Recent content by roadrunner1994

Thanks, I understand the chlorine will react with bath constituents which mostly gives us soluble compunds which will dissociate on ions, right? So we are back to square one and we have chlorides in the bath again. So is the decrease of the degreasing efficiency due to the fact that the bath...

Hi, Why contamination of the bath for electrochemical degreasing by chlorides or soap adversly affects degreasing of parts? What's the difference if chlorine is produced instead of hydrogen or oxygen? I thought that this gases only help the grease to break away from the part's surface. I'm...

I guess the center of mass will be affected by the force F=Matangential atangential can be calculated from Iα=Mg(3/2)L Therefore α=(9/14)g/L, and atangential=(27/28)g. But both forces: Mg and Matangential should be pointed downward, so why shoud they be subtracted to obtain correct result?

I got the result: F1=(54/31)Mg, and the correct answer is (1/28)Mg. I'm trying to solve the 1)st part :smile: Yes, I wanted to apply parallel axis theorem. It seems that I should rather use (1/12)ML^2 instead of (1/3)ML^2. In part 2) it gives the result: F2 = Mg + F(3/2)Lω^2 = (41/14)Mg, which...

Hi, I started with calculating the moment of inetria of the rod: I = ⅓ML^2 + M(3/2 * L)^2 = 31/12 ML^2 and I thought that the reaction force in the first case will be equal to centrifugal force: F1 = Mω^2*(3/2)L Angular velocity is calculated from the conservation of energy: Mg3/2*L=1/2 * Iω^2...

Thank you :smile: And for the subsequent subpoint: Show that W = 0∫trμMgVc(t)dt, where Vc is the velocity of the contact point relative to the supporting surface (the floor). The velocity of the contact point will be the following: Vc = ωR - v = ωR - v0 + μgt, right?

In this way? W = ΔK = ½Mv^2 + ½Iω^2 - ½M(v0)^2 since v = v0 - μgt ω = (5/2)μgt/R t = 2/7(v0/μg) the result is negative → W = -1/7[M(v0)^2] Is it correct?

One more question :smile: (3) Determine the work W that the forces of friction perform in the time from t = 0 to t = tr i.e. in the time where the ball was both rolling and sliding I suppose I can't just use the equation t0∫trFvdt, can I?

You are right, I think the solution will be: a = -μg α = (5/2)μg/R v = v0 - μgt ω = (5/2)μgt/R and at the moment when ball starts to roll v0 - μgt = R* (5/2)μgt/R t = 2/7(v0/μg)

Hi, I solved this prolem in the following way. I have started with the angular momentum theorem: Iα=fR As the force of friction vector is attached on point of contact of the ball and the supporting surface, the moment of inetria is: Ic= MR^2 + Icm = MR^2 + (2/5)MR^2 Ic*a/R=μMgR tr=7/5(v0/μg)...

Hi! Is anyone familiar with trivalent conversion coating on aluminum? I'm struggling with unrepeatable and too high values of contact electrical resistance of coating. I would really appreciate any advice how to deal with that coating. Thanks a lot!