Recent content by Very_Unwise

Alright alright, thanks for the help :D

Okay, so I(rod+snowball) should be Irod + Isnowball which is 1/3*M*L^2+ m*L^2 , m=1/9*M I(rod+snowall) = 1/3*M*L^2 + 1/9*M*L^2 = 4/9*M*L^2 Irod* ω1= I(rod+snowball)*ω2 1/3*M*L^2 * sqrt(3g/L) = 4/9*M*L^2*ω2 crossing out M*L^2 1/3*sqrt(3g/L)=4/9*ω2 ω2*4/3=sqrt(3g/L) ω2=sqrt(3g/L)*3/4

Yes that's true, substituting g=9.81 does make things quite messy. Thank you

Okay, so my previous calculations were wrong because the moment of inertia for the rod changes after collision? Meaning I(combined object) doesn't equal Irod? To find the new moment of inertia i have to find the new center of mass for the combined object, which is (0*1/9*M +1/2*L*M)/10/9*m...

Okay, so instead of m*r*v=m*r*v ill use I*ω = m*r*v? Substituting v=ω*L gives me I*ω = m*L^2*ω That gives me i*ω = 10/9*M *L*v v=ω*L i*ω1 =10/9*M*L^2 * ω2 , i=1/3*M*L^2, ω1 = sqrt (29.43/L), I=Irod I rod*sqrt (29.43/L) = 10/9 *M*L^2*w solving for w using symbolab gives me w = (1.627*sqrt...

Assuming no friction anywhere, no drag and perfect inelastic collision Using conservation of mechanical energy i can determine the rotational speed of the rod right before collision occurs. mgh=1/2*i*w^2 center of mass falls 1/2*L so we have: M*g*1/2*L = 1/2*(1/3*M*L^2)*w^2 Solving for w...