Pulley with two masses

[@manuboker1]

Wonderful lecturer that makes physics fun and understandable !!!

[@bindukurian3905]

your videos are really helpful, thank you.

[@sababatamanna2117]

Thank you so much, sir!

[@carstenschluter3446]

big up sir youre simply the GOAT

[@thelongjourney2671]

Hi Dr.Anderson, I am a high school student in Turkey. I don't understand this topic when i was listen in our physics lesson. I tried all kind of materials but i wasn't understand. Today i met with your lessons, Now i see most of our teachers has been memorize the formulas. You are the king of against rote system in education. This is the reason of why me and others in my ages hate physic. If I didn't met with your videos I could be hater of physic. I hope one day we will meet. Thank you for everything Loves from Anatolia E.Altintop

[@aarya1938]

yeah you are right

[@dilaradatl4710]

I feel you bro

[@Demi305]

Well said my friend!

[@debapamtalapatra3955]

Really your explaination is amazing

[@immaemi2565]

Thankk you wo much. Youre so good at explaining

[@yoprofmatt]

im maemi, You're very welcome. Glad you're enjoying the videos. You might also like my new site: www.universityphysics.education Cheers, Dr. A

[@ValentinA-nx9wm]

At 4:33 when m2a and m2g pull in the same direction, how can be both of them positive? seems wrong to me. I have m2a = T - m2g

[@thorragnarok9181]

Thank you! Very helpful!

[@arunfz]

sir , i have one doubt . how tension in both sides of the pulley is same ? then how does the pulley rotate here ? frictionless means the pulley surface and rope plays zero friction each other and the string just slides over pulley without causing it to rotate ? whats the importance of the mass of pulley here ? i felt that the difference in tension is causing the pulley to rotate .help me !

[@carultch]

The pulley is an idealized pulley, which means its mass is negligible, and the friction of its bearing is negligible. This means that the difference in the magnitude of tension on both sides of the pulley, is negligible compared to the overall tension in the strings. If you wanted to account for the mass of the pulley, you'd need to introduce Newton's second law in rotational form with the torque due to the difference in tension, and use the rotational inertia of the pulley to relate this to the angular acceleration of the pulley. You'd also set up a constraint to relate the linear acceleration of the rope and masses to the angular acceleration of the pulley, so the rope moves without slipping. Suppose the tension on block 2 were 100 Newtons, and the tension on block 1 were 99 Newtons. The 1 Newton difference that causes the pulley to accelerate is insignificant enough, that you can still get a decent enough answer by assuming a massless and frictionless pulley. Otherwise it introduces more equations to establish, and complicates the problem.

[@junaidmughal3806]

love u prof!

[@kaeez]

Dr. A, I've always wanted to ask this so here I'm doing it finally; do you actually write backwards or is the footage mirrored in post? I thought it had to be the latter but then it struck me that if you were mirroring it in post, then how would the students in your classroom whom you appear to interact with in many of your videos be able to decipher any of what's being written? I know you probably get asked this a lot and is tired of answering them but I would appreciate a reply. Thanks in advance.

[@w0nnafight]

the stutdents are given glasses that mirror what they see

[@iksanmaulana1295]

hello sir,how i can to calculate with out ignoring the friction force of the pulley?

[@carultch]

You would maintain two separate tensions on both sides of the pulley, and introduce additional equations to account for the friction and inertia of the pulley. You would also need the background of rotational mechanics to account for these factors, which aren't introduced to the class yet, at the time this lecture happens. If there were mass of the pulley, but not friction, we would have the following equation relating the torque on it to the angular acceleration: I*alpha = (T2 - T1)*R and angular acceleration to linear acceleration: a = alpha*R With friction on the pulley, we'd have the following equation, with B as the frictional torque in the pulley's bearing. I*alpha = (T2 - T1)*R - B The frictional torque B is given by the torque of the frictional force at the inner radius (r), which is equal to mu_k*N B = mu_k*N*r The normal force of the bearing pushing up on the pulley is the sum of the two tensions, plus the pulley weight (mp*g): N = T1 + T2 + mp*g Put it all together: I*alpha = (T2 - T1)*R - mu_k*r*(T1 + T2 + mp*g) a = alpha*R m2*a = m2*g - T2 m1*a = T1 - m1*g You would be given values for: R, the outer radius of the pulley, where the string acts on the pulley r, the effective inner radius of the pulley's bearing g, gravitational field masses m1, m2, and mp I, the rotational inertia of the pulley mu_k, the kinetic friction coefficient of the bearing You would solve these four equations, for the four unknowns of T1, T2, a, and alpha.

[@iksanmaulana1295]

@@carultch thank u, its been 3 tears, i appreciate it..

[@TaylorSaunderDrums]

Hi Dr. Anderson! i'm confused on one issue in the problem: shouldn't m1g be positive because it is accelerating in the upward y direction?

[@yoprofmatt]

Hi Taylor, Remember that m1g is a force. The direction of that force is opposite the acceleration (which is up), thus the negative sign on m1g. The equation "T-m1g = m1a" says the following: tension "T" is trying to accelerate the mass in the same direction as "a", while gravity "m1g" is trying to accelerate the mass in the opposite direction to "a". Hope this helps. Cheers, Dr. A

[@iksanmaulana1295]

sir i have a question,how i can to calculate the acceleration with out ignoring the frictional forces of the pulley?

[@iscoto4914]

@@yoprofmatt this helped!!

[@urvivallapareddy2206]

i love u merci merci

[@yoprofmatt]

Urvi Vallapareddy, You're very welcome. Glad you're enjoying the videos. You might also like my new site: www.universityphysics.education Cheers, Dr. A