General Physics Question - Rod moving about a pivot

Homework Statement

This problem involves a uniform rod of length L with a pivot at point L/4 (so that 1/4 of the rod is behind the pivot and 3/4 is in front of it). The rod is released from a horizontal position and drops down. You need to use conservation of energy to solve for the velocity when it is at the vertical position (when the energy is all kinetic). Then, using the idea of a physical pendulum, you have to calculate the period of oscillation if it is displaced slightly.

mgy = 1/2mv^2

The Attempt at a Solution

My idea for the solution is just to ignore the portion of the rod that is to the left of the pivot point (which is equal in length to L/4) and focus only on the remaining 3L/4 portion of the rod. I don't know, however, if this is the right way to approach the problem, or if there is something else i need to take into account.

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Doc Al
Mentor
That won't work--every piece of the rod counts. Hint: What's the rotational inertia of the rod about its pivot?

That won't work--every piece of the rod counts. Hint: What's the rotational inertia of the rod about its pivot?
The rotational Intertia would be the sum of MR^2, so you would treat the portion to the left of the pivot as one mass and the portion to the right as one mass, taking the length of each R (radius) from the center of mass. I got 9ML^2/108.

Can this be solved by summing torques and setting equal to (I)(alpha), solving for alpha (rotational acceleration) and using a rotational kinematic?

Doc Al
Mentor
The rotational Intertia would be the sum of MR^2, so you would treat the portion to the left of the pivot as one mass and the portion to the right as one mass, taking the length of each R (radius) from the center of mass. I got 9ML^2/108.
You'll need to redo this calculation. You can certainly treat the stick as composed of two smaller sticks joined together. But you cannot treat a stick as if all its mass is located at its center of mass. Instead, add up the rotational inertia of each smaller stick to find the rotational inertia of the complete stick. (Or make use of the parallel axis theorem.)
Can this be solved by summing torques and setting equal to (I)(alpha), solving for alpha (rotational acceleration) and using a rotational kinematic?
Use conservation of energy, as suggested in the problem statement.

You'll need to redo this calculation. You can certainly treat the stick as composed of two smaller sticks joined together. But you cannot treat a stick as if all its mass is located at its center of mass. Instead, add up the rotational inertia of each smaller stick to find the rotational inertia of the complete stick. (Or make use of the parallel axis theorem.)

Use conservation of energy, as suggested in the problem statement.
I found the moment of inertia, but I'm not sure exactly how to set up the conservation of energy. Usually its mgy = 1/2Iw^2 (where w is angular velocity at the bottom), but in this case i'm not really sure what "y" would be.

Doc Al
Mentor
Consider the change in height of the center of mass.

Consider the change in height of the center of mass.
the center of mass or the pivot point?

Doc Al
Mentor
The pivot point is fixed.

Thanks for all the help thus far, Doc Al.
The last part of the question is: The rod is brought to rest hanging in the vertical position, then displaced slightly. Calculate the period of oscillation as it swings.

Now, since this is a physical pendulum, I know that the period is 2(pi)(I/mgd)^(1/2), where I is the moment of inertia of the rod about the pivot. I would assume that d, the distance of the pendulum, is ONLY the portion of the rod below the pivot point, am I correct?

Doc Al
Mentor
Now, since this is a physical pendulum, I know that the period is 2(pi)(I/mgd)^(1/2), where I is the moment of inertia of the rod about the pivot.
Good.
I would assume that d, the distance of the pendulum, is ONLY the portion of the rod below the pivot point, am I correct?
In that equation, d is the distance between the pivot point and the center of mass of the pendulum.

By the way I didn't really understand what you said about the Inertia, i thought I had it but i reread your post above and it seems that you said 9ML^2/108 was wrong. My method for doing this was treating the rod as if it were composed of two smaller rods, one to the left of the pivot, and one to the right, then summing the mass*radius^2 for each of the two sticks to get the total moment of inertia. For each mass, i used a fraction of "M" (the total mass of the rod) and for the radius i used the distance between the pivot and the center of mass of each of the two smaller rods. Thanks.

Doc Al
Mentor
My method for doing this was treating the rod as if it were composed of two smaller rods, one to the left of the pivot, and one to the right, then summing the mass*radius^2 for each of the two sticks to get the total moment of inertia. For each mass, i used a fraction of "M" (the total mass of the rod) and for the radius i used the distance between the pivot and the center of mass of each of the two smaller rods.
You seem to think that the moment of inertia of an object is $MR^2$, where R is the distance from the pivot to the center of mass. NO! That's true for a point mass, but not for an extended object like a rod. To derive the moment of inertial for a rod you must integrate $dMR^2$ for each element of mass (dM) within the object--since each element has a different R. Of course, you can just look up the moment of inertia for common shapes--like rods, cylinders, spheres, etc.

Example: The moment of inertial of a thin rod about one end is $1/3 M L^2$--but using your (incorrect) method, you'd get $M(L/2)^2 = 1/4ML^2$.

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thanks, you just saved me alot of trouble. I found the correct answer to this problem, but was having trouble figuring out just how to get there. It makes sense now, a rod is not composed of a bunch of point masses. thanks

just one more question - over what interval would i integrate the function? L/3 of the rods length is to the LEFT of the point, and 2L/3 is to the RIGHT, so i would assume i integrate over -L/3 to 2L/3

Doc Al
Mentor
It makes sense now, a rod is not composed of a bunch of point masses. thanks
Actually, a rod is composed of a bunch of point masses--all at different positions, which is why you must integrate. But a rod cannot be modeled as a single point mass located at its center of mass.

just one more question - over what interval would i integrate the function? L/3 of the rods length is to the LEFT of the point, and 2L/3 is to the RIGHT, so i would assume i integrate over -L/3 to 2L/3
If you wanted to do the integration yourself to find the moment of inertia of the rod about its pivot, you'd integrate from -L/4 to 3L/4--since the pivot is at L/4.