Recent content by elsternj

hmm well the reason I am going about it this way is because the problem is in the section of the book where we are finding volumes by integrating the area of the cross section. I am not aware of another way to do this problem.

Homework Statement The base is the semicircle y = \sqrt{9-x2}(Square root of 9-x2.. i don't know why the formatting isn't showing up) where -3 <= x <= 3. The cross section perpendicular to the x-axis are squares. Homework Equations -3\int3 = A(y)dy A(y) = area of cross section...

Homework Statement Derivative of Sin22x Homework Equations dy/dx = dy/du * du/dx y=U2 The Attempt at a Solution Just want to make sure I am doing this right*. Do I let U = Sin2x or U = 2x? Let's say U = Sin2x y=U2 then y` = 2Sin2x * cos2x? Or if U = 2x. y =...

awesome thanks! that makes sense. I am now having trouble finding the amplitude if x = Acos(\sqrt{k/m}t) x = Acos(8.15*.771) however i still have 2 unknowns here. x and A. i can't see to find another formula in my book that will help with this. Everywhere I find A I also find x.

okay i see now. .10 is the time for a period. so .10 = 2pi \sqrt{m/220} multiply both sides by \sqrt{220} 1.48 = 2pi\sqrt{m} divide both sides by 2pi 2.43 = \sqrt{m} square both sides m = 5.43 which is still the wrong answer. is my math wrong? what exactly am i doing...

I tried .2 and .3. are any of these the right time for period? If so then my problem lies elsewhere. any insight? thanks

Homework Statement On a horizontal, frictionless table, an open-topped 5.50kg box is attached to an ideal horizontal spring having force constant 365 N/m . Inside the box is a 3.44-kg stone. The system is oscillating with an amplitude of 7.50 . When the box has reached its maximum speed, the...

would it be .2s? when it passes back through O and to where it started?

Homework Statement On a frictionless, horizontal air track, a glider oscillates at the end of an ideal spring of force constant 2.20 N/cm . The graph in the figure shows the acceleration of the glider as a function of time. Find the mass of the glider. Homework Equations T=...

Ah, I see it doesn't exert a torque because it is actually attached to the point of rotation, right? So... friction? -f(2R) = 1/2m(2R)2\alpha ? So oddly enough the answer to this question is 3.27 which kind of confuses me because I was expecting it to have to be in terms of M and R like it...

Homework Statement A uniform, solid cylinder with mass M and radius 2R rests on a horizontal tabletop. A string is attached by a yoke to a frictionless axle through the center of the cylinder so that the cylinder can rotate about the axle. The string runs over a disk-shaped pulley with mass M...

actually i just tried mr^2 / 2 (disk) + mr^2 (parachutist) and got the right answer. this the right way?

having trouble finding the moment of inertia once the parachutist lands then. I = \summr2 1120 = (140)(2)2 + (65)(2)2\omega and the answer still comes out to be 1.37. Wouldn't this be treating it as a 2 particle system?

okay that seems simple enough but still not coming up with the right answer. L = I\omega I = mr2/2 L = (140)(2)2/2(2) L = 560 560 = mr2/2(\omega) 560 = (205)(2)2/2(\omega) \omega = 1.37 Not the right answer I also used the equation L = r x mv (v = r\omega) for both and I got the same...

Homework Statement A large wooden turntable in the shape of a flat uniform disk has a radius of 2.00m and a total mass of 140kg . The turntable is initially rotating at 2.00rad/s about a vertical axis through its center. Suddenly, a 65.0-kg parachutist makes a soft landing on the turntable at...