Recent content by smruthi92

OOOOHHH I WORKED IT OUT! THANKYOU SO MUCH FOR UR HELP! :d:d:d:d

ok as I am assuming that they are rotating it at the end of the bar connected to the spring, and not say, the centre of mass? so if we include Vg, then its mg h, where h = the height you find forming a triangle?

sorry i made a mistake. um so what i have is. to find x I am creating an arc with 69 degrees + tan inverse (1.458/1.8), so then i would find the extension to be 2 x pie x 1.8 x 108/230. minus that from the original length and then use the eqn above. but this is all assuming, its rotated about...

oh awesome! so T would essentially just = 1/2 I (w2^2). so W = 0, 1/2 I (w2^2) + 1/2 k (x2^2-x1^2) = 0? i guess if it comes back to rest, x1 = 0. but i still don't know how to figure out that extended length using geometry? coz when ur rotating the rod by 69 degrees, could u maybe do...

pls see attached file for the question. so W = change in T + change in Ve + change in g T = 1/2 mv2 + 1/2 I (w1^2 - w2^2), but how do u find the velocity, shouldn't it be 0 since it starts/stops at v=0. for Ve = 1/2 kx2, how do u know how far it stretches once the moment is applied? :O :(...

ok thank u so much tim! uve helped me so much! :D

awesome! thank u so much, just one last question,. after finding the total torque at point A, how do u find the reaction force?

sorry that's right. I = mL^2/12 + md^2 a = alpha x r M = I x alpha M = T d T = mg +ma. but u know the a we find from a = alpha x r, is it the same a of the mass on the pulley?

im sorry i don't get it :( using M = I alpha I = mL^2/12 but that's all i understand. u know the tension, it produces a moment right? so using that i got: M = I alpha T = I alpha ma +mg = I x alpha

hey tim, i can't seem to find a to find the tension, that's my problem. :(

hey guys, please see the attachment for the question. i drew FBDs and everything. i want to know a few things tho. how do i calculate the tension in the pulley? if they'd given me acceleration i could have used T-mg =ma. but they havent. also once the Tension is found, do i just find the...

yeh but how do i calculate it with the data given? we haven't been given an equation to differentiate to find alpha. also I am still stuck on calculating w.

pls see attached file. so i thought i could use this eqn: aA = w2 AO en + alpha AO et but when i try to work out w, something goes wrong. i do: w = 7.16605 (normal velocity i found by having the given velocity as a y coordinate one, then resolving for normal/tangential) / 0.25. but...

ok let me show my working: normal acc = 0.009^2/0.066 tangential acc i tried this: s=r theta s = 0.066 times (66/180)*pie s = 0.07603 v = d/t t = d/v t = 0.07603/0.009 t= 8.4477 seconds therefore acc = v/t acc = 0.09/8.4477 then i squared this and the normal acc, but its not the right answer

pls see the attachment. so pretty much magnitude of acc would be root (at2 + an2). i figured out an by subbing it into v2/p. but i can't figure out how to calculate at! id really appreciate the help, I've been at it for 3 hours now! thanks!