I am a highschool physics teacher and I find video very helpful. You have earned one more subscriber

Currently in graduate school and this is really helping build my thinking skills in towards problem solving. Knowing exactly what questions to ask when analyzing a problem.

@DrBenYelverton

2 жыл бұрын

I'm glad it was helpful! I do like this problem as it seems simple at first but ends up requiring some creative thinking.

@AnshumanDVD

Жыл бұрын

Hi, would you try solving the case where for a given speed, the impediment height:radius is to be found for maximum post bump jump

Nice! 😁

thank you for your nice description... clearly, since the collision is "quick" gravitational torque about the climbing point is negligible and angular momentum about that point is still conserved. Afterwards, it's just a conversation of energy... would be interesting to see how the diminishing gravitational torque changes and ultimately capital omega becomes zero.

@daffavirwandy2322

Жыл бұрын

Wait , i thought the gravitational torque are implied on the conservation of energy? From the mgh part?

Really helpful sir, thanks for the video, hope you release more videos like this😀

@DrBenYelverton

Жыл бұрын

I'm planning to post many more classical mechanics videos in the future!

You make very good videos

@DrBenYelverton

2 ай бұрын

Thanks, I'm glad you're enjoying them!

Can you make a simillar video to slide ball out of a hole

Please explain why friction force in contact force between ball and step corner point is not considered? (I.e considered contact force = normal force)

Made some cases and just wanted to do a qualitative analysis of them, kindly tell if my thoughts are correct: i) Point of impact is frictionless -> The ball will continue to rotate forever, and only translational motion will be hindered.. The angular velocity will remain same as before collision, but translational velocity will become 0. ii) Point of impact has friction, but not sufficient to prevent slipping --> The ball will not be able to climb up the top, but rotate at the spot and thus eventually will stop due to the torque produced by kinetic friction at each point repeatedly. iii) Point of impact has sufficient friction to prevent slipping: The ball will be able to climb up to the top just like in the question.. * Lastly sir kindly clarify one thing, while solving the ques., did you consider: the translational velocity (just after collision) to become zero due to the impact such that the ball will be having only the angular velocity, whose rotational energy too will be converted to the potential energy, and thus finally ball would be at rest at the top.... to find the minimum speed possbile as per the question.

@DrBenYelverton

3 ай бұрын

This all sounds good to me, under the assumption that the coefficient of restitution is zero!

@asharani759

3 ай бұрын

Thanks for replying! Sir kindly check the last question too 😊

@DrBenYelverton

3 ай бұрын

Yes - that's a good summary of the logic I used in the video!

Thanks for the amazing video. I just started to study physics and I havent had any exams yet, do such difficult problems normally appear in the exams?

@DrBenYelverton

2 жыл бұрын

Glad you enjoyed the video. This sort of question could certainly appear in a university-level Physics exam, but it would probably be one of the more difficult ones and I would not expect the whole exam to be this hard!

Thanks for your video, it really helped me! I was wondering if the instantaneous center of zero velocity isn't the point where the wheel makes contact with the ground? (Instead of its center of gravity)?

@DrBenYelverton

2 жыл бұрын

Thanks for watching! Yes, in the frame of the step, the point on the wheel that's in contact with the ground has zero instantaneous velocity - this is where the v = rω condition comes from.

Really nice video! Besides, I think it's worth mentioning that there's also an upper limit for the initial velocity of the ball, because the ball mustn't lose contact with the step while climbing it.

@DrBenYelverton

Жыл бұрын

Thank you! You're right, this certainly relies on the assumption that it stays in contact. Presumably the upper limit you mention would be material-dependent and determined by parameters like the coefficient of restitution, which would complicate things quite a bit!

6:00 can someone explain why we took mv(r-h ) and what to do if the height is equal to the radius

Sir I have One doubt.how you can use law of conservation of angular momentum when sphere collide with small step?you could have done if there is no external torque about the axis on "h" meter height but did you notice the torque by weight of the sphere about the axis you took?

@DrBenYelverton

10 ай бұрын

This is fine because the collision happens essentially instantaneously. There's a torque due to the weight, but it produces a negligible angular impulse during the collision since the time interval is small, so we can equate the angular momenta immediately before and after the impact.

Sir, an addition to this problem is that for a given speed U, what should be the 'impediment height'/radius (which is

@DrBenYelverton

Жыл бұрын

That's an interesting question - can we assume that there's no friction between the step and the ball? If so, I can imagine approaching this by resolving the incoming velocity in the radial and tangential directions, then reversing the direction of the radial component (assuming no energy losses) and treating the ball as a projectile after the collision.

@DrBenYelverton

Жыл бұрын

I've just worked through this using the method outlined above and ended up with an answer of (2 - √2)/2, independent of the initial speed. I'll try to get a video recorded on this problem in the near future!

sir, if we are using conservation of angular momentum about point of contact then torque should be also balanced about point of contact, but there is no such force balancing torque by gravity could you please explain sir

@DrBenYelverton

Жыл бұрын

Good question - the angular impulse due to gravity can be neglected because we're considering the angular momentum immediately before and immediately after the ball collides with the step. Since these two times are separated by an infinitesimally small duration, the change in angular momentum is negligible even though the gravitational torque itself is not.

is that possible to do it using Lagrangian Mechanics?

@DrBenYelverton

4 ай бұрын

I've never used Lagrangian mechanics to handle collisions so I'm not sure! There is probably a way to do it but it won't be straightforward.

Sir, how to find maximum velocity?

@DrBenYelverton

Жыл бұрын

The approach would be a little different as it requires a different set of assumptions - mainly a nonzero coefficient of restitution. I haven't worked through this in full, but you could try resolving the motion into components parallel and perpendicular to the contact force, and write down the final velocity by considering the impulse exerted by the step. The result would depend on the coefficient of restitution (which in reality is velocity-dependent), and coefficient of friction as well if you wanted to include that. Something like the method I used in this similar problem: kzread.info/dash/bejne/poqKsMyncduzqtY.html

Had a small doubt while solving. I feel like saying this is wrong but I am not able to see why. pls correct me if this is indeed correct. "Because the the ball uses the step as a point of contact, the component of the velocity of the ball along the line joining center to the point of contact becomes zero. the only component left is the one perpendicular. so if the initial v = v_perp + v_para i can just make v_para 0 and use v_perp as the velocity of the center of mass right after collision" where is this argument wrong? or is it right? i say it's wrong because using this got me the wrong answer

@DrBenYelverton

7 ай бұрын

The parallel component should indeed become zero, but the perpendicular component changes. In this problem there must be a frictional force acting at the point of contact, otherwise the ball would just keep spinning with the same angular velocity. We make the assumption that there's enough friction to ensure no relative motion between the point of contact on the ball and the corner of the step. The friction will affect both the angular velocity and the tangential linear velocity. For an example of where that method does work, see the following video, where we assume that the ball bounces off the step without friction: kzread.info/dash/bejne/poqKsMyncduzqtY.html

How could angular momentum be conserved? Doesn't gravity act at the center of mass of the ball, thus inducing a torque when the ball rotates about the contact point?

@DrBenYelverton

Ай бұрын

It's not conserved while the ball is actually rotating about the contact point, but we're considering the angular momentum at two times separated by an infinitesimally small interval, before and after the moment of impact. So even if there's a non-zero torque, the angular impulse experienced during that time is essentially zero.

@Jacob.Peyser

Ай бұрын

@@DrBenYelverton Thanks for explaining!

I don’t think angular momentum is conserved. That’s because as the ball move upwards, it also loses velocity, and for the non-slipping condition to apply, it must lose angular velocity too (remember that v=wr), so there must be a torque (in this case there must be a friction force acting on the ball at the corner)

@DrBenYelverton

Жыл бұрын

There is indeed a frictional force, but it doesn't produce any torque about the point of contact with the step because it acts straight through that point (i.e. its perpendicular distance is zero). The angular momentum about that point is therefore conserved.

Sir please reply... We applied conservation of angular momentum. About the point of impact.. And we used only 2/5 mr^2 w.... Why didn't we write Moment of intertia about point of impact...

@DrBenYelverton

Жыл бұрын

That's because immediately before the collision the ball is actually rotating about its own centre of mass. So, the initial angular momentum has two parts, one due to this rotation about the centre of mass and another due to the linear motion. After the collision you do need to use the moment of inertia about the point of contact because it's now rotating about that point. The key idea is that the point it is rotating about is different before and after the collision.

@future_iitian24

2 ай бұрын

@@DrBenYelverton sir please reply. Shouldn't mv(R-h) be on the right hand side of momentum consv. equation? Momentum due to spin is about it's center, but mv(R-h) is about point of contact. So how can we add these two? They are written about different points! I think mv(R-h) should be right side of the equation.

And waht if the step was higher tahn radius is there a way with the same codition a ball could get up there ?

@DrBenYelverton

5 ай бұрын

I don't think so, because the impulse exerted on the ball would then be pointing directly to the left, causing the ball to either come to a complete stop or start moving backwards depending on how much energy is lost in the collision.

@kekso2205

5 ай бұрын

@@DrBenYelverton So even the agular momentum go to zero, after colision, would not be ball able to roll up the wall with the angluar momentum. (sorry for my grammer english is my second language)

@DrBenYelverton

5 ай бұрын

I suppose if there's enough friction it could roll upwards and make it to the top, but there would also have to be zero coefficient of restitution for it to grip on to the wall, and that would be hard to achieve in reality.

@kekso2205

5 ай бұрын

@@DrBenYelverton Thanks, in the example it statet that friction was high, I was not sure if it was possible, tanks for answer l.

I Have question, it seems at climbing, you didn't use conservation of energy because the energy isn't conserved on that case right? and you used angular momentum conservation instead. but why are you neglecting torque such as gravity?

@DrBenYelverton

Жыл бұрын

Yes, some energy must be lost in order for the ball to remain in contact with the corner of the step, which is why we can't use conservation of energy. For your second question - I'll paste here the same response that I gave to another commenter a while ago: the angular impulse due to gravity can be neglected because we're considering the angular momentum immediately before and immediately after the ball collides with the step. Since these two times are separated by an infinitesimally small duration, the change in angular momentum is negligible even though the gravitational torque itself is not.

@darwinvironomy3538

Жыл бұрын

@@DrBenYelverton i've been thinking about it, so that's why you are using the conservation of energy in the end right? i think this is enough for me to derive the opposite, ball climbing down a step. thank you.

@DrBenYelverton

Жыл бұрын

Yes, there's an instantaneous loss of energy at the moment of impact, but it's then conserved during the second phase of the motion when it's pivoting about the corner.

@darwinvironomy3538

Жыл бұрын

@@DrBenYelverton how do we know from the start that the energy isn't conserved there?, at first, i didn't figure it out why the energy isn't conserved.

@DrBenYelverton

Жыл бұрын

I'll give you two different ways of understanding this, from physical and mathematical perspectives: Physically, it's because the contact force between the step and the ball does work to slow the ball down during the collision, and therefore changes its energy. The work done is just given by (force) x (distance moved in direction of force). As the ball hits the step it has a component of motion in the direction of the contact force and continues to move a little in that direction as it compresses, which is why the work done is non-zero (though in the model we ignore the details of this compression and treat it as a hard sphere). Once the ball starts pivoting about the step, though, the contact force is not doing any work because the instantaneous direction of motion of the ball is always perpendicular to the force (the same idea as gravity in a circular orbit). Mathematically, requiring the ball's energy to be conserved would simply be inconsistent with the ball staying in contact with the step. In either case, we know that angular momentum must be conserved as discussed above. In the video we also made the assumption that the ball "sticks" to the step, and this assumption leads to one specific value of Ω. Introducing the idea of energy conservation would give yet another equation for Ω in terms of v, and the system would now be mathematically overdetermined and hence inconsistent.

Sir ,the angular momentum about the taken point of contact may not be conserved because the torque produced by gravity should be considered

@DrBenYelverton

Жыл бұрын

Although there's a gravitational torque, it produces a negligible change in angular momentum in this example because we're considering two times separated by an infinitesimally small interval (i.e. immediately before, then immediately after the collision).

@kkarthik797

Жыл бұрын

Understood sir

Why can't we include the translational kinetic energy into the equation?????????

@DrBenYelverton

2 жыл бұрын

This is because after the ball comes into contact with the step, its motion is purely rotational about the point of contact.

@zbrg7245

Жыл бұрын

@@DrBenYelverton Thank you for replying. Your comment helps me a lot, I understand it now. The reason we cant include translational kinetic energy is due to the fact its axis of rotation does not move while it is climbing the step. We could only include the translational kinetic energy when the axis of rotation is also moving with the object.

@DrBenYelverton

Жыл бұрын

@@zbrg7245 Yes, exactly - happy to help!