Recent content by marooned

sorry i meant distance to the axis of rotation, not the x-axis which is completely irrelevant if it is not the axis around which the object is spinning.. sorry. But yes i agree, the torque and angular acceleration vectors change so a scalar will not work. Thanks for your help, marooned

That sounds correct, the distance to the x-axis is the distance we are after, not the distance to the centre of the disc. However, if the moment of inertia of the disc was to be found with respect to its shaft, no matter what angle it is on, the distance from any point to the shaft is the same...

Thats very much Viet Dao, i got it out with no problems after your help, Thanks again, marooned

Are you suggesting the diagram as shown below? If so the relationship stated does not work, i am having trouble understanding where your H point is located, sorry to cause you so much work, i have trouble interpreting things sometimes... Thanks in advance, marooned [FONT=Courier New]C H...

I am just curious as to what the moment arm is; what is r1 and r2? Are they the distances from the ground where the ropes meet to their point of joining on the pole? No i am not mad at you, and thankyou very much for your help Viet Dao. I am also curious where you get the equation r1 = \frac{5...

I see now, thanks it is much clearer, I should be able to get it out now. I see the different angles mean the forces are not equal, so setting them to equal each other makes sense. (However clumsy that sounded I do understand). Thanks again Viet Dao, marooned

So it is the same equation as posed before, but with the sin relationship instead of the cosine relationship? That way the horizontal forces are resolved? But i don't see how the forces can equal each other, if they are at different heights on the pole, one force shall be bigger than the other...

Thank you thank you thank you. I can´t believe i didnt see that before, setting both horizontal components to each other, because they have to add to zero? I didn´t even think of disregarding the horizontal components because they are the same distance from the pole, right? That leaves only the...

But if the disc remains spinning on the same axis, the distance from the point masses to the axis remains the same. If the axis is then tilted the distance from the point masses to the axis is still the same, as the disc tilts with the axis. I realize this is probably wrong, but I can´t seem to...

On thinking about this problem overnight, i decided that the moment of inertia for the disc will be the same no matter what plane it is spinning in, as long as the axis around which it spins is constant. So that means the moment of inertia for the disc will be the same on an angle as it is...

I have been working on this problem to no avail... can anyone help? So far i have tried to find the horizontal component of the forces using the formula F1+F2=0 (since the pole is in equilibrium, the sum of the vector forces will be zero, as will the sum of the vector moments of force) so also...

to calculate the moment of inertia of a spinning disc (a spinning top for example) i know the formula is I=(mR^2)/2, but in order to find the moment of inertia of a spinning disc at and angle to the vertical, how would this be done? does the moment of inertia change if the disc is not spinning...

Sorry, here is the diagram... my description skills arent the best..

hello, i have a problem i have having trouble solving, it reads as follows: a vertical pole is supported by two ropes, each 5m from the base of the pole. The rope to the right of the pole extends 3m up the pole, and the rope to the left of the pole extends 4.5m up the pole. The ropes and...