- #71
Hak
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- 56
Yes, I agree. In this case, however, you have to admit that it is worth it, at least a little. I am sure it will bring good benefits.haruspex said:Not at all.... just a little time-consuming.
Yes, I agree. In this case, however, you have to admit that it is worth it, at least a little. I am sure it will bring good benefits.haruspex said:Not at all.... just a little time-consuming.
kuruman said:there is no normal force at that point in the model
No, in the actual problem the normal force will be >0.kuruman said:If in the model the initial velocity is in the negative y-direction, in the actual problem the normal force will increase above the static value of 12mg.
Yes, I have been queasy about this point. This thread has become the bane of my existence with all the mistakes I have made. There is another issue that bothers me. We have derived the expression $$N=\frac{1}{1+\beta}\left[mg-F(1-\beta)\right]$$ from the force equation and the torque equation relative to the CM. For ##\beta =2## this expression gives ##N=\frac{1}{3}(mg+F).##haruspex said:No, in the actual problem the normal force will be >0.
Consider F=0. In your model, that only leaves ##J_1##, a downward impulse, so P will have an initial downward movement, yet in the actual problem the normal force remains at ##mg/2##.
One has to be very careful taking moments about a point other than the mass centre when there are accelerations. Also, we have not considered horizontal acceleration at all. I have been assuming no friction, so the mass centre accelerates straight up, but the corners have horizontal acceleration.kuruman said:let's find the expression for ##N## with the torque equation relative to the point contact P as I started doing earlier. $$\begin{align} & F+N=mg+ma \nonumber \\ & F(2x)-mgx=\frac{4}{3}m(x^2+y^2)\frac{a}{x}\implies 2F-mg=\frac{4}{\beta}ma\implies ma=\frac{\beta}{4}(2F-mg).\nonumber \end{align}$$
1. Is @kuruman's model applicable in the light of the errors subsequently highlighted? That is to say: is it not applicable at all, or are certain adjustments and modifications possible that make the model valid and applicable?kuruman said:Below I explain why I think the model is applicable.
2. This assertion is not convincing and right for @haruspex. Indeed:kuruman said:Inequality (1) is a condition that must be satisfied for the normal force to drop below the static value.
As already mentioned, the inequality ##N>0## is not correct here. As seen in later posts, substituting (1) into (2) gives ##N<0##. Since you correctly observe that the condition ##N>0## is in agreement with one's intuition, the correct condition ##N<0## is in disagreement with one's intuition. How to solve this? How to make it clear and correct?kuruman said:We want to know what ##N## looks like when condition (1) is satisfied. If we replace ##F(1-\beta)## in the RHS of equation (2) with the smaller term ##mg## from inequality (1), then LHS > RHS: $$N>\frac{1}{1+\beta}(mg-mg)\implies N>0.$$It looks like the normal force will be positive and hence non-zero for all values of ##\beta##. All this agrees with one's intuition. As the stick is lifted off the surface, point P rotates about the CM and pushes down on the surface thus maintaining a non-zero normal force.
OK. 3. What does this imply? How do you change @kuruman's arguments so that the model is correct and applicable?haruspex said:No, in the actual problem the normal force will be >0.
Consider F=0. In your model, that only leaves ##J_1##, a downward impulse, so P will have an initial downward movement, yet in the actual problem the normal force remains at ##mg/2##.
4. I did not really understand the context of these equations. Or rather: you had already derived them correctly many posts ago. I don't understand how they can apply to the last model you set out (that of the rectangular block on a smooth surface and subject to the two pulses ##J_1## and ##J_2##). The problem of the normal force remains to be solved: its sign from your analysis appears to be negative, whereas we expect it to be positive. Which equations to bring in to support it?kuruman said:Yes, I have been queasy about this point. This thread has become the bane of my existence with all the mistakes I have made. There is another issue that bothers me. We have derived the expression $$N=\frac{1}{1+\beta}\left[mg-F(1-\beta)\right]$$ from the force equation and the torque equation relative to the CM. For ##\beta =2## this expression gives ##N=\frac{1}{3}(mg+F).##
Now let's find the expression for ##N## with the torque equation relative to the point contact P as I started doing earlier. $$\begin{align} & F+N=mg+ma \nonumber \\ & F(2x)-mgx=\frac{4}{3}m(x^2+y^2)\frac{a}{x}\implies 2F-mg=\frac{4}{\beta}ma\implies ma=\frac{\beta}{4}(2F-mg).\nonumber \end{align}$$Replace ##ma## in the top equation to get $$N=mg-F+\frac{\beta}{4}(2F-mg)=mg\left(1-\frac{\beta}{4}\right)+F\left(\frac{\beta}{2}-1\right).$$This is a different expression from the previous one. For ##\beta =2## this expression gives ##N=\frac{1}{2}mg.## I would choose the second expression as correct because the reference point for the torques etc. is at rest on the inertial frame of the horizontal surface.
5. OK, this expression is certainly correct, but it is just another way of calculating the same equation that was originally arrived at, no? How do we solve the problem of the applicability of the model and the normal reaction highlighted earlier?haruspex said:One has to be very careful taking moments about a point other than the mass centre when there are accelerations. Also, we have not considered horizontal acceleration at all. I have been assuming no friction, so the mass centre accelerates straight up, but the corners have horizontal acceleration.
The parallel axis theorem does not apply because the rotation is not about the corner P of the block.
We can treat the motion of the block as a mass centre velocity ##v## plus an angular velocity ##\omega##. Correspondingly, a linear momentum ##mv## and an angular momentum ##I_c\omega##. From the perspective of a fixed point at the initial location of P, the linear momentum of the mass centre adds an angular momentum ##mvx## for a total angular momentum ##mvx+I_c\omega##.
Differentiating, the rate of gain of angular momentum is ##max+I_c\alpha##.
The torque equation is therefore ##2Fx-mgx=max+I_c\alpha=max+\frac 13m(x^2+y^2)\frac ax=m\frac ax \frac 13(4x^2+y^2)##.
This leads to the same equation as taking moments about the mass centre. There is no fault in the Universe.
My position is that a) I do not understand the reasoning for the claim that the model is applicable, and b) it leads to answers that conflict with approaches that look sound to me.Hak said:How do we solve the problem of the applicability of the model and the normal reaction highlighted earlier?
I agree. For me, the striking fact is this: if there was a perfect match between arguments and answers, everything would have seemed to be in order, instead this could highlight the dangers and obstacles one might encounter in proposing problems such as this one formulated by the AAPT, which unintentionally created an exercise that was anything but simple and smooth. Tell me what you think. Thank you.haruspex said:it leads to answers that conflict with approaches that look sound to me.
As shown in post #11, allowing for some thickness is not that complicated, but it did produce a surprising result. (In fact, it should not have been that surprising. It is well known that there is a 'sweet spot' along a stick where a blow at right angles will not jar the hand holding the stick at the far end, whereas a blow in the middle would jar the hand one way, and a blow at the tip will jar it the other way. The threshold ratio found in this thread is similar.)Hak said:I agree. For me, the striking fact is this: if there was a perfect match between arguments and answers, everything would have seemed to be in order, instead this could highlight the dangers and obstacles one might encounter in proposing problems such as this one formulated by the AAPT, which unintentionally created an exercise that was anything but simple and smooth. Tell me what you think. Thank you.
Thanks for your point of view. We await @kuruman for clarification regarding his model.haruspex said:As shown in post #11, allowing for some thickness is not that complicated, but it did produce a surprising result. (In fact, it should not have been that surprising. It is well known that there is a 'sweet spot' along a stick where a blow at right angles will not jar the hand holding the stick at the far end, whereas a blow in the middle would jar the hand one way, and a blow at the tip will jar it the other way. The threshold ratio found in this thread is similar.)
The complications subsequently have come from trying different approaches in the hope of gaining greater understanding.
I agree with all of the above and concede a degree of carelessness probably due to lateness in my time zone. Here is the improvement.haruspex said:One has to be very careful taking moments about a point other than the mass centre when there are accelerations. Also, we have not considered horizontal acceleration at all. I have been assuming no friction, so the mass centre accelerates straight up, but the corners have horizontal acceleration.
The parallel axis theorem does not apply because the instantaneous centre of rotational acceleration is not at P.
No, you are still missing it.kuruman said:Since Q and P are essentially the same point, the parallel axis theorem applies because at the particular instant that we consider, the stick is about to rotate about P/Q.
Sigh. This means that the model in post #80 cannot be applied, but must be modified to be valid, right?haruspex said:No, you are still missing it.
Call the mass centre G. Point P of the block is accelerating horizontally, while G is accelerating vertically. (PG is constant. If P is not accelerating then G is accelerating diagonally up, implying friction at P.) This is not a state of affairs that arises after some interval, it is how things are as soon as acceleration starts.
It follows that the instantaneous centre of rotational acceleration, C, is on a horizontal line through G and a vertical line through P=Q. The parallel axis theorem clearly applies at C.
(Note that all forces have the same torque about C as about P, and there is certainly no reason for P to be a more appropriate axis to consider than C.)
Taking C as the axis and using the parallel axis theorem produces the same answers as taking G as the axis. Taking P as the axis also works if you work from first principles, as in post #74, instead of assuming the parallel axis theorem applies there.
Why the desire to fix the model? It is not needed.Hak said:Sigh. This means that the model in post #80 cannot be applied, but must be modified to be valid, right?
Yes, you're right, it is not needed. However, I have no desire to fix the model, but I would like there to be a reasonable one applicable to the physical situation. Three were proposed by @kuruman, all three did not work because of some oversights and wrong physical assumptions, but that does not mean there is not one. An applicable model, with all the correct considerations you pointed out, must exist in my opinion. In my opinion, @kuruman's efforts should be rewarded by this. It may well be that I am wrong.haruspex said:Why the desire to fix the model? It is not needed.
post #11.Hak said:a reasonable one applicable to the physical situation
Yes, obviously, I've already seen this. I was referring to @kuruman's efforts, but still nothing ... If @kuruman still feels like coming up with a convincing model that can fit the situation, that's fine, otherwise, if he wants to give up, we are content with this excellent discussion. Thank you very much.haruspex said:post #11.
I haven't checked, but #80 can probably be salvaged by using the correct moment of inertia, ##4x^2+y^2## instead of ##4x^2+4y^2##.Hak said:Yes, obviously, I've already seen this. I was referring to @kuruman's efforts, but still nothing ... If @kuruman still feels like coming up with a convincing model that can fit the situation, that's fine, otherwise, if he wants to give up, we are content with this excellent discussion. Thank you very much.
The above conditions, which you made in post #80, remain unchanged in post #88 below. In that respect, everything runs smoothly.kuruman said:Now let's look at the smallest possible value for ##N## which occurs when ##\beta = 0##, i.e. we have a vertical thin stick. Then ##N=F_0-\delta F.~## This becomes zero when ##\delta F=F_0## which means that ##F=2F_0=mg.## We knew that all along. As we pull on the vertical stick harder and harder, the normal force will diminish until it goes to zero and the stick is lifted off the surface. Points P, Q and the CM will all be lined up while this happens.
But what happens if ##\beta = 1##? I get: ##N = F_0##. This means that ##N## goes to ##0## when ##F_0## goes to ##0##, not when ##F=3F_0##, right?kuruman said:OK, let's redo post #80 using the torque equation relative to the CM. The force equation is the same and yields
$$N=ma+2F_0-F.$$ From post #11
$$F-N=\frac 13m( 1+\frac{y^2}{x^2})a=\frac{ma}{\beta} \implies ma=\beta(F-N). $$ Then $$N=\beta(F-N)+2F_0-F \implies N=\frac{1}{1+\beta}[(\beta-1)F+2F_0].$$ In the range ##1<\beta<3##, the RHS is clearly positive.
In the range ##0<\beta<1## we replace ##F## with ##F_0+\delta F## as in post #80 to get $$\begin{align} & N = \frac{1}{1+\beta}[(\beta-1)(F_0+\delta F)+2F_0]\nonumber \\
& =\frac{1}{1+\beta}[(1+\beta)F_0-(1-\beta)\delta F)]\nonumber \\
&=F_0-\frac{1-\beta}{1+\beta}\delta F. \nonumber
\end{align}$$This can be used to find for what increase ##\delta F## the normal force goes to zero.
Is the condition in bold preserved or not, given the new terms?kuruman said:What about a value of ##\beta## in the range ##0<\beta<2~##? Let's be specific and pick ##\beta =1##. Then ##N=F_0-\frac{1}{2}\delta F.## ##N## is zero when ##\delta F=2F_0 \implies F=3F_0.## In this case the contact is just lost and it doesn't matter whether the surface is smooth or rough when one considers the initial horizontal acceleration of point P.
Does this still happen in the new model? And if it happens, when does it happen? In the case ##F_0 >0##?kuruman said:What happens if ##\beta = 1## and ##F>3F_0~##? I am only guessing that corner P will be lifted off the surface but, because the CM is not directly above point P, there will be some kind of pendulum motion depending on how force ##F## is anchored on the stick.
I obtain: ##\delta F = \frac{1+\beta}{1-\beta} F_0##, or ##F = \frac{2}{1-\beta} F_0##. What does this tell me? Is there any particular information?kuruman said:This can be used to find for what increase ##\delta F## the normal force goes to zero.
Yes, and ##F_0## is just shorthand for ##\frac 12mg##, so, for ##\beta\geq 1##, N=0 implies m=0.Hak said:This means that N goes to 0 when F_0 goes to 0
It says that , for ##\beta\geq 1##, assuming N can be zero gives a silly answer, so it can't go to zero.Hak said:I obtain: ##\delta F = \frac{1+\beta}{1-\beta} F_0##, or ##F = \frac{2}{1-\beta} F_0##. What does this tell me? Is there any particular information?
Why for ##\beta \ge 1## and not for only ##\beta = 1##? Thanks.haruspex said:Yes, and ##F_0## is just shorthand for ##\frac 12mg##, so, for ##\beta\geq 1##, N=0 implies m=0.
Because for ##\beta>1## the equation you quoted has the form ##N=F_0+(positive factor)\delta F##.Hak said:Why for ##\beta \ge 1## and not for only ##\beta = 1##? Thanks.
Reviewing post #53, I cannot understand the above statement. Why the only way to make ##N > 0## and ##\beta = 1## is ##mg = 0##? Perhaps you meant to say ##N = 0##? I cannot understand it.haruspex said:It makes sense because otherwise you have made two contradictory assumptions: ##N>0, \beta=1##. The only way both can be true is mg=0.
Yes, good catch. Corrected.Hak said:Perhaps you meant to say N=0?