What is the Angular Speed of a Falling Rod?

[j2013]

Homework Statement

A rod of length L and mass M is pivoted about one end. The rod is released from rest in a horizontal position, and allowed to swing downward without friction or air resistance. When the rod is vertical, what is its angular speed w?

Homework Equations

Moment of Inertia, I = 1/3 ML^2

The Attempt at a Solution

I would like to apologize in advance if my notations are difficult to read.

Ki + Ui = Kf + Uf
0 + MgL = (1/2) Iw^2 + 0
MgL = (1/2)(1/3 ML^2)w^2
√6g/L = w

I think I'm supposed to use the rod's center of mass because my current answer is wrong and since the diagram in my book points out that the center of mass is at L/2 from the pivot point. I don't think they would point that out if it was unnecessary, but I don't understand why. Putting Ui as Mg(L/2) gives me the right answer though, according to the back of the book.

Any help clarifying this would be much appreciated.

 

[inky]

Homework Equations

Moment of Inertia, I = 1/3 ML^2

The Attempt at a Solution

I would like to apologize in advance if my notations are difficult to read.

Ki + Ui = Kf + Uf
0 + MgL = (1/2) Iw^2 + 0
MgL = (1/2)(1/3 ML^2)w^2
√6g/L = w

I think I'm supposed to use the rod's center of mass because my current answer is wrong and since the diagram in my book points out that the center of mass is at L/2 from the pivot point. I don't think they would point that out if it was unnecessary, but I don't understand why. Putting Ui as Mg(L/2) gives me the right answer though, according to the back of the book.

Any help clarifying this would be much appreciated.

Your I value is correct.
Ui is not MgL. Ui is MgL/2 because height is distance between initial CM point and final CM point. Perpendicular distance between two CM is L/2. So calculate again to find angular velocity.

 

[ehild]

I think I'm supposed to use the rod's center of mass because my current answer is wrong and since the diagram in my book points out that the center of mass is at L/2 from the pivot point. I don't think they would point that out if it was unnecessary, but I don't understand why. Putting Ui as Mg(L/2) gives me the right answer though, according to the back of the book.

Yes, the potential energy of a homogeneous rod of length L gets mgL/2 lower when it swings down from horizontal position into vertical.

Think of the road as a chain of N small pieces, each of mass Δm=M/N.

Initially, all pieces are at the same height. When the rod swings to vertical position the height of the piece at the pivot does not change, but the end moves down by L. The middle, which is the centre of mass for a homogeneous rod, moves down by L/2. If you take a piece at distance x from the CM, the piece above the CM has Δmxg higher PE than the CM, and the piece below the CM has PE Δmxg lower than the CM, the sum of the PE of both pieces is -2ΔmgL/2 with respect to the original horizontal position. Doing the same for all pair of pieces you get that the final PE= -NΔm*gL/2 = MgL/2 with respect to the horizontal position of the rod.

ehild