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...
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...
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...
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...