- 1. Rotational Dynamics Pulley Tension Atwood Machine Worked Example | Doc Physics
- 2. Intro to Integrals | Quick Calculus 3 of 6 | Doc Physics
- 3. Rolling Without Slipping - A sticky adventure in rotation and translation | Doc Physics
- 4. Rolling Without Slipping Worked Example (kinematics, too) | Doc Physics

I work through a healthy 2-mass Atwood Machine problem that my AP Physics 1 students just encountered. There's no friction, but the dang pulley has mass.

+Luke Martin the mass is 3m

Hey Doc! I love you and I love your videos, cz you are such an amazing
teacher. I was just wondering why you considered tension forces to point
downward then upward throughout the video and why you considered tension
forces to be causing torque while I thought it more intuitive that the
weights of the hanging masses were the origin of torque. Thank you in
advance!!ï»¿

+Doc Schuster Well, don't tell anyone but I ran out of the book, and came back to the Muggle world!Anyway, thank you very much for answering me that promptly and that clearly!!!

+Hermione Granger I'm so happy to finally talk to you online, because I must admit I thought you were fictional. Yahoo! Your intuition is fine, but I find it really useful to be very specific about what is causing each force. For instance, the hanging masses are never in contact with the wheel, so they should not be included directly. In the event of a rope that has mass, you'll find that the weight of the dangling rope actually changes the tension throughout it. Ick.

I think an error is present... if T_1-T_2 = (3/2)ma and T_1 = mg-ma and T_2
= 2ma+2mg, shouldn't (3/2)ma = mg - ma - 2ma + 2mg?
Note that 2mg is added, not subtracted....
Regardless, love your teaching style!ï»¿

+YoPoshi No. You need to add the opposite of T_2 as a whole.

1) why did you take T1 as pointing down when you pointed it up in freebody
diagram of m
2) do you always choose one direction of movement positive and make the
other negative?
sorry but these questions have confused me forever and i've never found a
solid answer..ï»¿

+I Zugec thanks man, after 3rd chapter of dynamics, I've pretty much cleared this one up.

+Raj Sondhi I'm going to try to answer your questions. So 1) It's because when you look a body m, the rope has "inertia" and therefore wants to stay in place, and that resisting force is tension and upwards (It also has to do with some things on molecular level but if you go on collage you'll go through that). So on that body diagram tension is upwards. However, By newton third law there is an equal and opposite force OF body ON rope. And THAT force is the one that interacts with PULLEY. So the crucial thing is to differ which body is of your interest.Â 2) Well, you have to actually look your system. Would you agree that here wouldn't be possible that both bodies go downwards?Hope I answered your questions. Cheers :)

I hope that you will begin seeing integrals everywhere after watching this. 'Cuz integrals are actually everywhere. Now you see them.

I am a mathematical and you hurt meï»¿

+Doc Schuster I have a degree in physics :D

+Jfkr Jfkr I LOVE THE MATHS! I have a degree in maths.

+Doc Schuster You have something against the maths. Do you?

+Jfkr Jfkr I thought mathematicals could not feel pain. No?

why is delta x= 0.5(v0+vf)t ??ï»¿

+Gunbnelch Maui It's a formula from kinematics in physics.

Mathematicitians know everything about everything! (17:05) The only reason
physics exists is because of math. ;-)ï»¿

the comment for which I was looking for

How fast does the axle of a bike wheel move? How fast does the BOTTOM of a wheel move? Editor's note - I found this 2012-vintage video on my hard drive.

Okay I love the video! Yet I get stuck I can't wrap my mind around the fact
of how the top is 2Vt yet we are brought up in my physics class
V=rw(romega) so when you ended the video with out clarifying on how the
equation works with the 2Vt instead of just Vt i was left uneasy and
confused. Can you break this down for me?ï»¿

+Doc Schuster so is that is why it's 2r*omega because in the specific example you are looking in on the system?

+sarah phillips Sounds fine. in the tire, you think tangential v is r*omega. Outside, you see that v_t is zero at the bottom, r*omega on the sides, and 2*r*omega at the top.

+Doc Schuster okay I think I figured it out I talked about it with one of my fellow students in my class after presenting him with the video as well, he then said that the velocity would be relative to the ground looking in on the system from two different perspectives. therefore he first perspective being in the system yourself; for example sitting inside a rolling tire. the second perspective being a person observing you rolling in the tire and the tangential velocity would seem different to each observer. Can you further conclude on these statement telling whether or not I'm over thinking things?

+sarah phillips That v from your class is the speed of the outside if the axle is still. Correspondingly, it will be the speed of the axle if the bottom is still, right?

+Refuse to Eat Shiit (tenacious pessimist) The bottom of the body is "instantaneously" at rest. It doesn't mean that some point on the wheel is stuck to the platform all the time.

+Eat Shiit (tenacious pessimist) Try looking at it as a simple motion equation (as he drawn it on the video). The velocity at the bottom of circle (or in real-world application, where the circle is in contact with the ground), is pointing toward the left for rotation, but pointing toward the right for translation. Assuming that both velocities are equal --> v(translation) - v(rotation) = 0. That is why the bottom of the circle is considered at rest.Just remember that this is only true for pure rolling without slippage.

sir i cant picture how the point of contact is at rest.I mean you started
from the zero mark and went on till some distance on the ruler.So the point
has covered some distance in some time then how can it be at rest. :-(ï»¿

+anuj balodi Try looking at it as a simple motion equation (as he drawn it on the video). The velocity at the bottom of circle (or in real-world application, where the circle is in contact with the ground), is pointing toward the left for rotation, but pointing toward the right for translation.

instantaneously it is. as position is changing every moment

A ball rolls (without slipping) up a ramp and then launches into the air. Rotational dynamics and 2D kinematics are both discussed at length. If you just want to ...

Hi Doc,
the solution present to the first problem is close, but not not quite
accurate. I'm referring to the fact that at the point when the sphere
launches from the top of the ramp, the center of mass of the sphere is
still slightly behind and below the point you assume in your analysis.
Have you heard this before, or should I elaborate?
-Doug.ï»¿

+Doug B Hi Doc, guess I caught you at back-to-school time. I was hoping you could take a look at my post and let me know what you think.Thanks-Doug.

+Doug B Sorry, the link to the diagram is https://drive.google.com/open?id=0B__hZELqJu-sS3V1TU4zWU5oVTQ

+Doc Schuster Hi Doc,no, this is not the case where the sphere barely makes it to the top of the ramp and tips/wobbles over the end. I don't think we have that situation in this case, although I have not been able to figure out how to prove it, so maybe we can come back to that one at some point.For this discussion, please refer to the diagram at https://drive.google.com/open?id=0B__hZELqJusS3V1TU4zWU5oVTQ. I was considering the normal force on the ramp and considering at what point it ends. And I believe it ends when the radius of the sphere gets to the top of the ramp (T), and the center of mass (COM) of the sphere gets to point A, normal to the top of the ramp (T) at a distance r. So once the COM of the sphere passes through A, the sphere is in free fall to the floor. So the question is what should be the initial value x0,y0 where the motion of free fall begins. Your analysis assumes an initial value of x0=0, y0=3, and you solve for t where the final value of y, yf=0. Now, of course, it is the COM of the sphere that follows the parabolic path from the top of the ramp to the final location where it will be r above the floor when the radius of the sphere touches the floor. So it seems like you are using a coordinate system with origin at r above the foot of the ramp. In this coordinate system, the point (0,3) is a point r above the top of the ramp, which I have called B, and the COM of the sphere will be at yf=0 when the radius of the sphere touches the floor, as required. I like the coordinate system you have picked, and I will use the same. However there is a problem with the starting point x0,y0=(0,3): the COM of the sphere does NOT pass through the point B. In case there is any uncertainty on this point, consider an arbitrarily fast moving sphere leaving the top of the ramp. I think you can convince yourself that in this case, the COM of the sphere will pass arbitrarily close to point C, (which is directly above T on a line parallel to the ramp through A) and not through point B. Now, nothing in your analysis precludes applying it to a fast moving sphere. So I think you can see it cannot be right to assume the COM of the sphere passes through point B and that this point is the start of free fall. In fact, as above, I believe it is point A which is the location of the COM of the sphere at the start of free fall, since this is the position of the COM of the sphere when the normal force ends. So we need to find the location of A in the chosen coordinate system. Well, I'm going to start by finding the location of A in the coordinate system relative to the base of the ramp at the floor, which I will call the primed coordinate system, and is shown as (0,0)' in the diagram, as that is easier to read off the diagram; and then transform it into the coordinate system described above. Note that the angle BTA is the same theta as the angle of the ramp. So in the primed coordinate system we can see from the diagram that xA' = -r*sin(theta)yA' = 3.0 + r*cos(theta)Now, we can transform to the coordinate system with origin r above the base of the ramp usingxA = xA'yA = yA' - rsoxA = -r*sin(theta)yA = 3.0 -r(1 - cos(theta))and use this for our starting point x0,y0 of free fall motion. So how much difference does this make in the example worked in the video? I'm going to start by re-solving the problem using the initial position of B at (0,3) as in the video, but using "the classic" starting immediately at the start of free fall, rather than the two-step method used in the video, as this should be more accurate since less steps means less rounding errors.Note also, vx0 = 7.61m/s * cos(25) = 6.90m/s and vy0 = 7.61m/s * sin(25) = 3.22m/s.So yf = y0 + vy0*t + 1/2at^2, or0 = 3.0 + 3.22*t - 4.9t^2Using the Quadratic Equation solver at //www.math.com/students/calculators/source/quadratic.htm with A=-4.90, B=3.22, C=3.0, I get t=1.177sec.andxf = x0 + v0x*t = 0 + 6.90m/sec * 1.177sec = 8.121m. Now using A as the starting point and findingxA = -0.2m * sin(25) = -0.0845myA = 3.0m - 0.2m*( 1 - cos(25)) = 2.981mand using x0 = xA and y0 = yA in "the classic" givesyf = y0 + vy0*t + 1/2at^2, or0 = 2.981 + 3.22*t - 4.9t^2Again using the Quadratic Equation solver with A=-4.90, B=3.22 c=2.981, I get t=1.175secandxf = x0 + v0x*t = -0.0845m + 6.90m/sec* 1.175sec = 8.023mSo the difference is about 10cm. So not a huge difference, but I would say not insignificant either. I think you could find other combinations of v0, r, and theta for which the difference could be quite significant. Does that make sense? Please let me know what you think,-Doug.

+Doug B Ok - excellent. I was thinking about this but then decided to ignore it. That will result, if there's friction on the ramp, in speeding up the rotation relative to the center of mass motion a tiny bit? Please elaborate!

Understood everything but the part at the end where wi=wf if friction is 0.
Why is the wi = wf if there is no friction? In the second case, since the
ramp is frictionless, isn't the ball SLIDING up the ramp since there's no
friction to cause it to roll? If it's sliding up the ramp, isn't it true
that it has no rotational motion and so isn't wf=o because the ball is not
rolling but rather sliding up the ramp??ï»¿

+Doc Schuster Just to confirm, is the correct answer for the final part that the sphere travelling up the ramp with no friction will have a lower translational velocity than the case where there is rolling w/o slipping as the KE lost as potential for mgh is taken fully away from the translational velocity and none is taken from rotational velocity whereas in the case of rolling without slipping energy is lost in rotation so less energy is lost in translation resulting in a greater final velocity?

No problem pal :)

+Jack Matthews I thought of that as I posted the question, thx btw â˜º

+David Loop They only have an effect on each other when friction is involved because it will stop the ball rolling as linear velocity decreases. No friction is a theoretical state (i.e you will never find two objects which have zero friction on each other) this means it won't follow the normal laws of physics so you have to use some initiative to visualise what could happen

+Jack Matthews linear velocity is related to angular velocity isnt it?

There will be an increase in gpe as it moves up the ramp but this only affects the linear velocity

+Jack Matthews will wi=wf? I thought there is gravitational potential energy.

+Jordan JonesÂ Happy to help :)

+Jack MatthewsÂ Ohhhhh that makes sense. Thank you!

+Jordan Jones By no 'net' torque he means that the angular velocity will 'not change' (from Newton's 1st law which states an object will remain at rest or continue moving with constant velocity unless acted on by a 'net' external force only with rotational analogues). Try to picture the situation, the ball is rolling with angular velocity 'w' before it meets the ramp, if the ramp has no friction then the ball will continue to rotate with the same angular velocity (it will not suddenly stop rotating because there is no net torque on the ramp to make it do so). Linear velocity will change as a result of the acceleration due to gravity acting down the slope but that is all.

+Doc Schuster I was wondering the same thing... You yourself said there is no net torque acting on the ball as it moves up the ramp; therefore, it does not rotate as it moves up the ramp but rather slides with no friction.

+Doc SchusterÂ Thank you so much ! You are just absolutely AMAZING! Best explanations I have ever come across. :)

+MASTER7able The only place there is no friction in the second problem is the ramp.

But if there is no friction, how is it rolling because what is the force that causes it to roll?

+MASTER7able Yes, in the second situation, it is sliding up the ramp. But not without rotation - in both situations, it starts out rolling w/o slipping on the flat ground.

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Luke Martincommented on 18 Nov 2015