Recent content by natasha13100

Homework Statement Hey. I am taking my first Real Analysis course at the undergraduate level and am not really sure where to start on a problem. I do have some ideas. I am not asking anyone to solve it for me, but I do need an idea where to start. Show that |f(x)-f(x0)|<|x-x0| if...

Yes, but I wanted to make sure I knew how to prove it if I ever needed to. It seems as I get farther into mathematics I forget about using the basics.

Thanks for the advice. It really helped. I can't believe I didn't think of such a simple solution. I ended using algebra to show xn>xn+1. Then I hashed out my thoughts about xn<4 and used the theorem I was trying to use.

Homework Statement Show that lim(n→infinity) 1/n*{2*4*...*2n/(1*3*...*(2n-1))}^2 exists without finding the limit. Homework Equations probably the following: Let {xn} be a sequence such that xn≥xn+1 and xn≥M for every n. Then the series is convergent. The Attempt at a Solution I know...

Homework Statement A pendulum consists of a 300 gram mass attached to the end of a string of length 75 cm. • a) Using the small angle approximation(sinθ ≈ θ), derive an equation for the angular acceleration of the pendulum, assuming that there is no friction. • b)] The pendulum is given a...

okay. thanks :)

so out of curiousity, what would I have done if I was looking at the rod when it is horizontal? I can't say the angular acceleration is zero there. In fact, angular acceleration would be at its highest. In that case mgL/2=Iw^2/2=mL^2w^2/6 g=Lw^2/3 w=sqrt(3g/L) I would guess t=rfsin(theta)=mgL/2...

With uniform circular motion the tension in the string or in this case the force exerted by the pivot is the net force acting on the object. The magnitude of the velocity doesn't change but its direction does. Since gravity is negligible very close to the bottom of the swing, the rod can be...

Would I be able to consider this as uniform circular motion? I believe so because the force due to gravity doesn't make much difference near the bottom of the swing. In this case, it would simply be a=v^2/r

I did obtain zero angular acceleration, but zero acceleration at the free end of the rod when it is in the downward position was incorrect. The horizontal component was 0 but the vertical component is nonzero. I am guessing this is kind of like a pendulum where the string keeps going after...

theta is the angle between the horizontal line extending from the pivot and the rod with counterclockwise being positive and clockwise being negative

I ended up figuring it out on my own based on some thought (#5) and the conservation of energy ended up agreeing once I got my signs straight (#6). However, the second part is what's tripping me up. I thought it would be easy once I understood the first part, but I'm missing something. The...

oops. I'll put the diagram in now.

conservation of energy: if I use the pivot as the origin, mgL/2=Iw^2+mgL/2*cos(theta) Iw^2=mgL/2*(1-cos(theta)) w is maximized when cos(theta) is maximized which occurs when cos(theta)=-1 which is at the bottom

Well kinetic energy is maximized when potential energy is minimized. I think angular acceleration increases up until the bar is horizontal because t=rFsin(theta) and sin(theta) increases from theta=90 to theta=0 (assuming the horizontal is theta=0). Then the angular acceleration decreases until...