Need to find velocity and aceleration of B as the function of fi. I have no fucking idea how. God, I hate kinematics.

add the binas formula's and i might be able to figure it out/ But draw it out and see what parts you're missing usually works the best with physics

i don't miss anything. i just can't figure out how to connect the angle and the velocity of that crap with U=2 m/s. I just can't.

If you could provide the statement, I could help u, I don't understand very well the exercise. Is "a" a longitude? I don't see the m/s² so I doubt it is acceleration which is supposed to be "a" (or at least, that's the letter used for acceleration) also what's "u"? I don't know what's m/c or is it m/s? kinda confusing. And the image quality doesn't help :/

The first mass causes a force due to velocity U. Therefore F=dp/dt which is equal to mU/dt. Multiplied by cos(b) (where b the angle between the above string the the bar and using trigonometry from our starting state-using the base triangle-you can get that it's about 0.05 degrees*), it gives us the centripetal force of the system (since it rotates around the axis connected to the bar). That is also equal to Mv^2/r where M is our other mass (which is hanging), v the tangent velocity (which is what we are looking for) and r the radius of the circular movement which in this case is L. You can now combine these as equal to find v. You do have an issue though: m and M are not the same. Therefore you use your data about max φ to calculate a point of static balance between the weight of M and the force caused by the moving mass m. That is done by multiplying the weight Mg with cosφ since you want the weight force vector that is parallel to the centripetal parameter of our first force (with cosb included of course). You once again return to our previous equation and simply replace M with a function of m that also includes cosφ. And you shall get velocity v. If my own calculations are correct, you'll find: v(φ) = 6.26*sqrt(cosφ) m/s The answer makes sense on first sight for a movement like this, velocity is at maximum when φ=0 at its balance position and at minimum when it reaches its maximum angle. *To be more precise: Initial base triangle has a right angle in the far end, and its two legs are a (0.2m) and a+L (4.2m). You divide the first by the latter and get tanb. From there, hit shift tan for the answer on your calculator and the angle shall emerge as about 0.05 degrees. Spoiler: DISCLAIMER I'm not 100% sure that angle b remains static and I have suspicions about whether there is any balance at all at maximum angle φ, but the main idea is this: velocity U causes the force, that force is translated into a force that pulls B towards the center as centripetal.

Don't u read? It's kinematics not dynamics. There are no forces and everything is massless. That thing A hanging at the edge of a stick essentially doesn't do anything and can be ignored. And your answer is incorrect coz it doesn't match my numerical solution.

That doesn't mean you don't use forces at all. At least from my experience. Especially if you are working on a circular motion that involves polar coordinates. You could have stated that everything is massless though, there is clearly a mass hanging on from point B that is detrimental to the movement of the bar. Of course it is, I wrongly assumed there is mass and therefore weight on point B. Anyway, I'll scrap it and re-do my math. I'll post a new answer if I find anything.

u're so thanksless :/ Is it accelerated uniform circular motion? If so, you need to integrate to get acc and int again to get vel iirc. If not, u have to calculate avg. speed and then calculate the acc with the value u get of avg speed. (since u don't have time, the formula is v²/r iirc) I don't remember the rest of the process ;/ I also hate kinematics a bit

That won't work, he needs to use U geometrically to get a function of φ and then get the velocity and acceleration. He doesn't really have any mathematical function yet to perform anything on it.

Orly. Like I don't know that on my second year. ---------- Post added at 01:20 AM ---------- Previous post was at 01:19 AM ---------- The initial task is to calculate for B but it's the same as for A.