Recent content by DavidAp

Hi there! I know it's been a while since I've posted this but I've been scratching my head and I can't figure out what PE is. I was thinking that, after taking another look at the problem, the PE for the center rod is mg(l/2) because that's where it's center of mass is, but I'm not sure what...

A rigid "H" falls rotating about one of its legs. What's its angular velocity? A rigid body is made of three identical thin rods, each with length L = 0.340 m, fastened together in the form of a letter H, as suggested by the figure here. The body is free to rotate about a horizontal axis that...

Thank you so much, not just on this problem but on future problem to come! I never thought of visualizing it that way!

I don't know how to draw 0 <= y <= x. I think that might be part of my problem... I did something wrong though when using pi/2! The answer is 3/64 pi^2 but I keep getting 3/16 pi^2! Why is my denominator 4x less than the answer? Here's my work. ∫(1 -> 2) ∫(0 -> pi/2) θr drdθ 1/2 ∫(1 -> 2)...

The problem and my work is shown in the image below. However, I feel like I did something horrible wrong but I'm not sure where! I'm sorry if my handwriting is illegible. If you're having difficulties please leave a comment and I will not hesitate to type it out as a response. Any...

O.O! I love you forever, thank you!

∫∫cos(x^2 + y^2)dA, where R is the region that lies above the x-axis within the circle x^2 + y^2 = 9. Answer: .5pi*sin(9) My Work: ∫(0 ->pi) ∫(0 -> 9) cos(r^2) rdrdθ u = r^2 du = 2rdr dr = du/2r .5∫(0 ->pi) ∫(0 -> 9) cos(u) dudθ .5∫(0 ->pi) sin(u)(0 -> 9) dθ .5∫(0 ->pi)...

So the centre of mass of the beam, disregarding the package on top, should be the center of the beam. I did some quick integration and it was right. Xcm = ∫(0,l) xρdx / ∫(0,l) ρdx = ρ(X^2 / 2)(0, l) / ρ(X)(0,l) = (l^2 / 2)/(l) = l/2 Once I have that do I use a series to calculate the center...

Figure 12-52a shows a horizontal uniform beam of mass mb and length L that is supported on the left by a hinge attached to a wall and on the right by a cable at angle θ with the horizontal. A package of mass mp is positioned on the beam at a distance x from the left end. The total mass is mb +...

Thank You! I solved it after a triple glance at the board! I knew something had to be wrong so I change my first integral to be 0 -> 1 and it worked! I'm just putting this out there because I don't know how to delete a thread, but thank you anyways forum!

The double integral xcosy is bounded by y=0, y=x^2, and x=1. I was able to integrate almost wholly through; however, toward the end I was unsure what to do when i was asked to plug in x^2 into x^2. What do I do?! Here is an image of my work on the white board. Please, if my hand writing is...

A man (weighing 746 N) stands on a long railroad flatcar (weighing 3000 N) as it rolls at 16.8 m/s in the positive direction of an x axis, with negligible friction. Then the man runs along the flatcar in the negative x direction at 4.46 m/s relative to the flatcar. What is the resulting increase...

Block 1 of mass m1 slides from rest along a frictionless ramp from height h = 3.1 m and then collides with stationary block 2, which has mass m2 = 5m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction µk is 0.35 and comes to a stop in distance d within...

Block 2 (mass 1.10 kg) is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 144 N/m. The other end of the spring is fixed to a wall. Block 1 (mass 1.70 kg), traveling at speed v1 = 3.60 m/s, collides with block 2, and the two blocks stick...

Two objects have unequal masses, m1 > m2. If their kinetic energies are equal, which has the greater momentum? This is the way I approached this problem. I know that, Momentum = mass*velocity K = 1/2mv^2 So, I solved for v since it is the only unknown in these two equations. k = 1/2mv^2...