Recent content by mspaic

anyone know how to calculate the moment of inertia from the information given

ok i got a i gues they wanted torque

Hey I am doing some reading on inertia and stuff and am having problems with these questions: --- An m = 11.9 kg mass is attached to a cord that is wrapped around a wheel of radius r = 12.3 cm The acceleration of the mass down the frictionless incline is measured to be a = 1.91 m/s2...

thanks a lot man

anyone...

x^2 (M1)(M2) + x ( (K1+K3)(M1) + (K1+K2)(M2) ) + (K1)(K2)+ (K1)(K3) +(K2)(K3) =0 there is my simplification by the way

ok I'm pretty sure i figured out how to prove that they are negative. however, can someone please do similar to what he did above (just for my simplification) to prove that the roots are distinct. I'm sure my simplification is right because i confirmed it with numerous classmates.

can someone else factor it please see if i have it right it shouldn't take u long

the term with m1m2(k1 +k2)(k2+k3) why would u need the mi and m2 when you multiply you alread have them on the bottom so when you common factor why do u need it up there

x^2 (M1)(M2) + x ( (K1+K3)(M1) + (K1+K2)(M2) ) + (K1)(K2)+ (K1)(K3) +(K2)(K3) =0 look at this you know the first and the last term will be positive right since k and m are always positive therefore, the middle term has to be negative, for the equation to =0 thus, x has to be negative

ehe ok finally i'll just explain to him why x will always be negative though

but underneath the root you have like terms you can't simplify and for x I'm supposed to get a number ...

i just put it into the quadratic and it ain't going to work errrr like the stuff underneath the square root gets way too complicated

is it possible to do a quadratic formula with these values like i don't have values for the masses or the k's . i will try it

does anyone know it.. or is it even possible the teacher assistant i think hinted that it was possible to find the two values of x