Recent content by tylertwh

The velocity of an object at t=0 seconds is 40 m/s directed due North. The velocity of the same object at t=5 seconds is 30 m/s directed due east. Find the magnitude of the average acceleration of the object over the time interval from t=0 to t=5 seconds. Also, show the direction of the...

I found my mistake for part A. The answer to part A is 1.486

Homework Statement A 550.0g bird is flying horizontally at 2.20m/s , not paying much attention, when it suddenly flies into a stationary vertical bar, hitting it 25.0cm below the top. The bar is uniform, 0.730m long, has a mass of 2.20kg , and is hinged at its base. The collision stuns the...

figured all of it out! no more help needed

well... I used all my attempts for that one.. the answer is (6/19)v/L Could you possibly explain why?

wait. r => (1/2)r... so the answer would actually be 6v/L = ω

Iω(initial) = Iω(final) mv2ω(initial) = (1/3)mv2*[v/r] (mv3)/r = ((1/3)mv2ω ω = 3v/r = 3v/L

how would that work when the bullet hits the rod at a 90 degree angle? I don't know what I would put for the ω(bullet). Unless somehow I incorperated ω=v/r, but even then I am not sure what I would put for the r? Iω(initial) = Iω(final) mv2ω(initial) = (1/3)mv^2*[v/r] ω(initial) = ?

Homework Statement 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...

The answer was correct! Thank you very much!

Yeah I noticed that when I made the reply. I edited my previous reply! Thank you so much!

(m/L) * (1/3)L3 - (1/3)03 which when you simplify is M*(1/3)L2 Thank you so much for your help. I will post if the answer is correct or not shortly

L = Iω I(bug) = Mr^2 ω(bug) = v/r L(bug) = Iω = Mr^2ω = Mrv I(bar) = 1/3*Mr^2 ω(bar) = v/r L(bar) = Iω = 1/3Mr^2ω L(bug) = L(bar) M*r*v = (1/3)*M*r^2*ω 0.013*0.12*0.025 = 0.02167*0.12^2*ω 3.9*10^-5 = 3.12*10^-4ω ω = 0.125

Why is it r2 and not r3? because I thought you would take the integral of mr2 and in order to get it outside of the \Sigma you had to take the anti-derivative which would be 1/3mr3?

So if I change the moment of inertia of the bar to 1/3mr^3 then I should get the correct answer?