Ok I did it again making sure to get the powers correct and I got M = 44k/5.
I calculated My to be 9k/2.
When I divided My/M to get xbar I got 88/45 which can't be right!
ok I watched a couple videos on youtube and this is what I came up with:
M= \int_{x=-1}^2 \int_{y= 2-x}^{4- x^2} ky dydx
⇔ k\int_{y=2-x}^{4- x^2} ydy
= k (y2/2) = k [(4-x^2)/2 - (2-x)/2]
= k/2 [4-x^2 - 2+x]
= k/2 [-x^2+x-2]
then
M= \int_{x=-1}^2 (k/2 [-x^2+x-2]) dx
= (k/2)[(-x^3)/3 +...
Ok I'm in my first calculus class and some of unfortunately I don't even recognize some of your notation. But I do understand somewhat and will try to use this. Thanks very much.
ok i am trying to figure this out. but my density function is ky, where k is some constant. it doesn't have an x value
Homework Statement
Set up integrals to find M, Mx, and My for the thin plate if d(x)=ky
bounded by y=2-x and y = 4-x^2 or x = sqrt(4-y)
Homework Equations
M = \int dm = \intd(x)dA
Mx = \intydm
My = \intxdm
The Attempt at a Solution
I broke this down into three equations. x1 =...
Homework Statement
Trying to find all Max and Min.
F(x) = x^(2/3)(x^2 - 4)
Homework Equations
I know to use the product rule
The Attempt at a Solution
I tried and got this answer:
x^(10/3) + (2(x+2)(x-2)) / 3x^(1/3)
torque = moment of inertia * angular acceleration
so, since net force = 1n, torque = r * F, so torque = .1N*m.
Angular acceleration = t / I = .667 ms/^2
is that right?
Homework Statement
Two forces are applied to a pulley with a moment of inertia I = .15kg*m^2.
The pulley is mounted so that its frictionless axle is fixed in place.
1) What is the angular acceleration of the pulley?
2) If the pulley starts from rest, how long does it take to undergo...
ω is constant, but it is uniformly changing direction. So there has to be an angular acceleration.
edit --
but since angular acceleration = tangential accel. / r -- would angular acceleration = 0?