Recent content by sun18

Oh right of course, my mistake. I was forgetting that μ absorbs G and M. Luckily it should end up cancelling out anyways. I had a feeling that the directionality was non-trivial. The first part of the problem had the rod colinear with the displacement vector ρ, so it didn't matter then. So I...

Just to point out, the units of torque are N*m, and you're writing kg*m. In the end, it won't matter because the factor of g will cancel out, but if you had to show your work, you would need to include (mass)*(gravity)*(length)

Imagine that the board is completely massless, except for one concentrated mass at the center of mass of the board. Then, the only masses in the system are the orange cat on the left, the mass of the board on the left, and the blue cat on the right.

That's right, so now think about the problem: you have 2 masses on the left of the pivot point, and one on the right. Now try to balance the torques and solve for M.

You are trying to balance torques about the pivot point of the seesaw. What you have written is mass*length, which does not have units of Newton-meters. You are close to the right idea, but the second term on each side of the equality is not correct. As mukundpa says, find the center of mass of...

Hm I hadn't thought about using the angle. I will give that a try tomorrow and report back. Thanks for the responses by the way, I really appreciate it.

There are no external torques in this problem, therefore angular momentum is conserved. h = Iω

You are correct on all three accounts: ρ is the position vector from the Earth's center (mass M), to the infinitesimal mass element dm. ρc is the distance from Earth's center to the rod's (mass m) center of mass ρG is the distance from Earth's center to the rod's center of gravity. And yes, the...

Homework Statement Consider a slender rod satellite in a circular orbit. Show that the position of the center of gravity, ##\rho_G##, can be written in terms of the position of the center of mass, ##\rho_C##, as: ##\rho_G = \rho_C (1+\frac{l^2}{4\rho_C^2})^\frac{1}{4} ##, where ## l## is the...

Generally speaking, center of mass and center of gravity are not in the same location, even with uniform mass distributions (assuming a non-uniform gravity field). This is because gravity pulls harder on things that are closer to the attracting body. So for the case of a vertical rod, the...

We haven't covered phasors yet but I read ahead and it makes sense now. Thank you for the response.

Are you adding the two vectors together? Is that the problem? I'm going to assume so. In terms of notation, it is best just to choose an x-y coordinate system to place your vectors. Also, try drawing them both originating from the same point. After you set up a coordinate system, decompose each...

Homework Statement I'm supposed to find the current in a circuit with a voltage source, capacitor, and resistor in series. The voltage source is described by V=V0ejwt. Here, j is the complex number j2=-1, and i is the current Homework Equations I=C*dv/dt The Attempt at a Solution I have...

You have already solved this problem! You are looking for an answer in rpm's. You have a solution in radians per second. There is only a conversion to do from here.

Thanks so much for the response gneill. I guess I was overthinking it instead of concluding the obvious.