Recent content by dkgojackets

For U as potential energy, wouldn't I multiply it by the 4R above the Earth's surface?

Homework Statement Find the energy required to launch a satellite from Earth into circular orbit at radius 5Re in units of joules. Msat = 1000 kg Me = 5.98e24 kg Re = 6.37e6 G = 6.67259e-11 Homework Equations F = GMm/(r^2) The Attempt at a Solution I'm thinking somehow...

Got it. I was just combining the two approaches, using the parallel axis theorem to find MoI in the first equation.

I know I can substitute v/r for w. I have mgr = .5mv^2 + .5I(v/r)^2, with v being velocity at the center of mass. I have r and m, I just don't know how to adjust the moment of inertia for it rotating around the edge.

Homework Statement A uniform solid disk of radius (r) 6.71 m and mass (m) 38 kg is free to rotate on a frictionless pivot through a point on its rim. If facing the disk, the pivot is on the left side. The disk is then released. What is the speed of the center of mass when the disk reaches a...

OK here's what I did. v = wr t (of one rotation) = (2pi)/w t (total) = (62pi)/w I also used Vf = Vi + at ... so 0 = v - 9.8t, t = v/9.8 Set those ts equal (62pi)/w = v/9.8 substitute w = v/r and cross multiply (9.8)(62pi) = (v^2)/r Now would r be the length of the stick...

v=rw, so velocity at center of mass would equal .135w? I also figured that the time of one rotation would be 2pi(.27)/w, so for all 31 rotations it would equal 62pi(.27)/w. Am I on the right track?

Homework Statement A massless rod length L has a small mass m attached to the center and another mass m attached at one end. On the opposite end, the rod is hinged to a frictionless hinge. The rod is released from rest at a horizontal position and swings down. What is the angular velocity as...

Homework Statement A girl throws a stick of length .27 m and mass .18 kg into the air so that the center of mass rises vertically. At the moment it leaves her hand, the stick is horizontal and the speed of the end of the stick nearest to her is zero. When the center of mass reaches its...

Yeah I got it. The first three parts of the question were the same value.

Homework Statement Tarzan (mass 98.3552) stands on a 4.45 m cliff next to a river. He grabs a vine of length 30.1 m whose point of support is directly above Jane (mass 57.5336). He grabs Jane at the bottom of his swing and has just enough speed to make it to the other side. How long does the...

still need help on frictional force please

You have two vectors at a right angle to each other, and one connecting them, pointing straight down. Just draw it and solve for the right triangle.

I got it now. The bullet into the block gives the block a kinetic energy, which is immediately transferred into the spring. I know the spring potential energy and used it to solve for the velocity of the block immediately after the collision, which I put into conservation of momentum equation.

I get that momentum is conserved during the collision, and mechanical energy after. But why would conservation of energy not work if I have just one unknown in the equation? .5(.009)(300^2) = .5(.009)(v^2) + .5(904)(.02^2) is my energy conservation equation.