A slender, uniform, rod of mass \"M\" is pivoted without friction on a horizonta
ID: 1414475 • Letter: A
Question
A slender, uniform, rod of mass "M" is pivoted without friction on a horizontal about an axis through its midpoint and perpendicular to the rod. A spring of force constant "k" is attached to one end with the other end attached to a rigid support. When the spring is in its unstrethed position, the rod and support are parallel. The rod is displace by a small angle and released. Given [M,k], Determine:
a. The period of the rod for small amplitude oscillations.
b. Repeat part a for the case where the pivot is at the far end of the rod rather than at the center.
Explanation / Answer
a) I think we're assuming that the amplitude is small (so that the lower end of the rod stays at more or less the same vertical level all the time).
Let R be the distance from the axis to the end of the rod (i.e., R is half the length of the rod).
Now say the rod rotates by a small angle . That causes the spring to stretch by an amount x = R, and to exert a force F = kx = kR on the rod. This results in a torque equal to = FR = (kR)R = kR².
Now, we know that = I (torque = moment of inertia times angular acceleration). This means:
I = kR²
And, for a slender rod rotating about its center,
I = ML²/12 = M(2R)²/12 = MR²/3
Sub into previous equation:
(MR²/3) = kR²
or:
M/3 = k
Now, since is the 2nd derivative of (with respect to time), this is just a differential equation:
(M/3) = k
You can do the math from there, but the upshot is that this is 100% analogous to the "standard" SHM equation, "mx = kx"; except that "M/3" takes the place of "m". That means, where the period of the standard SHM equation comes out to T=2(m/k), you can replace "m" with "M/3" to get:
T = 2(M/(3k))
b) In this case, we need to change some assumptions. First, the Moment of Inertia will be different. Here, The moment of inertia about the end of the rod can be calculated directly or obtained from the center of mass expression by use of the Parallel axis theorem. Therefore, Iend=(1/3)ML^2=(1/3)M(2R)^2
Iend=(4/3)MR^2
Also, the torque will be different. This means that = F2R = (k2R)R = k2R².
Thus, k2R^2=(4/3)MR^2*a or -k=Ma(2/3). From this, you can repeat the same process as in part a and obtain the new expression for T