Find final angular velocity of swinging meter stick

[makks]

Homework Statement

A meter stick is suspended vertically at a pivot point 25 cm from the top end. It is rotated on the pivot until it is horizontal and then released from rest. What will be its maximum angular velocity (in radians/second)?

Homework Equations

I began by setting up a conservation of energy formula since the earth-stick system is not influenced by any nonconservative forces.
U_grav(i) = K_rot(f)
Mgh = (1/2)Iw²

The Attempt at a Solution

When I try solving for w² I become confused on what to use for h and for I on either side.

I'm thinking that h should be the vertical distance between the center mass at its initial and final points, which would be h = 1m - .25m = .75m

and for I: I = (1/3)ML² where L is supposed to be the length of the rigid object; however, I do see that the pivot point is not at the end, so I do not know how to adjust this accordingly.

 

[kuruman]

You need to use the parallel axes theorem to find the moment of inertia about the pivot. If you set the zero of potential energy at the pivot, then h is the distance from the pivot to the center of mass.

 

[makks]

You need to use the parallel axes theorem to find the moment of inertia about the pivot. If you set the zero of potential energy at the pivot, then h is the distance from the pivot to the center of mass.

I am having trouble reasoning whether to put the center of mass at the .5m mark (middle of the meter stick) or whether the center of mass is the midpoint between the pivot and the end of the meter stick that is swinging downward.

 

[kuruman]

The center of mass of a uniform stick is the midpoint of the stick. This comes straight from the definition of center of mass which has no reference to a pivot. In other words, if you hang a stick from a pivot, the stick's center of mass will not suddenly shift to a new position. This is true for any extended massive object. Once you find its CM, that's it unless you change the distribution of the mass.

 

[rwisz]

Your approach is correct. The height (h) should be the vertical distance between the initial and final positions of the center of mass. In this case, it would be 1m. As for the moment of inertia (I), you are correct that it should be (1/3)ML^2. However, since the pivot point is not at the end, you will need to use the parallel axis theorem to adjust for the distance (d) between the pivot point and the center of mass. The moment of inertia then becomes I = (1/3)ML^2 + Md^2.

Plugging this into the conservation of energy equation, you should get:

Mgh = (1/2)[(1/3)ML^2 + Md^2]w^2

Solving for w, you get:

w = √(2gh / [(1/3)L^2 + d^2])

Substituting in the values, you get:

w = √(2*9.8*1 / [(1/3)*(1)^2 + (0.25)^2])

w = √(19.6 / [0.3333 + 0.0625])

w = √(19.6 / 0.3958)

w = √49.466

w = 7.03 radians/second

Therefore, the maximum angular velocity of the swinging meter stick would be 7.03 radians/second.