Impulse for pendulum motion help

[Robert123]

Today in class the teacher and I was having an argumentive discussion over where the maximum vertical impulse would be on a pendulum. The question goes like this...
Impulse is defined as change in momentum. At what point during the boat oscillation is the vertical impulse at its maximum.
For the motion of its oscillation, it's swings up to an angle of 60 degree from the vertical (so it doesn't create at right Angle with the horizon) and then swings back down.
Her reasoning was that the greatest change in momentum would be at the top of its oscillation due to the velocity changing direction as well as most of the momentum would be in the vertical instead of the horizontal.
Mine was that it will be a constant change in momentum throughout its oscillation. This is due to the fact that when it does change direction at the top of its oscillation, the speed is relatively slow and that it will even pause for a  split second. Also, due to the forces in the vertical direction, there is always a constant weight force in action, the only force that will vary would be the tension holding the boat (and normal force but I don't believe that will affects its motion).

So, what is the right answer and why? She was trying to explain her logic to the answer for about 15 minutes but even then it didn't change my understanding of why.

Thanks for any help, I haven't done any pendulum motion study before and I know it isn't a part of the study design but it would greatly help my understanding of motion

[availn]

OK this is really confusing me. Firstly, what is a "boat oscillation"? Secondly, how can you have impulse at a "point"? Impulse is change in momentum, or force by change in time, so how do you measure it at a point?

[Robert123]

OK this is really confusing me. Firstly, what is a "boat oscillation"? Secondly, how can you have impulse at a "point"? Impulse is change in momentum, or force by change in time, so how do you measure it at a point?

Sorry that's my bad wording, the object that is oscillating is a boat (this is a question about a ride at luna park) that is swinging in a pendulum motion. The question could be rephrase as 'what stage of its oscillation would the impulse in the vertical plane be greatest'

[availn]

Sorry that's my bad wording, the object that is oscillating is a boat (this is a question about a ride at luna park) that is swinging in a pendulum motion. The question could be rephrase as 'what stage of its oscillation would the impulse in the vertical plane be greatest'

Ah, that makes more sense. But still, you cannot measure impulse at a stage or point. Impulse is defined as the change in momentum between two points, or the area under a force-time graph between two points. You can't just replace △t with dt either, because then you just end up with the momentum, when what we want is the change in momentum.

[Robert123]

Ah, that makes more sense. But still, you cannot measure impulse at a stage or point. Impulse is defined as the change in momentum between two points, or the area under a force-time graph between two points. You can't just replace △t with dt either, because then you just end up with the momentum, when what we want is the change in momentum.

With this question though, it is not a caluclation one but rather a conceptually reason one. So using motion concepts, you need to determine where the greatest momentum change in the vertical will occur.

Sorry about the arbitrariness of the post.

[EspoirTron]

Okay so the question you are asking has no calculated answers?
This is how I would break the situation down.
Impulse is defined as the change in momentum, which is delta p/ delta t (sorry I don't know how to insert the symbols).
Well the thing here is as avalin said we cannot calculate the instantaneous change in momentum as that is not possible as we are asking for a change in, which is given by pf-pi. You mentioned oscillating, well if this pendulum was 'oscillating' perfectly it would be uninterrupted motion, so I would presume constant velocity. So the Impulse of this pendulum over this period of time would be the same value over and over again, depending on what you picked for the time value.

[availn]

With this question though, it is not a caluclation one but rather a conceptually reason one. So using motion concepts, you need to determine where the greatest momentum change in the vertical will occur.

Sorry about the arbitrariness of the post.

OK, I'll try this your way, but I'm not sure about my answer, take it with a pinch of salt.

If you want to try to find an instantaneous point where impulse is greatest (which I will say again, I don't think you can calculate), you would have to find where the momentum is changing (the fastest), because impulse is change in momentum. Velocity is directly proportional to momentum, so where the acceleration of the the bob in the horizontal plane is greatest, the change in momentum (per second!) would also be greatest.

What is important to note is that the rate of change of momentum is NOT constant! This is because the boat is oscillating, and all oscillating objects will have a different force on them at different positions. At some point between the highest and lowest point of the pendulum, the net vertical force is 0. When the bob is below this point, the net vertical force is directed upwards, and when the bob is above this point, the net vertical force is downwards. The force will have to be largest at either the highest point or the lowest point, so we'll need to figure out which one it is.

At the highest point, the net force can be found through trig. First you find the net force acting on the pendulum, which will be perpendicular to the string.

F_net = mg·sin(θ_max)

Now we need to find the vertical component of this force.

F_netv = mg·sin²(θ_max)

This force is directed straight down.

At the bottom, the net force is straight up, and is the centripetal force.

F_c = mv_max²/l

The maximum velocity for a pendulum of θ_max < 10deg I think is equal to θ_maxl·root(g/l) so

F_c = mgθ_max²

As we have already restricted θ_max to small angles, we can use small angle approximation on F_netv to equate sin(θ) with θ. Through this we find that:

F_netv = mgθ_max² = F_c

I guess that this means that the vertical force acting on the bob at both the bottom and the top of the pendulum swing is equal for small θ. So rate of change of change of momentum would be greatest when the pendulum is in the vertical and its highest positions...

Keep in mind that this is only for small θ, and that this is where the rate of change of momentum is greatest. This is not the impulse! Impulse is just the change in momentum and, once again, requires two times! If θ becomes larger, I get the feeling that the rate of change of momentum will be greater at the bottom, than at the top.

[Robert123]

Thanks you so much for that
It helped clear up her explanation of why it should be up top

[lzxnl]

If you approximate pendulum motion as angular simple harmonic motion, i.e. motion of a spring, you can see that the acceleration is actually proportional to the angular displacement.
Using torques, we have net torque=mg*length*sine angle made with vertical=rot. inertia (just a constant) times angular acceleration
Multiplying both sides by the length of the pendulum, we have mg*length^2*sine angle=rot. inertia times linear acceleration
Hmm. It's a lot of constants isn't it? If we bunch them together, as the lengths and the inertia are all constants, we have mg*constant*sine angle=linear acceleration

This means that the magnitude of the linear acceleration depends on the sine of the angle and so the larger the angle, the larger the sine value and hence larger the linear acceleration. Therefore, your teacher was correct.