i did not get the logic why we are taking small displacements in the second question... maybe analytically if the answer satisfies small displacements then it is valid for large displacements as well.. but to judge that the small displacements case will suffice how to do intuitively see that in the beginning of the question?
Small suggestion: Make the thumbnail more bright and large so that diagrams are clearly visible Will help to attract more students to such wonderful sessions
How did you take the exit velocity to be V2 in the derivation of impulse equation and then claimed the outlet velocity to be equal to inlet velocity? Please clarify.
It's just continuity equation Av= constant. In the problem A is not varying so v remains the same at all cross sections. With the derivation I wanted to point out it's outlet velocity - Inlet velocity that way it looks cleaner.
@@Phyx_IITM If outlet velocity is equal to the inlet velocity,then shouldn't the impulse be zero? Sorry if I am being too oblivious, also l have been watching your videos for a while, keep up the good work.
Bhaiya how can we say that that small dm mass would have constant velocity for dt time as that body is accelerating, I mean is there any general statement for it
Its just that when the vanes speed is u let's say after dt sec it's speed will be u + adt. We are adding a differential term to a finite quantity right so we can neglect that term for the small time dt . Like adding 0.0000001 to 5
