Recent content by integra2k20

just one more question - over what interval would i integrate the function? L/3 of the rods length is to the LEFT of the point, and 2L/3 is to the RIGHT, so i would assume i integrate over -L/3 to 2L/3

thanks, you just saved me a lot of trouble. I found the correct answer to this problem, but was having trouble figuring out just how to get there. It makes sense now, a rod is not composed of a bunch of point masses. thanks

By the way I didn't really understand what you said about the Inertia, i thought I had it but i reread your post above and it seems that you said 9ML^2/108 was wrong. My method for doing this was treating the rod as if it were composed of two smaller rods, one to the left of the pivot, and one...

Thanks for all the help thus far, Doc Al. The last part of the question is: The rod is brought to rest hanging in the vertical position, then displaced slightly. Calculate the period of oscillation as it swings. Now, since this is a physical pendulum, I know that the period is...

the center of mass or the pivot point?

I found the moment of inertia, but I'm not sure exactly how to set up the conservation of energy. Usually its mgy = 1/2Iw^2 (where w is angular velocity at the bottom), but in this case I'm not really sure what "y" would be.

The rotational Intertia would be the sum of MR^2, so you would treat the portion to the left of the pivot as one mass and the portion to the right as one mass, taking the length of each R (radius) from the center of mass. I got 9ML^2/108. Can this be solved by summing torques and setting...

Homework Statement This problem involves a uniform rod of length L with a pivot at point L/4 (so that 1/4 of the rod is behind the pivot and 3/4 is in front of it). The rod is released from a horizontal position and drops down. You need to use conservation of energy to solve for the velocity...

Homework Statement Sir Lost-a-Lot dons his armor and sets out from the castle on his trusty steed in his quest to improve communication between damsels and dragons (Fig. P12.20). Unfortunately his squire lowered the draw bridge too far and finally stopped it 20.0° below the horizontal...

i think i pretty much figured it out. i played it on my DVD player at 2x and 4x, and at 4x time appeared to be "correct" as in at the normal speed. i am assuming that the entire scene takes place at 1/4th of the real speed, so in this way i can figure out the real times

OK, so which do you think would work. the problem with time is that both are in slow motion. for the purpose of my proof i was trying to measure time in "slow motion" seconds. For the second one, i have no clue what to estimate the weight at. in the movie, the knives were 14" long and looked...

movies like "star wars" and other movies that are not supposed to be based in reality are not allowed, unfortunately...

the knife (14" long or so) was thrown very straight at the guy,hit him, lifted him up off the ground and back 1-2 meters. he did not "stumble" but was literally thrown back by the thing.

"Bad" Physics in movies. OK, so, I have to do something for my physics class involving disproving the physics of something that happens in a movie. I have come up with several ideas to make this work, including scenes from: * Transporter 2 (A man jumps out a window about 3 stories high...

oh wow, never mind, i made some type of calculation mistake. for some reason i got 140 when it shoudl have been 92. my last question is about the last part, it says to find the total kinetic energy evaluated from (sum of)mi vi^2/2.