after five days of watching nonstop (with breaks for eating and sleeping and stuff OBVIOUSLY, plus pausing to take notes, but otherwise i havent been doing much else).... only ~6 hours of ur ap physics 1 playlist left.... thank u for ur service king im ur number 1 fan for real
Thank you so much! I bought the ultimate review packet and the RUclips videos combined with the URP make my learning much easier! Especially concerning how my teacher rarely teaches in detail.
And we are only a month away from the exams and I still do not understand many topics my teacher didn't teach in detail. This really helped me review most efficiently and effectively!
You're amazing. I'm telling everyone I know about your channel. So insightful, you were answering the questions I was having right as I was thinking about them.
I think u have more than enough videos already for even jee ....as i believe ap physics have almost same syllabus....... So i think u must add them too in ur playlist of jee ......untill u ll make dedicated videos about them...... As it ll help alot to students like me......
The two pulleys would share a common angular velocity (and angular acceleration) by virtue of being part of the same rigid body, but have different tangential velocities and tangential accelerations of the spooling string, that are each proportional to their respective radii. You would also need to set up a relationship between the two angular accelerations based on the ratio of pulley radii. As an example, suppose mass 1 is 10 kg, and is connected via a massless string to the left side of pulley #1 with a radius r1 = 8 cm. Mass 2 is 5 kg, and is connected via a massless string to the right side of pulley #2, which has a radius r2=4 cm. The two pulleys are rigidly attached to a shared axle mounted on frictionless bearings, and the entire assembly of the axle and pulleys has a rotational inertia of 0.012 kg-m^2. Calculate the unknown accelerations, tensions, and angular acceleration. N's 2nd law on m1, assuming it descends: m1*g - T1 = m1*a1 N's 2nd law on m2, assuming it ascends: T2 - m2*g = m2*a2 Torque on the pulley, and rotational version of N's 2nd law: T1*r1 - T2*r2 = I*alpha Relation to alpha of a1 and a2 a1 = alpha*r1 a2 = alpha*r2 5 equations, 5 unknowns (alpha, a1, a2, T1, and T2). Solutions: a1=5.6 m/s^2, downward a2=2.8 m/s^2, upward alpha=70 rad/s^2, out of the page T1=42 N T2=63 N
They do, but indirectly. The masses are connected to the pulley through the string segments. Since the tensions in the string segments are what act on the pulley, we don't need to further account for the masses, when summing torque on the pulley. We already did that by accounting for the tensions, each applied by these masses.
@@carultch Well, to be more precise, the total torque on the pulley about the central axis is due to the static friction between the rope and the pulley. When we sum these forces, mathematically, it's equivalent as if it is the tension force that is producing the torque. Unfortunately, this is omitted from most if not all high school textbooks.
@@Test-ph1vk True, it is static friction that is really causing the torque. This is probably ignored from most high school level textbooks, because it is an extra detail that doesn't really contribute to the main substance of the problem. This is usually introduced in freshman-level engineering classes in college. There is an equation called the capstan equation that governs the upper limit of tension ratios on opposite sides of a string wrapped around a cylinder, that is derived from static friction in this example. Named the capstan equation, because a capstan on a sailboat is an application of this formula. Where you would want to model this detail in this problem, is if you were asked to determine limits to the hanging masses, before the string would skid on the pulley and roll-without-slip is no longer valid.
Yes. Unit 5: Basic concepts of rotational motion; moment of force, torque, ... moment of inertia. You can find it in my playlist here: ruclips.net/p/PLPyapQSxH6mYV7H8o6y44tOOe7-BS-HbV
Are you assuming that the string is massless? Cuz then only F(T1) acting down on the pulley = F(T1) acting up on the mass. Cuz if it weren’t the case, a little piece of string would have an infinite acceleration
after five days of watching nonstop (with breaks for eating and sleeping and stuff OBVIOUSLY, plus pausing to take notes, but otherwise i havent been doing much else).... only ~6 hours of ur ap physics 1 playlist left.... thank u for ur service king im ur number 1 fan for real
Wow. I've got a lot of videos. I think there are 50+ hours of AP1 videos. I hope you learned a lot! (And "liked" every one of those videos. 😇)
Explaining every step and even predicting students mistakes are so helpful. Thank you so much
You are welcome.
i love this channel!! The hard works shows my man!
Thanks!
Thank you so much! I bought the ultimate review packet and the RUclips videos combined with the URP make my learning much easier! Especially concerning how my teacher rarely teaches in detail.
And we are only a month away from the exams and I still do not understand many topics my teacher didn't teach in detail. This really helped me review most efficiently and effectively!
That is wonderful to hear. Tell all your friends and more!!!
This is so entertaining and I'm actually laughing while learning physics! Thanks for explaining everything so clearly.
I cannot think of a higher compliment. Thank you!
Any chance you could help me out by doing what I ask people to do in this video? bit.ly/2y4tOCA It would be a great way to show your appreciation!
You're amazing. I'm telling everyone I know about your channel. So insightful, you were answering the questions I was having right as I was thinking about them.
Thanks for the love!
Great video, this helped me with a problem I was struggling with. thank you
I too enjoyed learning with you
Thank you for your video! I am taking the AP Physics 1 exam tmr, this video helped me so much 🤩
Good luck tomorrow!
how was it
i have mine in a few days, im doing the june makeup
I think u have more than enough videos already for even jee ....as i believe ap physics have almost same syllabus....... So i think u must add them too in ur playlist of jee ......untill u ll make dedicated videos about them...... As it ll help alot to students like me......
And in the description of playlist just let us know which topic of the chapter is left....... That's it
Thank you
Thanks for this one!
You are welcome!
Hey could you make a video on how to linearize data and graphs so that you can use the slope of the straight line to find a value
I do something similar to that in this video: www.flippingphysics.com/graphing-power.html
What if pulleys with different radius share the same rotational axis?
Thanks.
The two pulleys would share a common angular velocity (and angular acceleration) by virtue of being part of the same rigid body, but have different tangential velocities and tangential accelerations of the spooling string, that are each proportional to their respective radii. You would also need to set up a relationship between the two angular accelerations based on the ratio of pulley radii.
As an example, suppose mass 1 is 10 kg, and is connected via a massless string to the left side of pulley #1 with a radius r1 = 8 cm. Mass 2 is 5 kg, and is connected via a massless string to the right side of pulley #2, which has a radius r2=4 cm. The two pulleys are rigidly attached to a shared axle mounted on frictionless bearings, and the entire assembly of the axle and pulleys has a rotational inertia of 0.012 kg-m^2. Calculate the unknown accelerations, tensions, and angular acceleration.
N's 2nd law on m1, assuming it descends:
m1*g - T1 = m1*a1
N's 2nd law on m2, assuming it ascends:
T2 - m2*g = m2*a2
Torque on the pulley, and rotational version of N's 2nd law:
T1*r1 - T2*r2 = I*alpha
Relation to alpha of a1 and a2
a1 = alpha*r1
a2 = alpha*r2
5 equations, 5 unknowns (alpha, a1, a2, T1, and T2). Solutions:
a1=5.6 m/s^2, downward
a2=2.8 m/s^2, upward
alpha=70 rad/s^2, out of the page
T1=42 N
T2=63 N
If the pulley is frictionless, why will it rotate? Won’t the strings just glide over it?
A "frictionless" pulley has a frictionless axle only.
The mass of a string is so small relative to everything else as to be negligible.
why does the pulley have to be frictionless
The addition of friction in the pulley would add a torque caused by friction to the net torque equation which complicates things quite a bit.
thankyou so much!!@@FlippingPhysics 💛
why do the hanging masses not contribute to the torque ?
They do, but indirectly. The masses are connected to the pulley through the string segments. Since the tensions in the string segments are what act on the pulley, we don't need to further account for the masses, when summing torque on the pulley. We already did that by accounting for the tensions, each applied by these masses.
@@carultch Well, to be more precise, the total torque on the pulley about the central axis is due to the static friction between the rope and the pulley. When we sum these forces, mathematically, it's equivalent as if it is the tension force that is producing the torque. Unfortunately, this is omitted from most if not all high school textbooks.
@@Test-ph1vk True, it is static friction that is really causing the torque. This is probably ignored from most high school level textbooks, because it is an extra detail that doesn't really contribute to the main substance of the problem. This is usually introduced in freshman-level engineering classes in college.
There is an equation called the capstan equation that governs the upper limit of tension ratios on opposite sides of a string wrapped around a cylinder, that is derived from static friction in this example. Named the capstan equation, because a capstan on a sailboat is an application of this formula. Where you would want to model this detail in this problem, is if you were asked to determine limits to the hanging masses, before the string would skid on the pulley and roll-without-slip is no longer valid.
Professor is this for jee?
Yes. Unit 5: Basic concepts of rotational motion; moment of force, torque, ... moment of inertia. You can find it in my playlist here: ruclips.net/p/PLPyapQSxH6mYV7H8o6y44tOOe7-BS-HbV
Are you assuming that the string is massless? Cuz then only F(T1) acting down on the pulley = F(T1) acting up on the mass. Cuz if it weren’t the case, a little piece of string would have an infinite acceleration
Yes, we do assume a massless string.
Wait, are guys wearing headbands a thing?
If I'm doing it, yes.