I think I gave the wrong impression here. This is not a homework, I'm a graduate student as a matter of fact. This was a problem I saw while searching for an equation in an old textbook of mine. I just worked out the problem at my office and got an answer which turns out to be half the provided...
http://img824.imageshack.us/img824/5831/friction.jpg
The coefficient of friction between surfaces and masses are shown in figure. Here the block M3 is pulled by the force F and for a particular F friction between m2 is twice as that of m1. So what is F? I'm keep getting half the value of...
I've tried out the two body problem and tried to work out the trajectories with respect to their center of mass frame(located at the origin) as follows (it worked!:smile:):
particles = m1 and m2
Fm1 = force of m2 on m1
Fm2 = force of m1 on m2
a = Gm2\hat{r} / r2
\hat{r} is the unit vector...
The selection of theta is what is causing the confusion here. Theta lies in between the vectors r and F.
In the case of seesaw the radial vector is pointing from the center towards any of the mass and the corresponding force is pointing downwards or upwards. so in that position the angle...
The net force being zero means you shouldn't feel it. But one feel the force of gravity while making movements. even the slightest of movement in muscles require energy to do work against this force. This is why feel it.
Another way to describe this : force should create a change in momentum...
By the way, I need a help in doing a project based on this(the disc's strange behaviour and similar objects like tippy-top). How can i include a sample calculation that shows how long the disc will stay in its stable position, how much time will it take to flip the hole position. How can i make...
But, after going through the tippie top's theory
:http://ckw.phys.ncku.edu.tw/public/pub/Notes/GeneralPhysics/Powerpoint/Solutions/Tippy%20Top/Gray-Nickel_AJP68-821(2000).pdf
I think a similar case can be applied here, except that the moment of inertia in 3 axes would have to be modified to...
W = integral of (torque x dθ) = Iω^2 / 2
Considering change in torque is difficult i guess. since Radial vector changes and the dθ is infact a tilt to backwards... so I guess we need a much complicated differential equation, isn't that so?
Change in torque is rather uncommon in such cases...
I knew it would be hard, sorry. Let me phrase it in a simple way.
There is a disc with a hole (I recently posted here asking for its moment of inertia). It is spinning about an axis passing through its own diameter. Due to the shift in point of contact and center of mass the disc scratches the...
Friction causes a torque that opposes angular momentum. It gets reduced. how can we derive a relation connecting these. friction starts with maximum and becomes zero. angular velocity and its corresponding momentum decreases maximum in the beginning and increases towards the end. the radial...
The diameter of both, since they line up. That is the line passing through their centers.
so is the answer
I = 1/4 M R^2 - (1/4 m r^2 + m b^2)
where the 2 became 4
Let me do it another way.
I = 1/2 m r^2 + m b^2 , where b is the distance between the center of rotation and the center of mass of m.
If you cut this out of a disk of mass M and radius R, you are left with
I = 1/2 M R^2 - (1/2 m r^2 + m b^2)
what about this time