at what speed do you turn your cart? :)

[mamobono]

thanks for the answers.
It will be for the year that we have in common but I like + your solution that uses the energy budget, since we are talking about a dynamic system.
I did for the truth, believe me, I realized that the mass appeared on both sides but, as I explained, it seemed to me to be precise to insert the rotational inefficiency energy of the cart given the premises I had done (loss of energy caused by the shock in relation to the position of the impact point).
Surely this will decrease the energy available to the tipping and increase the minimum speed to the impact.
manage a dynamic system by comparing static pairs and work albeit from the dimensional point of view it is correct I do not like it, but it is my point of view perfectly not shared.
the two methods find a balance at very different speeds one double of the other 0,8/1,78 m/s
which is 2.88 km/h or 6.4 km/h which makes a lot of difference.
in any case the speeds are such that there would be corrections to the load to make the system more stable.

greetings

 

[ceschi1959]

thanks for the answers.
It will be for the year that we have in common but I like + your solution that uses the energy budget, since we are talking about a dynamic system.
I did for the truth, believe me, I realized that the mass appeared on both sides but, as I explained, it seemed to me to be precise to insert the rotational inefficiency energy of the cart given the premises I had done (loss of energy caused by the shock in relation to the position of the impact point).
Surely this will decrease the energy available to the tipping and increase the minimum speed to the impact.
manage a dynamic system by comparing static pairs and work albeit from the dimensional point of view it is correct I do not like it, but it is my point of view perfectly not shared.
the two methods find a balance at very different speeds one double of the other 0,8/1,78 m/s
which is 2.88 km/h or 6.4 km/h which makes a lot of difference.
in any case the speeds are such that there would be corrections to the load to make the system more stable.

greetings

the correct result is what I posted to post 25 i.e. 2,57 m/s and comes from the principle of virtual works l=mr* from the fact that it must be exceeded by the kinetic energy possessed by the system. think about the real case, think instead of the motorcycle cart to the standing cart against an obstacle and push it where the horizontal force is applied, as soon as you win the resistant moment you have to lighten the thrust otherwise this turns over. In fact if you continue to push with the same force the arm of the horizontal force increases while it decreases the vertical component of the applied weight and therefore the resistant torque.
to finish www the '59 il

 

[mamobono]

hi, but in the calculation you do to the post 25 the work of the couple does not multiply it for dalfa 0.39 rad (which would be 25° = sin-1(730/1758) but use the full pair 560 nm which then equals the kinetic energy etc.
v=(560*39*2/200)^0.5=1.47 m/s
value quite similar to that calculated to
post 47 v=(122*2/76.7)^0.5=1.78 m/s

I would say that now I am satisfied, 2 methods which come to similar results
Hello www the '59

 

[reggio]

Good morning guys, I wanted to thank and do the point
You will forgive me if I miss some doubt but given the amount of answers I will need next weekend to read them and understand them all and make them mine
above all I care to understand why of the difference of result between @marco ferraroni _ e @pollo .. above all the first two, in my littleness, I think I have understood them and convince me ..

However, the force of the man I indicated it to simplify (I hope I did not unleashed someone) or, not knowing in reality the type of plan on which the wheels run I put a high value (when I will come to understand and transform in formula excel everything, I will enjoy to "play" with the various numbers) in order to consider that the cart also flows on rough ground (in effect 200kg on wheels with coefftivo).

when the man realizes to impact a rock/muretto/gradino, he will certainly stop pushing, it is that moment I wanted to analyze, to determine at the end,
- if the distance 730mm can be lower or should be greater, considering a soil also not smooth (fs=50kg).
- if the speed (as you have explained to me: deceleration) - (doubt: but if the man hits the rock at 5km/h instead of 2km/h, the "rebalt" deceleration will not be different?) of 4km/h type of man step (we assume that the man does not rush 200kg) is excessive and causes the tipping or is acceptable and does not make the cart.

kg or newton: it is right to remember 9,81n=1kg - in my case the approximation to 10n=1kg served only to make the calculation more "absorbable/ understandable" - but of course, better not to round

I hope that these info can help me, it would be nice to draw sums with a final summary including formulas + calculations, so I could be able to understand everything even myself;-p
ps: for all those who have had fun with this bizarre question, I would have one that I consider even more "bizarre", which always requires the cono-science of physics as in this case but also contemplating the rigidity of a falling body ... (flying straps)

I will shortly open a new post about this further doubt, I hope to find you there too, all determined to make me the a-z pass to simplify the problem, understand and resolve

 

[mamobono]

Good morning, guys.

Hi.
the determined speed is the one that determines the impenancement of the cart without overturning it.
If you convert it to km/h you realize better.
So I would say that you have to or change the geometry of the cart by increasing 730 or lower the center of gravity, I would say that if the push makes a man the measure you have put is the most ergonomic and therefore cannot be touched.

I also did not regret with the differences found then, as from post 63 baskets1959 solved the doubt, two different methods giving similar results.
But in my opinion, the shock with the rock will lose part of the energy but if you read the message 48 I explain, but basically the speed is this.
I liked this discussion, so I did a little bit of resumption. :

Hi.

 

[mamobono]

added
everything is calculated as from your indication with fixed obstacle.
If the rock should fly away obviously we should understand how much energy it absorbs for the shock and how much it remains for the tipping.
ps in changing the geometry of the cart is to be evaluated the possibility to put wheels of diameter + large that will be impured with less ease.

 

[biz]

the correct approaches are the energy ones and that consider the moment of rotational inertia transposed as you can not consider the moment of main inertia. calculating straight work is not a correct approach since we talk about a rotting rigid body.

 

[biz]

I exhume this discussion as no one has made clarity, including me.
the limit speed is easily found with the preservation of the angle moment.
in practice:
1_I find the angle moment compared to the tipping point
I'm sorry.
2_for the conservation of angular momentum
l=i omega
with the moment of inertia compared to the point of tipping and omega angular speed.
calculation of kinetic energy
3_I find the gravitational potential energy considered when the cart rises and the center of gravity is located on the vertical with the tipping point.
epot = m g h
4_I equal the potential energy to rotation energy and finally find speed.

 

[ceschi1959]

I exhume this discussion as no one has made clarity, including me.
the limit speed is easily found with the preservation of the angle moment.
in practice:
1_I find the angle moment compared to the tipping point
I'm sorry.
2_for the conservation of angular momentum
l=i omega
with the moment of inertia compared to the point of tipping and omega angular speed.
calculation of kinetic energy
3_I find the gravitational potential energy considered when the cart rises and the center of gravity is located on the vertical with the tipping point.
epot = m g h
4_I equal the potential energy to rotation energy and finally find speed.

I no longer intervened in the discussion after the author's clarification to post 64. I don't want to get into controversy but when they say "when the man realizes to bump a rock/muretto/gradino, he will certainly stop pushing, it is that moment I wanted to analyze, to determine at the end.. ."then the problem has no solution; In fact, we do not know how long the man has pushed, and in the absence of friction, a millisecond is enough to make him reach the speed of light! Anyone who's experimented with frictionless rails knows.