Search results for query: *

so I need to use the eq.: omega_pr = ##mgr / L_s## with r = 6cm. But from how I imagine/see it, the rod pierces the disk through the middle so how does this create precession if the weight will not cause a torque then and thus a hortizontal change in angular momentum? Thanks in advance!

should've examined it more properly basically.. the problem is not reallly difficult if I would've thought a bit simpler about it. i.e. from the basic principles up.

okay right.. it's the support forces then, correct?

well.. that is exactly what I don't understand.. I didn't understand why the solution has used two integrals in such a way.. I haven't used a volume integral or smtn. I did differently by using a mass element and summing over that with one definite integral.

exactly that is what I didn't do.. should've seen that there still is gravity as well.

It's gravity, isn't it? That the one that creates the torque. If I were to take the 'rotation axis' in f.e. disk2, the gravitational force of disk 1 would still exert an angular impulse

yes apologies

well.. a torque? but internally isn't it?

Well, friction will occur. But I can't see how this creates non-conservation of AM Or.. is it the torque/moment that the force of your hand will apply to the disks when smooshed together that creates an angular impulse?

Here is question + drawing.

this is the solution. not mine :)

Yep, that is what I did I expressed in terms of y^2 and then went ahead with that. But I just didn't understand how this solution came about basically.

I don't really understand what the 2 integrals (dx and dxdy) for I_x represent. Could I get some explanation here please? Thanks in advance.

yeah apologies it should have been 2 times the symbolic equation, just forgot to type it out. Omg... totally forgot that... thanks a lot... I needed to use the moment of inertia about the center of mass basically..,.

so calculated, the moment of inertia for a rod about an axis at the end of the rod is I = 1/3 * M * L^2 here for case 1: arms to the side I is calculated to be ##I = 0.224## for case 2: arms stretched ## I = 1 / 3 * M * L^2 + M * d^2 ## with L = 0.6 m (length of rod) and d = 0.2 (dinstance from...

Hey guys, Can someone help me understand how to understand this problem intuitively please? How I understand is that I need to look the acceleration relative to the lift as if it were f.e. on another planet with a different acceleration. this gives me a = g - 5. But then again if I didn't look...

thanks guys I understand it now I think. ##\Delta \theta_1 = \Delta \theta_2## which basically makes the angular velocity equal.

apologies, went on vacation. well.. that is the only thing that I don't understand. How do we determine that the angular velocities of the two starts are the same??

do you think that using vectors will fix the problem? And what do you mean by using angular velocities? like v = omega * r or?

let me show you what I have done here: for m1: newton's law: ##Gm_1*m_2/(r_1+r_2)^2 = m*v_1^2/r1## --> ##v_1 = square root of m_2*G*r_1/(r_1 + r_2)^2## for m2: netwon's law: ##Gm_1*m_2/(r_1+r_2)^2 = m*v_2^2/r2## --> ##v_2 = square root of m_2*G*r_2/(r_1+r_2)^2## then ##vcm = (m1v1 + m2v2) /...

I tried relating vcm to v1 and v2 + using the newton's 2nd law then putting it together in ##T = 2pi*r_(cm)/v_(cm)## but I don't see how that is going to help unfortunately

well no that is what I'm puzzled about. Analyzing the orbit around a planet I do understand, but looking at two planets orbiting about its center of mass.. don't know. I know that two planets will both orbit around the center of mass and not f.e. one planet around the other but that's about it.

Hello guys, Would it be possible to get some help on how to approach this problem? I don't really understand it. do I need to look at the orbital motion of the center of mass here or? If so how should I start? Thanks in advance.

Could I get some input on how to fix this problem please? Thanks in advance! The exercise for which I need the model. The simulink model together + the error that needs to be fixed + Matlab code used: Note: Gain 1 = 1/Ti

Hello, so we have two potitions right, if we take ##\theta = 90## as the first position (i.e. both rods are flat) and then the second position at ##\theta = 0##. I totally understand the exercise, not difficult. The only issue I am having is the torsional spring... it says that it is uncoiled...

or.... do we consider all of the metalhydroxides (that do not hydrolise) to be strong bases because they will form OH- ions which is basically considered a strong base?

So I know that I can get whether an acid is strong or not from a table where if ##pK_a## < 0 than it is a strong acid, but there is no such table for bases, so how do I determine whether a base is strong or weak? Also.. for metalhydroxides MOH, which will be strong and which are weak bases...

yes ? so what do you mean by that, I already said that I am taking into account the moment of ineratia of the whole plate right? because if I were to look at it in 2D it would just look as if it's a rod which it is not no?

Will the moment of intertia be the same then? and yeah but I can only use principle of work and energy here, haven't seen conserv of energy for rigid bodies yet.

hello guys, I wanted to ask whether I can just consider/think about this as being rotation around a fixed axis in a plane representing it as if it was 'just' a rod. This is mainly so that for the kinetic energy in the second position is where if we think about it in just a plane. Is this...

This is what i have done

hey, thanks a lot! I guess I'll just always be checking with ##\psi##. but.. I have another question about this. Because in HIbbelers book engeneering dynamics he uses ##\psi = arctan(r/dr/d\theta)## so that is what I use, but.. the problem here is that I get -72.23 degrees which is different...

so this is what the FBD is.... but to be fair, to me this one looks as if the normal force in the direction of the radial line, yet it isn't???? here in the solution, it's not along the radial line, whys that???

I get - 60 degrees which indeed makes it 30 degrees now..... thanks a lot!! Need to watch out on those two mistakes I just made here.

oh dear......

how come is it not?

##r = cos(\theta)## and for the given, r = 0.866 and for rdot --> ##rdot = sin(\theta) * \theta(dot)## = -1

oh, that is what they mean.., okay thanks a lot. I thought that it was downwards or something.

nope... I am getting -40.89

btw @erobz, this is in feet, but look at the FBD, haven't they forgotten to include the gravity force...

actually you might be right.. I got something in the 40's, but I had the wrong equation for r=f(##\theta##) one sec let me try again.

well, so if you look at this example the angle psi, the formula I mentioned above is used for the normal force angle.

yes yes of coursre that I know, and that is why I am asking why I can't use the formula to determine the angle between the radial line r shown and the normal force. Instead they use 30 degrees here, but how do you determine that?

what do you mean? the path's center is not in the pin of the arm guide right? and the force F is the force exerted by the arm guide on the can/ball or whatever.

for more clarity: here, 30 degrees is used, but why is that?

so I was wondering. there is this normal force on the can from the path. And there's this formula to find the angle between the radial line and the tangent or also between the normal force and either the radial or theta axis. the formula is ##\psi = r/dr/d\theta##. The thing is that here they...

thanks a lot! but I think you were totally correct in your previous comment. :) Knowing that in the extreme cases it's not horizontal anymore gives us an indication that there actually is an angular acceleration for the middle link.

Hello that is indeed what I did. I had 9 unkowns, 5 forces with 5 eqn's of motion from the 3 bodies and the rest from the kinematics. but.. the thing is. I took AB's angular acceleration as an unknown and I indeed got a nonzero value. And it says in the picture that only the angular velocity is...

well let's put it like that, so on two of the links there's no angular acceleration given. The following is where I had my doubts: So for link AB for the moment equation, the doubt was whether the angular acceleration will also be 0 at that point then or whether I should take it as an unkown...