Recent content by angelina

17...17 1/2 at christmas day*

now i get the formula I = mr^2 [n2/(n1+n2)] [(gt^2)/(2h) - 1] where m = mass of the hanging wieght; r = radius; n1 = no. of revolutions before the weight reaches the floor; n2 = no. of revolutions before the flywheel stops; t = time taken for the weight to reach the floor; h = height of...

math is a beatiful art, really, but it takes a long time for me to realize it. and by the time i realized, i hv dropped it...*sigh*...must pick it up again when I'm to go to university

very hopefully, still in the field of physics

for the case of the spinning skater, if friction exists, which axis will its torque act about? the same vertical (y-axis) or the z-axis or both??

then my question will become - is it true that as longer as there's external torque, no matter this torque is acting about the same axis as the rotation or about a different axis, angular momentum is not conserved??

*total* angular momentum means the system of the skater alone or the system of the skater + ice floor? also, say the spinning direction is on the x-z plane, then the friction acting on will be providing a torque along the x-direction. my question is, how does this frictional torque affecting...

when talking about conservation of angular momentum of a spinning ice skater, the contact surfaces are assumed to be frictionless. why?

seems that men get a better ability in math and phy but i really don't want to admitt that (for I'm a girl as u can see...)

gained a comment of "quite good" in one of my papers which i scored only 28 out of 70

(a) angular speed = 2pi/period = 2pi/(91 x 60) = 1.15 x 10^-3 rad/s (b) acceleration = w^2 r = (1.15 x 10^-3)^2 (210000 + 6370000) = 8.71 m/s^2 (c) magnitude = 4.1 (8.71) =...