0.30 Phy1300: angular momentum -- rotational kinetic energy and  (Page 7/11)

Iend = (1/3)*(1kg*(96inches)^2) + (1/12)*(1kg*(3.75 inches)^2)

Plugging this expression into the Google calculator gives us:

Iend = 1.98 m^2 kg

(Remember, however, that this isn't an accurate absolute value because we aren't using the actual mass of a piece of 2x4 lumber.)

Icen = (1/12)*(1kg*(96inches)^2 + 1kg*(3.75inches)^2)

The Google calculator gives us

Icen = 0.496 m^2 kg

And the ratio is...

If I formulated the problem correctly before plugging the expressions into the Google calculator, the ratio

Iend/Icen = (1.98 m^2 kg)/(0.496 m^2 kg) = 3.99

Therefore, it should have been about four times as difficult to swing the 2x4 like a baseball bat than to spin it at its center. (The downward torque causedby gravity probably made it seem even worse that that.)

The dumbbell scenario

Among other things, this scenario illustrates the parallel axis theorem .

Consider a dumbbell, or a barbell, whichever you choose to call it. This object consists of two identical solid spheres, each with mass M. The centers ofmass of the spheres are separated by a distance L. The radius of each sphere is R.

The spheres are connected by a thin rod with mass m of length d. Thus, the length of the rod is L-2*R.

Find the rotational inertia of the dumbbell about an axis at the center of and perpendicular to the rod.

Solution:

The total rotational inertia of the dumbbell about the chosen axis consists of the sum of three parts:

1. The rotational inertia of one sphere about that axis.
2. The rotational inertia of the other sphere about that same axis.
3. The rotational inertia of the rod about that axis.

There are really five items in the sum

We learned from the parallel axis theorem that the first two items in the above list are each made up of the sum of two items:

1. The rotational inertia of a sphere about an axis passing through the sphere's center of mass
2. The moment of inertia of the center of mass of the sphere, treated as a point particle, about the chosen axis.

We learned from the earlier Examples of rotational inertia that the rotational inertia of a solid sphere about an axis through its center of mass is

Isphere = (2/5)*M*R^2

We learned in Rotational inertia that the rotational inertia of a point mass rotating about a chosen axis is

I = M*r^2, or in this case

I = M*(L/2)^2

Thus, the rotational inertia of each sphere about the chosen axis is the sum of those two, or

Isphere_axis = (2/5)*M*R^2 + M*(L/2)^2

The total rotational inertia of the dumbbell will be twice this value plus the rotational inertia of the rod about the chosen axis. We learned in Examples of rotational inertia that the rotational inertia about the chosen axis for the rod is

Irod = (1/12)*m*d^2

This, the total rotational inertia of the dumbbell about the chosen axis is

Itotal = Irod + 2*Isphere_axis, or

Itotal = (1/12)*m*d^2 + 2*((2/5)*M*R^2 + M*(L/2)^2)

Because the length of the rod, d is

d = L-2*R

We could substitute this expression for d giving us

Itotal = (1/12)*m*(L-2*R)^2 + 2*((2/5)*M*R^2 + M*(L/2)^2)

which reduces the expression down to include