Recent content by mitch987

of course! Forgot that the particular solution cannot equal the homogeneous solution. Thanks for your help.

Homework Statement y'' + y = -2 Sinx Homework Equations The Attempt at a Solution finding the homogeneous solution, is simple; yh(x) = C1 Cos(x) + C2 Sin(x) for the particular solution, I let y = A Cos(x) + B Sin(x) thus, y' = -A Sin(x) + B Cos(x) y'' = -A Cos(x) - B...

For the MOI about the centre of mass, take take the radius as 4.72m and the centre x=0. Hence, centre of mass, x_{cm}= \frac{2.5\times -4.72 +9.1\times 4.72}{2.5+9.1} =2.68m MOI, I_{cm}=\frac{1}{12}(M_{system})(x_{cm})^{2}=\frac{1}{12}(11.6)(2.68)^{2}=6.94 \omega=\frac{v}{r}...

To find the correct value of I, you would need the moment of inertia about the centre of mass?

yeah, but that only encompasses magnitude not direction. anyway i figured it out by using the various laws. if, u · v = u · w then, u · (v-w) = 0 if, u x v = u x w then, u x (v-w) = 0 therefore, v-w is both orthogonal and parallel to the non zero vector u, hence v-w = 0 therefore v=w

Thanks, i completely forgot bout the angles not being equal. however, going with that i can only simplify it down to \upsilon \cdot \upsilon = \omega \cdot \omega

Help!? Linear algebra proof Homework Statement Suppose that u,v,w are geometric vectors such that u\neq0, u\cdotv=u\cdotw and uxv=uxw Prove that v=wHomework Equations The Attempt at a Solution So far, I'm not sure if this is correct u\cdotv=u\cdotw |u||v|cos\theta=|u||w|cos\theta...