Recent content by Vasili

I realized that I had to isolate omega, as is obvious by the fact that I put in my posts precisely that: What I did not know is how to begin simplifying this equation. There is a difference between gently nudging someone toward the right answer and being deliberately vague and unhelpful.

6(MgL) = (\frac{1}{3} ML^{2}) \omega ^{2} + (2M(d)^{2}) \omega ^{2}+ Mg L + Mg(L-d) Divide 2M(d^{2}) out: \frac{3gL}{(d^{2})} = \frac{1}{6} \frac{L^{2}}{(d^{2})} \omega ^{2} + \omega ^{2}+ \frac{g L}{(2(d^{2})} + \frac{gL}{d}... Like I said, I don't know how to simplify this.

Ah. 3(MgL) = \frac{1}{2}(\frac{1}{3} ML^{2}) \omega ^{2} + \frac{1}{2}(2M(d)^{2}) \omega ^{2}+ Mg\frac{L}{2} + 2Mg(L-d) Now is there anything *else* that's wrong with the formula? :frown:

Okay, so you get: 3(MgL) = \frac{1}{2}(\frac{1}{3} ML^{2}) \omega ^{2} + \frac{1}{2}(2M((L-d)^{2})) \omega ^{2}+ Mg\frac{L}{2} + 2Mg(L-d) But to be honest, I don't know where to begin simplifying this to solve for omega. Where do I even start?

Are they expecting me to write the distance d as \frac{d}{L} instead of L - d, though, to get their formula? Would that work? Here is the new equation with the moment of inertia of the particle: 3(MgL) = \frac{1}{2}(\frac{1}{3} ML^{2}) \omega ^{2} + \frac{1}{2}(2M(\frac{d}{L}^{2})) \omega...

Am I really expected to include the moment of inertia of the ball in this? The question doesn't say whether the ball is solid or not. I assume that the axis of rotation goes through the center of the ball, but again the question doesn't say. So there are three possible moments of inertia that...

It seems to me that you should be able to represent the motion of the rod as potential energy before = rotational kinetic energy after + potential energy after. Does this formula make sense? E_{i} = 3(MgL) E_{f} = \frac{1}{2}I \omega ^{2} + Mg\frac{L}{2} + 2Mgd I =\frac{1}{3}ML^{2} 3(MgL) =...

Yes, I seem to have problems calculating the centre of mass. The formula for the centre of mass I'm using is: COM= \frac{m1 X1 + m2 X2}{m1 + m2} Which in this case is COM= \frac{M \cdot \frac{L}{2} + 2M \cdot d}{3M} The masses cancel and it leaves L/2 + d. Does that seem more...

Homework Statement A thin, uniform rod of length L and mass M is pivoted from one end. The rod is released from rest in a horizontal position, and allowed to swing downward without friction or air resistance. Suppose a small metal ball of mass m = 2M is attached to the rod a distance d from...

Oh, I got it. Sorry, I keep doing this with my posts here. XD

Homework Statement 1. Use the method of Dimensional Analysis to show that the unknown exponents in Equation (1) are l=-1/2, m=-1, and n=1/2. Homework Equations Equation (1) is \lambda = k \mu ^{l} f ^{m} T^{n} Where: \lambda is the wavelength; f is the frequency of the sound; T is...

Oh, right. Square root is the same as taking it to the 1/2 exponent. I keep forgetting that, thanks.

Well, the equation is a lab equation and they always give gravity along with its uncertainty. But is the rest of it right?

Well, if the ball is released at the height of the basket it means that you have to find the initial velocity that will result in the ball returning to its original y position over a horizontal distance of 14 metres.

Homework Statement Derive the error expression \delta v_{}x from the equation v _{x}=\frac{s}{sqrt(\frac{2h}{g})} The Attempt at a Solution I've derived error expressions before, but I have a history of getting the calculations right and the error equations wrong. So, if possible I'd...