So, I was studying about general properties of matter and topics like surface tension. I came across the phenomenon of water rising along a glass plate like in the picture. I looked for some mathematical interpretation of this on the internet and in some books.
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My approach:
Let us take two orthogonal axes: x, parallel to the racket's plane and y, perpendicular to it. For the ball to not spin, the components of initial velocities of the racket and the ball along x-axis must be same. Also, as the line of collision is along the normal to the racket's...
Sir, what are the finally correct equations? Are they
$$\frac{d}{dt}(\frac{\partial L}{\partial {\dot x}}) - \frac{\partial L}{\partial { x}}=F$$
and
$$\frac{d}{dt}(\frac{\partial L}{\partial {\dot y}}) - \frac{\partial L}{\partial { y}}=F$$
? Please clarify this
Then my second equation of motion is $$
\frac{d}{dt}(\frac{\partial L}{\partial {\dot y}}) - \frac{\partial L}{\partial { y}}=F\cdot \frac{\partial {(x+y)}}{\partial y}$$? Is it correct now?
Sir, What is the problem if the displacement of m2 depends on both x and y? How can I modify my equations for that? Is there any problem with my generalized force calculation?
<<Moderator's note: Moved from a technical forum, no template.>>
Description of the system:
The masses m1 and m2 lie on a smooth surface. The masses are attached with a spring of non stretched length l0 and spring constant k. A constant force F is being applied to m2.
My coordinates:
Left of...
Should I conserve momentum along the string? But how can I? The string prevents a free movement along it, but I can only conserve momentum along an axis which allows free movement along it.
I think this is the scene just when the string becomes taut. Should I now break the velocities into orthogonal components? I can not figure out any conservation law to apply. Please help.
Homework Statement
Homework Equations
$$
J \space (\text{Impulse})= \Delta (mv) = F \times \Delta (t)
$$
The Attempt at a Solution
As much I interpreted, we have to calculate the impulse caused by the tension till the relative velocity of approach along the string becomes 0. T to this, I...
This is a new system.
All things are same, only in this case, the rod is attached to an ideal string with very small length.
Here, the tension acts on the rod parallel to it(along it, to be more specific). So, no force acts along the α axis, and my math works here, does it not?
I did not write my post #9 based on your #8. My α axis is on the plane and not normal to it and perpendicular to the rod. And I balance forces along this axis on the COM.
I have marked both the masses along with the center of mass of the rod. α axis is an axis defined by me as shown in the diagram. It is perpendicular to the rod.
Now, my question is, if we balance the forces on the COM along the axis and ensure that there is no movement of the COM along the α...