Question

http://s3.amazonaws.com/answer-board-image/628ae3fdb906a2b8b95fb53923eb34c7.jpg A rigid T consists of a long rod glued perpendicularly to another rod of length l that is pivoted at the origin. The T rotates around in a horizontal plane with constant frequency ?. A mass m is free to slide along the long rod and is connected to the intersection of the rods by a spring with spring constant k and relaxed length zero (see Fig. 6.27). Find r(t), where r is the position of the mass along the long rod. There is a special value of ?; what is it, and why is it special? http://s3.amazonaws.com/answer-boa

Answer 1

the centripetal force directed towards the centre once the mass goes along a circle
it experiences centripetal force directed INSIDE so the spring conected to the mass gets compressed by some amount x but in response the spring crerates a reaction force=kx directed OUTWARDS the circle this is gonna be equal as the centripetal force imparted to it by the mass
so
mv^2/r=-kx
so we can find omega
as
omega=v/r
then when the compression of spring is minmum (stretched)
i think that is the case tehy are asking for

Answer 2

To what reference point are we referring the position vector, r(t), to? the origin (pivot point) or a point on the long rod? I'm assuming the reference point is the origin. Ok, the position vector of the mass at any time t will be made up of two components. \[r(t)=l(t)+x(t)\]where l is the vector of constant length equal to the length of one of the arms of the T bar and x(t) is the vector with magnitude equal to the length of the spring at any given time. We expect x(t) to be changing in magnitude as well as direction as the motion of the system progresses. Of course vector l never changes magnitude but is still written as a function of t because it's direction depends on t. Let's deal with vector l first. Ok, it's magnitude is already determined to be a constant = l. It's direction is simple to describe as well. The angle that l makes with the polar axis (I'm using polar coordinates here) is \[\theta = \omega t\]So we can describe vector l as
=<l,wt>
where l is the (constant) magnitude and wt is the angle it makes with respect to the polar axis at any time t. Now we need to describe the second vector component x(t). Here not only is the direction changing, but the magnitude is changing as well as the spring extends and compresses in response to the forces exerted on it by the mass, m. Let's describe the direction first. Notice that the direction of x(t) is just \[\theta + \frac{\pi}{2}\]where theta is the angle vector l makes with the polar axis as described above and we are adding 90 degrees (pi/2 radians) to it. So, in terms of t, the direction of vector x(t) is\[\omega t + \frac{\pi}{2}\]The magnitude of x(t) is the last piece we need. Once we have the magnitude of x(t) at all times t, we will have an expression for r(t), since r(t) is just the vector sum of l(t) and x(t). I'll leave this one for you and others to discuss for now.