Recent content by abercrombiems02

in cylindrical coordinates the integral of the density gives the distribution. In this case the problem requires integrating over an area thus we have a double integral. In polar form J = int(int(f(r,theta)*r*dr)*dtheta) With the appropriate limits. Then J = int(int(1/(sqrt(x)+2),x) from 0 to...

first of all I am somewhat confused in terms of the notation typically c is the damping constant and k is the spring constant I will use this notation in my discussion here... the EOM can be written as y'' + (k/m)*y = F/m*cos(wd*t) here k is the spring constant, m is the mass (w/g)...

The solution turns out to be a first order differenetial equation Since V = a*sqrt(s) where s is the distance traveled we can relate s to the radius and the angle subtended s = r*theta (just basic geometry here) so V = a*sqrt(r*theta) From basic kinematics V = r*thetadot...

You are neglecting the kinematics of this problem. Your work is perfect to the point where you have found the force of friction. Which can sustain a maximum acceleration of u*g. (2.94 m/s^2) Now think of the in plane components of acceleration. The problem states that he is initially...

I get the following expression for my kinetic energy, there is a sign difference in the last term, did you maybe mean to include a plus on the last term? 1/2\,m \left( {{\dot{x}}}^{2}+{{\dot{y}}}^{2} \right) +1/2\,m{\omega}^{2} \left( {x}^{2}+{y}^{2} \right) -m\omega \left(...

So L = T - V I know d/dt(dL/dxdot) - dL/dx = something and also I know d/dt(dL/dydot) - dL/dy = something On the right hand side is there where I place the ficticious forces (for each corresponding equation)?

The ultimate tensile stress is independent of the force and dimensions of a particular material. It is relatively a constant determined bythe material. For example, a piece of Alumnim - T6061 with a diameter of 12 cm has the same ultimate tensile stress as a similar piece with a diameter of...

Your work looks correct F = Gm1m2/r^2 In order for the book to be right you must have Gm1m2/.08^2 = F This means 6.67e-11*1*1/0.0064 = F Evaluating gives... F = 6.67e-11/0.0064 = 1.0422e-8 N This number is both ugly and way too small for 1kg.wt to be equal to 1.0422e-8 N. I...

How does one go about computing the locations of the 5 largrange points? I know where they are but do not know how to derive the equilibrium equations. I know you will get a quintic polynomial which can be solved numerically depending on the masses of the two large bodies, but using forces, how...

You need to start this problem defining two sets of unit vectors, a set that is inertially fixed and a set that is fixed to the rotating disc. In this case if we are in a horizontal plane we can neglect the effects of gravity since this disc is just traveling on the horizontal surface. I've done...

Google the words... concave mirror focal length radius of curvature Just by reading the text below the first link you should be able to find the answer. I'll give you a hint, all you need to do is multiply your 10 cm by a single number!

How would one go about finding an analytical solution to the following ODE note that we are trying to find y(x) subject to... (x*y'')^2 - (1 + (y')^2) = 0

for the first problem i also get an acceleration of 2 m/s^2 since sum F = 0, then it should follow that F = m*a = 4 N The book must have a typo for the second problem it also looks like you solved the problem correctly. While speed typically does travel around 340 m/s at standard...

solved it in 5 mins, enjoy the function! here you go! function out = mat2col(A) [row col] = size(A) for n = 1:col out([1+(n-1)*row:n*row],1) = A(:,n); end

You did everything right, you probably just didnt interpret the answer right Fit1 is a vector of the polynomial coefficient in descending order In this case its telling you the best linear fit for the data is time = -0.0185*year + 47.6837 One easy way of plotting this thing would be...