coordinate systems for the forces. In the case where it is rotating clockwise (yours) then |alpha| = -alpha, and you get the right signs.
However, if you have knowledge of which way the system is rotating, you wouldn't approach it this way at all. Since you know m2r2 is greater then m1r1 in this case, you would just calculate torque2 - torque1 knowing the result will give you the magnitude of the acceleration, and if you needed the vector description of acceleration, you could assign it after the fact using the right hand rule.
I did not appreciate, since this problem had specific numbers, this allowed choices that cannot be applied easily if the system switched direction due to different numbers. If you had to do the general case, where you did not know which way it would rotate, you would want to set up the problem differently, and chose one coordinate system for the forces and rotation and let both a and alpha be positive or negative. I've been wanting to redo this video for some time (so slow! too many "uhhs" and "umms"!). This will probably give me the motivation to do it. Mark, I greatly appreciate the comments and feedback.
Can you plz explain how to find the rotational speed of the system
free moroccan Once you have the acceleration, you can find the speed using constant angular acceleration kinematics. (final angular speed = initial angular speed + angular acceleration*time. However, starting from scratch, it may be easier using conservation of energy.
this. some. good. shit. if. i. say. so. myself.
You can! And you did! Thanks for the kind words.
isnt gravity acting on the pulley too?
woops you fixed it lol nice nice
OLaYBoTs Channel got to watch the whole video before i comment lol
I don't get 1.37 rad/sec using the numbers in your problem and your final answer formula
aworthtx yep, calculator error. Added a comment, but I'll probably just redo the last couple minutes.
money shot @ 21:20
Length of the video makes it more confusing
Home
There is just one problem doc you have very slow expression :)
only one?! Seriously, though, you are correct. It takes some practice getting good in this medium, especially if your experience is in the traditional format. As I continue, I am working hard on speaking more quickly (while not making errors) and being more concise.
I hope, you will make it because you seem so good in your job
AbrahamPhysics just be natural I think the speeds ok
Long video... You could honestly have done this in less time.
Can't you just use the equation that the net torque equals the moment of inertia times the angular acceleration?
m1*9.8*r1 - m2*9.8*r2 = I*a
Then just solve for a. I kept getting the wrong answer on my homework and I don't get why.
The net torque is the moment of inertia times the angular acceleration, but the tensions are not m1g and m2g. You have to find the tensions by applying newton's second law to the falling masses.
yeah my teacher did the same thing for the answer sheet of the hw and so i end up getting .42 doing the way of video and she gets .49 doing it the way you wrote and I'm confused