Anna Werner
Here is another problem that has me scratching my head. It goes as follows: "A uniform rod of length 1.23 m is attached to a frictionless pivot at one end. It is released from rest from an angle θ = 17.0° above the horizontal. Find the magnitude of the initial acceleration of the rod's center of mass."

Because the rod is uniform, I should be able to use the formula I = (1/3)m(L^2), correct? For this formula, I am not certain how to find m based on the given information.
Robittybob1
QUOTE (Anna Werner+Mar 12 2012, 04:51 PM)
Here is another problem that has me scratching my head. It goes as follows: "A uniform rod of length 1.23 m is attached to a frictionless pivot at one end. It is released from rest from an angle θ = 17.0° above the horizontal. Find the magnitude of the initial acceleration of the rod's center of mass."

Because the rod is uniform, I should be able to use the formula I = (1/3)m(L^2), correct? For this formula, I am not certain how to find m based on the given information.

I don't know the formula but it looks like you have no way of finding out the mass from the information you were given, but if you know another formula for calculating "I" containing Mass as well you can make both equal to each other for they both calculate "I" and you will find the mass factor cancels out.

It could be as simple as F = MA. A is for acceleration and that might be as good as I if I is initial acceleration. Does this give you a clue?
boit
I don't know what formula you'll use so let me throw in some few additional information perhaps you may pick the useful bits.
i) The rod is uniform therefore the centre of mass is at the geometric centre
ii) How high is the centre of mass of the rod above the ground? Sine theta = oppdosite/hypotenuse. We know the hypotenuse is 1/2 the length of the rod. That's is .615 m.
iii) a=W/t
Sorry I had to write w and note the Greek letter omega that stands for angular velocity.
boit
Not the best source but it gives a link you'll have to manualy copy.