Does the official kvpy mock test on their website not show marks? I just attempted it and at the end of 3 hrs it just showed the summary of attempted questions and upon clicking "next" it just went back to the login page where I could attempt the test again.
MOIs are additive, so if we divide the sphere into two equal hemispheres then the MOIs of the hemispheres will be half the MOI of the whole sphere. Yes?
There are two separate things: 1. the zero offset 2. the error in the zero offset
The offset means when you close the gauge with nothing in it the gauge doesn't read zero. For example suppose it read +7, that means when you make a measurement you need to subtract +7 from your reading to get the true value. OK so far?
@JohnRennie If sum of any two vectors is 0. Let the vectors be OA & OC. OA makes angle 37 with X axis & OC makes angle 45 with Y axis. Can we say Resultant of OC & OA is 0 ?
@S.M.T The only way two vectors can sum to zero is if they are equal and opposite. That is, they must have equal magnitude and point in opposite directions.
@Wolgwang the perpendicular axis theorem is applicable only for laminar bodies i.e 2D bodies, be careful. See the derivation on wiki, then it should be fairly obvious why the theorem doesn't work here.
@YashAgrawal ah I did the same question yesterday I got it wrong too, although I was getting (b).. my logic was that the linear velocity of the com just after the impulse is simply given by J = mv, so taking into consideration the rotation is irrelevant, although that is probably wrong. Why are you taking rotation into consideration in the first place..?
oops yes. If the rod was on a smooth floor it would tend to rotate about it's com, but since it can't rotate about its com at first there is a normal impulse too.
well tbh I'm only getting more confused. I do agree that the body would still seem like it's rotating in the com frame, that's for sure. But yeah I'm still a bit unsure as to why my solution fails.
while doing in practically with a pen , if I give the pen less impulse it remains intact with the floor but if I give it more ,it just fly off my head xD
@AshishAhuja there is nothing as linear or rotational KE
By work energy theorem , ∆KE = work done by all forces , so here only mg do the work and initial kinetic energy is given by 1/2(I)(w)^2 and final is zero
@YashAgrawal why? Once it's in air there is no torque about about the com to stop the rotation.
@YashAgrawal because I don't believe the rod will stop rotating at it's highest point. I remember a youtube video which did a bullet block experiment, let me find it.
@YashAgrawal that is only to figure out w after the impulse. Once the rod is in the air rotation about A is going to become incredibly complicated I guess because the torque about A once the rod is in the air is non-zero.
@YashAgrawal yes I used to always think my physics was the strongest subject, turns out it's not :D I found kvpy chem a breeze, the math/phy were relatively much tougher than the chem I guess
No, because the COM is moving upwards. Suppose immediately after the impact the angle is rotating with angular velocity ω, then the COM has an upwards velocity ¹⁄₂Lω.
@YashAgrawal if you don't mind me asking, how are you doing in the past year kvpy papers? I've been getting 60-65, but I get a bit scared because usually more than half of that is from only chem.
@AshishAhuja well I'm attempting papers in bits and pieces (because my revision/syllabus is not completed) , so no idea , but one thing for sure I'm hardly able to get 10/20 in physics section 2 :(
@AshishAhuja Can you tell is there good no. of questions from s,p,d,f blocks ?
ah ok. yeah there was one paper which I thought I was acing and I attempted 8/10 question in phy sec 2, but lol in the end I got 3.5 marks in phy sec 2 :-|
@YashAgrawal there are question but usually the logical ones, plus stuff you just naturally pick up. I haven't done any of those chapter properly either except p block and I've been able to correctly solve >80% of those questions, although many were just educated guesses.
there are many bits and pieces of mine that are incomplete as well, but the vast majority of them you can still solve so it's never too big an issue I guess.
Im working out the exact calculation, you can try it to. Set up a coordinate system, find the coordinates of A and COM, differentiate them to get the velcoity vectors
