Recent content by VMP

Okay, I've been a bit vague. Imagine you've got an origin ##\mathcal{O}## at the center of the stick and there is a vector ##\vec{a}## that is given by the contact point P of the spring and the origin. This vector ##\vec{a}## is always perpendicular to the vector ##\vec{r}'##. Vector...

P.S. I'll try to model the spring "constant".

Okay, I think I have a solution. If we watch time interval (t,t+dt) then length of the spring at time t+dt is: r'(t+dt)=r'(t)-dr'-a\cdot d(\theta-\alpha)\;,\;r'(t+dt)-r'(t)=dr'\;,\;\alpha=arctan(\frac{\sqrt{r^2-a^2}}{a}) \\ \;\\ r'^2=r^2-a^2\Rightarrow...

Yes I did and no success. I have a hunch that the spring is culprit of the problem, so I'm trying to model the spring in following way: The speed of the spring near the contact is approximately 0 therefore ds=L-dr'\;,\;dr'=\frac{rdr}{\sqrt{r^2-a^2}} where ds is the amount of the spring wrapped...

Homework Statement In the beginning a point mass is rotating in a circle of radius L. The spring is providing the centripetal force (\vec{F}=-k\vec{r}) and the mass rotates with constant speed. At some point in time, a stick of radius a (a<<L)lands near the center of the circle in such a way...

If for the i-th day day U_{i}=i+\frac{1}{7}(m-i-\sum_{j=1}^{i-1}U_{j}) then U_{n}=n+\frac{1}{7}(m-n-\sum_{j=1}^{n-1}U_{j}) since \sum_{j=1}^{n-1}U_{j}=m-n It is easy to see that U_{n}=n follows the mentioned pattern. I've already checked the result in the beginning but forgot to mention it...

Interesting, it's a good convention in that context. Anyhow, here we go: Solution which should prove uniqueness: Let U_{i} be defined as following: U_{i}=\left\{\begin{matrix} 1+\frac{1}{7}(m-1),\;i=1\\\\ U_{i}=\frac{6}{7}(1+U_{(i-1)}),\;i=2,...,n \end{matrix}\right. For an arbitrary i>1...

Note that if i=1 the sum is undefined. Did you guess the answer or do you have a systematic approach? Cheers!

Let's say m=8 then the number of medals given on day 1 is U_{1}=2 Problem is to find an elegant condition for U_{i} such that every U_{i}\epsilon\mathbb{N},\forall i,(i=1,2,...,n). P.S. The condition where m=-1+13k is wrong, same for n.

Hello everyone, I have an issue solving the following problem: You're on a mathematical Olympiad, there are m medals and it lasts for n days. First day committee gives U_{1}=1+\frac{1}{7}(m-1) medals. On the second day U_{2}=2+\frac{1}{7}(m-2-U_{1}) medals, and so on... On the last day...

In system of equations marked by (1) \mu is one function which is defined differently on different intervals. For clarity sake: \mu(\theta)=\left\{\begin{matrix} (\,\mu_{k}=const.\,),[0,\theta_{0})\bigcup (\angle,\pi]\;(*)\\ \mu_{s}(\theta),[\theta_{0},\angle]\end{matrix}\right. Also, note...

Time derivative of equation (3) in post #84 is 2\dot{\theta}\ddot{\theta}=\frac{10g}{7R}cos\theta\dot{\theta}. You'll have to be more specific with your last \mu related question.

No, first term is kinetic energy of CoM, second term is rotational kinetic energy. Recall, if the ball is rolling without slipping, then v_{cm}=R\dot{\theta}=r\dot{\varphi}.

Hey, as I've said, I came back to redeem myself. I totally neglected the nature of \mu and thus blundered spectacularly. The solution to initial question "What should be the minimal friction coefficient at angle θ so that the ball won't slip?" I will mark with (*) for clarity. Now I will focus...

Yea, I just realized i screwed up the differential equation solution, sorry. The energy equation should also contain the plus sign. I'll try to redeem myself when I get the time. Cheers