Anoninho

@Fima Hello fima... it seems to be the right equation, you see... The force only starts to push the block when it "wins" the Friction Force (Ff). So, its only after 2 seconds has passed. Using your numbers: Ff = m*g*mu = 2*10*0.2 = 4N... F=bt^2 = 1 * (2)^...

@Fima, You've said: B) vf=5.33 But this v_f is only for t=4, that used at question C, at question B is used t=5. Check it out.

@fima, Have you submited that deltaE value? I thought it had to take in account Kinect Rotational Energy of the Merry-go-round... But the energy loss is about the same (974.234 without it, 951.129 considering it.) Just wanting to confirm.

@Joe, I've tried something like you've said... But I don't know if I'm right... I found velocity at t=125 and t=124.999, and found 113.139 and 113.140. So instant acceleration would be somthing abount 0.9312 m/s^2 (roughly, 1 m/s^2)? I could understand yo...

Hil, I haven't tried this one yet... but I don't know if a is an "average" cause a changes as function of mass... you see? As the rocket goes foward, it loses mass, but it keeps throwing mass out, so it has an "average" thrust... so, thrust won't change....

Sorry Damon, I'm trying to make by myself, and your notes are beeing really helpful, but I could't understand 1 thing, at here: ax = d^2x/dt^2 =-.075 sin T (dT/dt)^2 + .075 cos T alpha I see that a_x it's the second derivative of x = (l/2)*sin(theta)... B...

@Phy, a) Integrate beta*t^2 from 0 to 5... as it is the only force, you shall get someting like: p_f = beta/3 * t_f^3... b) Its more complicated, need some click on your mind. The block will only start to move when F > Ff... So it won't start to move unti...

Oh, I see... So, you must derive function "sin/cos" and theta, as both change with time. Then (without l/2, for simplicity) x = sin(theta); dx/dt = cos(theta)*d(theta)/dt; d^2x/dt^2 = a*db/dt + b*da/dt; a*db/dt = cos(theta)*d^2(theta)/dt b*da/dt = (d(thet...

@Kumar Trying to "massage" alpha equation, I've got: 3*g*sin(theta)/(2*l) Wich that for fima values, got alpha = 50, as she said. Try to evaluate it again.

@MegaM Maybe I'm wrong, but I think that your alpha should be: 3*g*SIN(THETA)/(2*L) Not cos(theta)... Try to check it out, maybe I'm wrong. (:

Hello guys, After some massage got an function for f_y, using f_x: f_y = (2 * I_cm * alpha)/(l*sin(theta)) + f_x * cotangent(theta) For I_cm used 1/12 * m * l^2, so made it throught the center of mass... For m=150g and l=20cm I've got 0.958, and an green...