# Three identical balls, with masses M, 2M, and 3M are fastened to a massless rod of length L as shown. The rotational inertia about the left end of the rod is: That's the layout below. Would calculus be needed in this problem (intergration) because then Im in trouble. I know the rotation at the end of rod is I=ML^2/3. Could I use that formula.

3M-----L/2----2M----L/2-----M

It is not clear in what plane the rotation is. If the rod is not rotating about its axis, but in fact about a perpendicular through its left end, then just add the I for the outer two masses (1/2 mr^2) for each, and that makes the composite I. No, you don't need calculus. The formula you proposed is for a rod with mass.

I came up with an answer of 3ML^2/2

does that look right. Thanks

## Lencho

## 3m

## To find the rotational inertia about the left end of the rod, you can consider each mass individually and then sum up their contributions.

Let's consider the first mass (M) at the left end of the rod. Since it is rotating about its center of mass, the rotational inertia for this mass is given by (1/2)M(L/2)^2.

Next, let's consider the second mass (2M) located at a distance L/2 from the left end of the rod. Since it is rotating about its center of mass, the rotational inertia for this mass is given by (1/2)(2M)(L/2)^2.

Lastly, let's consider the third mass (3M) located at a distance L from the left end of the rod. Again, since it is rotating about its center of mass, the rotational inertia for this mass is given by (1/2)(3M)(L/2)^2.

Now, sum up all the rotational inertias for each individual mass:

Rotational inertia = (1/2)M(L/2)^2 + (1/2)(2M)(L/2)^2 + (1/2)(3M)(L/2)^2

Simplifying this expression:

Rotational inertia = (1/8)ML^2 + (1/2)ML^2 + (9/8)ML^2

Rotational inertia = (3/4)ML^2 + (9/8)ML^2

Rotational inertia = (6/8)ML^2 + (9/8)ML^2

Rotational inertia = (15/8)ML^2

Therefore, the correct expression for the rotational inertia about the left end of the rod is (15/8)ML^2.

## Yes, your answer of 3ML^2/2 is correct for the rotational inertia about the left end of the rod.

To derive this result, you need to consider the rotational inertia contributed by each ball separately and then add them up.

The rotational inertia of a point mass rotating about an axis through its center is given by the formula I = mr^2, where m is the mass of the point and r is the distance from the axis of rotation.

Considering the first ball with mass M at a distance L/2 from the left end of the rod, its rotational inertia would be (M * (L/2)^2) = ML^2/4.

Next, considering the second ball with mass 2M at a distance L/2 from the first ball, its rotational inertia would be (2M * (L/2)^2) = 2ML^2/4 = ML^2/2.

Finally, considering the third ball with mass 3M at a distance L from the left end of the rod, its rotational inertia would be (3M * L^2) = 3ML^2.

Now, you can add up these individual rotational inertia values by simply summing them:

ML^2/4 + ML^2/2 + 3ML^2 = (ML^2/4) + (2ML^2/4) + (12ML^2/4) = 15ML^2/4 = 3ML^2/2.

Therefore, the rotational inertia about the left end of the rod is indeed 3ML^2/2, as you correctly calculated. No calculus or integration is required for this problem.