Recent content by Shifty

Got it: P_{initial}=m_rvL/8 Then it works. Thanks, I realized that the angular momentum of the bullets should have been the same, your suggestion worked :).

P_{net}=m_rv/4=m_rL^2\omega/3+m_b(L/2)^2\omega=m_rL^2/3+m_rL^2\omega/16 That gives me: P_{net}=19/48*m_rL^2\omega\Rightarrow\omega=12v/19L^2 Not quite there yet.

Yeah I did get that Moment of Inertia

Before: m_{B}=m_{r}/4 v=L\omega/2 \omega=2v/L P_{net}=m_{r}L^{2}(0)/3+m_{B}\frac{L}{2}\omega=m_{r}L^{2}v/4 After: I_p=I_{rod}+m_BL^2/4=19/48*m_rL^2 P_{net}=19/48*m_rL^2\omega Solution: m_rL^2v/4=19/48*m_rL^2\omega\Rightarrow\omega=12v/19 I got close but I mest up somewhere...

A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes...

Thanks for the reply :), but I already tried that and calculated the change to be -203 J, unfortunately it didn't yield a correct answer. \Delta U=-mgy=203 y=23m/4 EDIT!: looking at this reply I realized the total mass would only be half since only half of the rope was lowered. That solved...

A uniform 3.60- kg rope 23.0 m long lies on the ground at the top of a vertical cliff. A mountain climber at the top let's down half of it to help his partner climb up the cliff. What was the change in potential energy of the rope during this maneuver? -This in a moment of inertia assignment...

Well I couldn't get the answer, and a different part asked me to input the direction of the rock and it's not the same to the bullet's initial direction. This means the bullet can't be going in the opposite direction, because in order to change the direction of the rock, the final direction of...

Thank you for the reply, sounds reasonable :). What confuses me is the bit of "right angles" to the original direction. Is it talking about 2 right angles adding to 180 degrees? Either way, horizontal gives it away I guess. I would have understood understood it better if it just said its...

Hey guys, I'm having some trouble understanding the highlighted text below. Just need some quick clarification on the right angle stuff, don't get how to draw my diagram. Problem: 0.100 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 310...