Recent content by JHCreighton

Jeez, this problem has just worked me over, and for no good reason. OK, simply substituting everything into the angular displacement equation, then converting back to revolutions, I get 14 rev. I can't tell you how much I appreciate the help, especially since you took the time to go step by...

OK, so let's see. 420 rev/MIN is 7 rev/s, which is 44 rad/s. Rearranging the kinematic equation to be \alpha = vf /t and substituting, I get the angular acceleration to be 22 rad/s^2. Then, if the initial velocity is 0 rad/s, the final velocity is again 44 rad/s. This converted to rev/s is...

1) So, 420 rev/s would be 840(pi). (I can't find the Pi character) because 1 rev=2(pi) rad. Correct? 2) The equation should be \omega = \omega 0 + \alpha t

Homework Statement An airplane engine starts from rest; and 2 seconds later, it is rotating with an angular speed of 420 rev/min. If the angular acceleration is constant, how many reovlutions does the propeller undergo during this time? Homework Equations \theta = \omega 0 t +1/2 \alpha...

Hey, that's great! You're right, it is pretty simple. I almost feel foolish for not thinking to solve like that. Thanks for the speedy response. JHCreighton

I am just curious as to how this fits in. If momentum P=mv, and kinetic energy KE=1/2mv^2, how would one combine, derive, switch and swap (whatever the process is called), these two equations to end up with the formula m=P^2/2KE. It seems like a no-brainer, but I can't seem to make sense of...