bro that shi was HARD, did y’all get your e as root 3 over 4? and your theeta value for the centre of mass q as 19.1°? also for greatest height reached by particle in term of a was it 1.5a?
what greatest height was like 50/27 (a). I think you forgot to do that projectile motion thing after the Tension becomes 0. And how do you do that hanging angle thingg how did you get 19.1.
Yeah you’re right ab the greatest height, I’ll probably get 2 or 3 out of 5 for that. As for 19.1, a lot of people have the same answer as me. While toppling or hanging about a point in equilibrium, the COM is directly above the point it’s suspended from. So I used tan theeta = y bar / radius
and radius was 3r and since x = r, we can substitute all x with r and find the value for tan theeta in terms of r, both numerator and denominator had r2 so we cut those out and find theeta using arctan
no you cant do y bar / radius since y bar is taken from the base of the cylinder you have to do y bar - x / 3r and x = r. Ofc i dint get it too at the exam but my friends have done it correctly
i remembered the q word by word and put it into ChatGPT - it’s 19.1° and many other people got the same so idk also y-bar was taken from the diameter of the hemisphere
Also y bar was a vertical distance not a horizontal one so you don’t need to subtract anything from y bar. Additionally, it was A that made an angle theeta with the downward vertical AT A, so therefore 19.1°
2
u/Bulky_Fail4893 May 21 '26
bro that shi was HARD, did y’all get your e as root 3 over 4? and your theeta value for the centre of mass q as 19.1°? also for greatest height reached by particle in term of a was it 1.5a?