Recent content by haruspex

Yes, except that, as I wrote, the force is not constant. What you have calculated is the average. Is that what you are asked for?

Ok, so can you do what I suggested in post #8? Do I need to explain more?

I agree that is confusing, but since the question asks for a tension there was, presumably, some mention of a rope in the original text. Thus, F might be illustrating the force on the rope, not on the cylinder.

Yes, I think we all read it that way. @jbriggs444's point is that we have no info re the surface area of the reservoir, other than that the level is not shown as having risen. Maybe it's not circular - infinite into the page perhaps 😉.

You just have to draw the before and after pictures. You can consider the water as having gone from the volume that is occupied in the before picture but not in the after picture to the volume occupied in the after picture but not in the before picture.

Yes, but I am giving you another way to figure this out that's clearer. I asked how much water disappears from the top if the cylinder only descends a small distance ##\Delta y##, not the whole 0.5m. There is a reason for that, as you will see. Looking at where extra water appears at the...

Yes, but sketch where that water has come from. And where it has gone to. What is the change in PE? Btw, the force will not be constant.

You might find it clearest to think in terms of energy. If the cylinder descends a distance ##\Delta y##, what is the change in PE? Draw the region where the water goes from and where it goes to.

ok, thanks for checking it all out.

The answer looks rather different. Strange that theta does not appear in the final equation for T. This is troubling: "1. Angular momentum is conserved only about P." but later: "We now calculate the acceleration of P on the rod in the direction of the string" So is P accelerating or fixed...

I see no benefit in introducing V. You are to find a range of values for U First, identify the points of the trajectory which will determine the minimum and maximum speeds and what the constraints are on the speeds at those points. Then use energy conservation to relate those to U. As...

Correctly, the analysis considers the highest risk of losing contact is at A and B. F=ma is being applied along the normal, so at 45° to the vertical.

Yes, as I wrote in post #21, you can (only) elect to switch the sign of the tangential acceleration arbitrarily, but if you choose not to you must arrive at ##\dot{\omega}=0##. As in, the given differential equation accurately represents the dynamics, so applying it in a simulation should...

Ok, so it is the difference between a solution that satisfies a differential equation and one that corresponds to a physical process which the differential equation represents. If ##\dot x=f(x)## is cause and effect then once ##x## satisfies ##f(x)=0## then it is stuck there. This never struck...

It's not based on a solution, merely the fact that an explicit equation like ##\dot\omega=\sqrt{b^2-\omega^4}## (positive square root implied), together with the initial condition ##\omega=0##, should guarantee unique development. Moreover, 1) ##\omega## must be non-decreasing and bounded above...