Recent content by kuruman

Sure, but the screenshot with the statement of the problem in post #1 specifies "at any instant" and we have to go with that.

In what way is it limited? I would argue that ##~ J_{cm}=I_{cm}~\omega ~## is a special case of the more general equation in post #68, $$\mathbf J(x)=I(x)~\boldsymbol {\omega}~;~~~I(x)=\frac{1}{12}mL^2+mx^2+m\left(\frac{L}{2}-x\right)^2$$ where ##x## is the distance between the midpoint of the...

It might also be worth mentioning that the spin angular momentum ##J=I_{cm}\omega## may be calculated about a point on the body that is moving relative to the lab frame. I will illustrate with an example specific to the rod mentioned in this thread. We are looking at the case after the bullet...

I interpret the two numbers 1 m2 and 0.8 m2 (left top) to be the areas of the cylinders. One may calculate radii from these. Units are useful after all. One would think that a detailed figure like this would be drawn to scale. There appears to be no change in the fluid level in the basin (see...

Indeed. This problem was educational for me in that it jarred me out my complacency of "having seen it all."

This problem is from the Indian National Physics Olympiad – 2020. A "tentative" solution appears here on page 15.

What is in a symbol? If U is the speed at the lowest point of the motion, you are looking for its maximum and minimum values, call them ##U_{\text{max}}## and ##U_{\text{min}}## such that ##U_{\text{min}}<U<U_{\text{max}}.##

Yes, I ignored that it's the point where the centripetal acceleration changes direction.

I think that would make thing difficult on the other side of the equation. To begin with, the potential energy at the top of the arc between B and C should negative because the top of the arc is below point D. Secondly, the vertical distance between D and the top of the arc might be a bit...

Yes, sorry I confused your problem with the example. In your solution, why do you say that the potential energy at A is mg(20)? Where do you take the potential energy to be zero?

Please show some consideration to those who are trying to help you. We are not mind readers. What is the statement of the problem? Also a diagram would be helpful. What are points A, B and C?

We cannot guide you in the right direction without knowing how and where you got lost. Please post you attempt at a solution and the specific problem you were asked to solve.

The real problem with wording is in the statement of the question. The masses cannot be unequal and have equal speeds in opposite directions. An observer at rest with respect to one mass sees the other mass orbit around him with some frequency ##\omega##. The same can be said about an observer...

Yes, I foolishly thought that the CM velocity need not be added. That agreement vis-à-vis what the problem was asking is another point I missed. Thanks for the clarifications. I think I understand now.

I have (had) a different approach to this problem, which I have not gotten to work, and I think I know why. So please hear me out. Step 1. Initial considerations Referring to the original diagram, we are looking for the tension ##T(\theta)##. This tension has a vertical and a horizontal...