And perfect penmanship

Bow tie carried

+Michel van Biezen If there is not a friction between string and pulley then string must slip over the pulley. That means pulley must not rotate instead of that string is slipping over the pulley. It is not clear how tangent force of the string which has a tension acts on pulley?. Again it is clear if tangent is acting on pulley then friction force must act between pulley and string. Other wise how tangent force acted

Thank you for your kind words.

Glad you found our videos! 🙂

You are welcome. We started doing that on our newer videos.

thislittlemiggy Glad to be of help. Thank you for the comment.

Glad you liked it! 🙂

Because we only look from the pulley's point of view. Had we instead taken a look at the forces acting on m1, the direction of the tension T1 would be upward.

The professor used a pulley as a system. So the tension in pulley's point of view must point down.

because tension is pulling force

Glad you found our videos and found them helpful! 🙂

+pepsi1241 There are lots of examples in this playlist: PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS

You can do it!

T1 and T2 will be different because the pulley has mass and therefore a moment of inertia.

We have uploaded many videos on this topics.

That is correct.

Yes, if the pulley does not have mass and there is no friction, then T1 = T2

It's our pleasure

No, since the blocks move as a system. The block on the left moving down is the same motion as the block on the right moving up since they are connected. That is why the technique in the video is used.

@@MichelvanBiezen i still dont get it

@Mert Kaplan The Tension in that equation is from the perspective of Box 1, not the pulley. You can tell because the masses are not marked "P". So for m1, m1a = m1g - T1, solve T1, T1 = m1g - m1a

Thank you. Glad you found our videos. 🙂

It is better to ignore the internal forces of the system (the tensions) and only concentrate on the external forces acting on the system and causing the whole system to accelerate. Since m1 is larger m1g will be aiding the acceleration (acting in the same direction as the acceleration of the system) and m2g will be opposing the acceleration (acting in the opposite direction as the acceleration of the system) and a can be found by: a = (m1g - m2g) / (m1 + m2)

I was looking into that too for acceleration ( a = (m1g - m2g) / (m1 + m2) ) (Atwood's Machine 'n stuff) While using that equation, the acceleration comes out to 3.27 m/s^2. which is 7% larger than the value calculated in the video. Which result is closer to true value of the systems acceleration? Also, thanks for the FAST reply and video to help students like myself understand physics!

Since the pulley in this example has mass, you must also include the moment of inertia and the energy required to get the pulley spinning, thus the actual acceleration will be less.

+Luis Alva To show yourself that this is correct, draw a free body diagram and use F net = m1 a

Thank you for your comment. We are glad you found our videos.

Yes, our older videos were recorded in mono sound (not stereo) before we figured out what we were doing.

+Bryant Flores Where you given the radius and moment of inertia of the pulley? If yes, you'll have to use the principles seen in the videos of this playlist: PHYSICS 13 MOMENT OF INERTIA APPLICATIONS If not, the you can find the force by taking the torque and dividing it by the radius of the pulley. Then add the force to F net = m total * a

+Aaa Aaa The tension T is only equal to mg when the objects are motionless (or when they are moving at a constant velocity). Let's pause for a moment and think about it. When an object is hanging from a string and not moving, the tension T must be equal to the weight of the object (mg). Now imagine that you pull harder on the string in order to accelerate the object upwards. The additional force must be equal to: F = ma. Therefore the total tension T must now be the sum of the two forces: mg + ma

www.kwantlen.ca/science/physics/faculty/mcoombes/P1101_Solutions/RotDynamics/P08_00006.gif ^^^This image is a similar representation of my question.

Thomas. Not yet. It takes a while to cover every possible topic in physics. I am currently going back and adding additional examples on specific topics. (currently working on conservation of momentum and impulse).

Michel van Biezen Understood. Thanks for the videos they help a lot.

+Gugu Nyambe That depends on the initial conditions when you cut the strings. 1) What are the positions? 2) What are their velocities? Both of those conditions affect how long it takes for each block to hit the ground when the string is cut.

The friction between the rope and the pulley allows the rope to apply a torque (which would not be possible without friction), but the friction does not add torque.

Anyways, excellent video. Especially with how rigorous the explanation was.

The tension will be the same only if we don't take the mass of the pulley into account. The pulley will have a moment of inertia which will cause the tension on both sides to be different.

@@MichelvanBiezen awesome,thanks

You can use Newton's 2nd law. See the videos in this playlist: PHYSICS 5 APPLICATIONS OF NEWTON'S LAWS

Awesome, thanks!

Yes, unless it is a statics problem

sorry i mean anticlockwise as positive

It is better here to only use the magnitude of the torque and not worry about the direction of the torque. We only need the magnitude to find the acceleration.

You are very welcome

Does the pulley have moment of inertia?

no sir.

It is easy to check. Once you have calculated the acceleration, for the larger mass accelerating downward: T = (10)(9.8) - (10)(a) and for the smaller mass accelerating upward: T = (5)(9.8) + (5)(a)

Yes indeed, that is a good method.

+rurutchi The moment of inertia depends on the shape of the rotating object. For a solid disk the moment of inertia is (1/2) mR^2 For a hollow disk it is mR^2 Take a look at the playlist: PHYSICS 12 MOMENT OF INERTIA

Fundamentally, moment of inertia is Σm*r^2. Or in its continuous form, ∫r^2 dm. When you account for the distribution of mass, you often will get a formula for moment of inertia equal to X*M*R^2, where X is a fraction that you get after carrying out this integral. For a thin uniform hoop, X=1. For a solid disk or cylinder around its circular axis, X=1/2. For a sphere, X=2/5. For a thin spherical shell, X=2/3.

Welcome!

Welcome!

The moment of inertia is mR^2 if all the mass is located at distance R from the point of rotation, but for a solid disk the moment of inertia is (1/2) mR^2

Free online Physics courses for CBSE n ISC school's.... Join Now...kzfaq.info/get/bejne/rdmbnth2m762Y30.html

There is no frequency in this type of problem.

The friction (or lack of friction) referenced here is the friction on the bearings of the pulley.

Glad you enjoyed it.

To find the acceleration of the system, this is the best method. To find the position and the velocity of the masses you can use the conservation of energy. There are many examples like that in the playlist. PHYSICS 8 WORK, ENERGY, AND POWER kzfaq.info/sun/PLX2gX-ftPVXWtv0wrHIHTCUrZHUkVDJiZ

Thank you sir... :)

Most welcome

Since the pulley causes the direction of the tension to change it is better to think about it like this: F net = F aiding the acceleration - F opposing the acceleration = (total mass) x ( a)

If I understand this correctly, wouldn't that mean that T1 = ma - mg ---> (ma = T + mg => T = ma - mg)? Both tension (T1) and mg are pointing in the downward direction, which is the direction that the acceleration is pointing, so the Forces are positive (aiding acceleration). By getting T1 = mg - ma, it seems that your original equation was ma = mg - T.

You need to add the subscripts to your masses and T in the equations before I can answer your questions. Also T1 = T2

I suggest you watch several examples in the playlist, including those with the inclined plane and a pulley to see how to work these types of problems.

Sorry, from above: m[1]a = T[1] + m[1]g => T[1] = m[1]a - m[1]g Both T[1] and m[1]g are pointing in the downward direction, so that's why I set up my above equation like I did. Since you got T[1] = m[1]g - m[1] it seems that your original equation was m[1]a = m[1]g - T[1] And I'm currently on the Torque playlist :)

Torque is defines as F x perpendicular distance. However if the pulley is smooth then there is no way to apply a force on the pulley and therefore you cannot apply a torque.

So if pulley is smooth , the pulley will not rotate, the rope just slips on pulley with some acceleration ..im I right sir?

We are making as assumption that the pulley is massless and frictionless, which of course is impossible. (Like we assume no wind resistance, which is also impossible). All pulleys have some mass and some friction. Thus in real life, there is a small amount of torque and a small amount of friction with all pulleys. That said, if there was a perfectly smooth pulley, then the pulley would not rotate, but then if the pulley had no mass, and minute amount of friction would make it rotate.

+TRICKYY11500 Only if the pulley is a solid disk with moment of inertia = I = (1/2) m R^2. The equation can be different with a different moment of inertia.

The moment of inertia of a solid disk is (1/2) MR^2

Those two equations are always correct.

Glad you liked it. 🙂

Because the pulley's mass is not negligible.

As long as there is friction in the pulley or the pulley has mass such that it has a moment of inertia the tension will be different.

Michel van Biezen Since the tensions are different, how is it that the acceleration is the same for the blocks. Ofcourse it's obvious that the acceleration of both blocks must be same, it's stupid to ask but if you consider the constraint relation , where total work done by the string is 0 i.e T1•x+T2•x=0 (dot product) T1x-T2x=0 (cosπ=-1) T1x=T2x T1a=T2a (differentiate w.r.t time) T1 = T2 But we know that T1 is not equal to T2, so what's happening here? I'd understood that tensions must be different and acceleration the same, but then I'd thought of the constraint relation, and now I'm confused If I had to reason I'd say maybe the constraint relation doesn't apply here because the pulley has some mass and exerts frictional force, so total work done by string is not 0?

Only if the objects are not accelerating. If the objects are accelerated you have to add the force required to accelerate the object.

@@MichelvanBiezen Thank you so much