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I found a few Phys.SE threads in which an impulse is used to exert a torque on a rod, is it possible to get a realistic description of the force required to open a door?:

If we have a free 'door' ($l = 1m$, $M = 12 Kg$, $I = 4$) and we exert force of 120 N on the CM for 1/10 sec, the impulse will be 12 kg m/s but will the acceleration/velocity be: $a = J/m =v_{CM}= 12/12 = 1 m/s$? (that is no sure, see point 1.)

enter image description here

If the rod is pivoted at one end and we apply the same force on the CM the angular acceleration/velocity will be:$$\Delta \omega = J* \frac{d}{I}= 12*.5/4 = 1.5 rad/s = .75 m/s$$

The answer did not clarify the main problems:

  • 1. however short the time of application, the rod will start to rotate and the impulse will not act perpendicularly

, instead of 120 N ... apply .... 12,000 N on the CM for 1/1000 sec, etc.

it is not realistic that one can exert a force exceeding a ton (for such a short time)**. Do you get the same conclusions considering J as the impulse from a collision (a push or a kick?)

  • 2. how can we find the angular acceleration $\omega$ if we do not know the mass of the rod at each point?

From this, we can calculate the change in angular velocity: $\Delta > \omega = d J / I$. - Michael Seifert

if we use the same formula at each point, we are considering that its effective mass is the same (I = 4 kg) at each point all along the 'door', so,

  • 3. the same impulse at CoP will produce: $\Delta \omega = J * d/ I = .66 * 12/ 4 = 2 rad/s = 1 m/s$ (and at the tip: $\omega = 1.5 m/s$)

But then it takes less force/impulse/energy to get the same angular acceleration, at CoP it takes an impulse $J=9$ to get $\omega=1.5 rad/s$

  • 4. how do we determine the reaction force R at the hinge? if $$R = J - m \omega \frac{l}{2}$$

    at CM $R = 12 - (12*.75) = 3 $

Next, we note that the linear momentum of the center of mass immediately after the impulse is delivered is mωl/2; this quantity must be the total linear impulse delivered to the rod.

and at CoP $R = 12 - (12*2*.5= 12) = 0 $,

I'm not sure "why" it should be this way other than by the above chain of logic.

this logic doesn't seem correct, for example: at the Tip $R = 12 - (12*3*.5= 18) = -6, $ does this negative value make any sense? can you explain this? Should I start a new thread or can someone answer this basic

question:

can we correctly deduce the opposing force[s] at the hinge with this rather simplistic logic?

  • a) the moments of inertia of a F[ree]/ H[inged door] are different (I = 1; I = 4), b) the effective masses at each point of F/ H are different, c) the very effective mass of H can't have the same value at all points (in this case: m = I = 4),
  • is it valid to extend the 'test' at CM to other points? hitting the CoP, F will both translate and rotate, at any point (but CM) there are two opposing forces at the hinge, which combine in a different way,
  • if we consider the impulse J =12 as the product of m = 12 * v= 1 as in a collision, we get different results, but aren't they more reliable, since they are inexorably governed by conservation laws? at CM we have:

  • F: v = 1, H: $\omega = 1.71/I$ ; at CoP:

  • F: v = .86, $\omega = 1.71/I$ ; H: $\omega = 1.71/I$

That seems more plausible, can someone correct some eventual mistakes?

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  • $\begingroup$ Hi user78234. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$
    – Qmechanic
    Jun 7 '15 at 7:50
  • $\begingroup$ @Qmechanic, hi, my question asks to clarify some unclear aspects of the linked question, which has no homework tag, so, how can it be considered a homework question? I even borrowed the image from there, I only pointed out some conclusions that do not seem realistic or plausible, since my comment there, after a month, got no reply. $\endgroup$
    – user78409
    Jun 7 '15 at 8:10
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To answer your questions:

  1. Yes, the rod will start to rotate; and if the force continues to act in the same direction and the wheel rotates for long enough, then the torque will change. However, we can take the limit where a very large force acts for a very short time, in such a way that their product is constant. (That is, instead of 120 N on the CM for 1/10 sec, apply 1200 N on the CM for 1/100 sec, or 12000 N on the CM for 1/1000 sec, etc.) In this limit, the actual angular rotation of the rod goes to zero, and we can assume that the impulse is all exerted at one spot and at one time. This is the only limit in which the rest of your questions are easily addressable, so I'll assume this is true from now on.
  2. $J$ is the linear impulse delivered to the door by the applied force, and it's the time integral of the applied force. In our case, $J \approx F \Delta t$. This implies that $m \Delta v = J$. Similarly, we can define an angular impulse $J_{ang}$ as the time integral of the torque, and we will then have $I \Delta \omega = J_{ang}$. Moreover, if force that acts with a constant lever arm $d$, then $J_{ang} = (F d)\Delta t = d (F \Delta t) = d J$. So if the force would have delivered an impulse $J$ to the "free" object, then it delivers an angular impulse $d J$. From this, we can calculate the change in angular velocity: $\Delta \omega = d J / I$.
  3. No, because the lever arm $d$ would be different.
  4. $R$ in this case appears to be the impulse delivered by the hinge (i.e., the time-integral of the reaction force imparted by the hinge.) To find this, we first note that the impulse required to get the rod moving from rest at angular speed $\omega$ is $J = I \omega / d$ (rearranging the equation from point #2 above.) Next, we note that the linear momentum of the center of mass immediately after the impulse is delivered is $m \omega l/2$; this quantity must be the total linear impulse delivered to the rod. Since this total impulse was the difference of the two impulses $J$ and $R$, then we conclude that $J - R = m \omega l/2$.
  5. Your calculations here are correct, although you would say that you want to have an angular velocity of 1 rad/s, not an angular acceleration of 1 m/s. As noted above, the angular impulse will be different if you apply the same linear impulse at a different location.
  6. The units of $R$ are those of impulse, which are the same as momentum: kg m/s, as you say. I'm not sure "why" it should be this way other than by the above chain of logic.
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  • $\begingroup$ Responding to a couple of the questions: (a) whether you call it a "collision" or not is up to you. Basically, you need to have the final angular velocity $\omega$ times $\Delta t$ be negligible, but "negligible" depends on how accurate you want your answer to be. (b) As far as actually giving you the numerical answers, I'd rather not, as we try to make the answers here be more general and less about specific situations. But your calculations for $\omega$ are correct in both cases, so it looks like you know how to do it. $\endgroup$ Jun 5 '15 at 15:51

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