Question

The sledgehammer shown consists of a head (with a high mass) and a handle (with a much, much lower mass). Points P, O, L, N, and M represent possible axes of rotation that are perpendicular to the screen and evenly spaced. Point O is at the exact middle o

Screen Shot 2023-03-22 at 12.30.55 PM.pngThe sledgehammer shown consists of a head (with a high mass) and a handle (with a much, much lower mass). Points P, O, L, N, and M represent possible axes of rotation that are perpendicular to the screen and evenly spaced. Point O is at the exact middle of the head, and point L is at the exact middle of the handle. Treating the head of the hammer as a point mass and the handle as a straight, thin rod, rank the rotational inertia () of the hammer about the various axes.

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Answer #1

To rank the rotational inertia (I) of the sledgehammer about the various axes, we need to consider the parallel axis theorem, which states that the moment of inertia about an axis parallel to and a distance "d" away from the axis through the center of mass is given by:


I_parallel = I_cm + m * d^2


Where:


I_parallel is the moment of inertia about the parallel axis.

I_cm is the moment of inertia about the axis through the center of mass.

m is the mass of the object.

d is the distance between the two axes.

Let's rank the rotational inertia of the sledgehammer about the given axes, starting with the highest inertia and going to the lowest:


Axis P (furthest from the center of mass):


The distance between axis P and the center of mass (O) is the longest among the axes.

The moment of inertia about axis P is the highest.

Axis N (closer to the center of mass than P):


The distance between axis N and the center of mass (O) is shorter than for axis P.

The moment of inertia about axis N is lower than that for axis P but higher than the rest.

Axis O (through the center of mass):


The axis through the center of mass is the axis of rotation for the hammer's head.

The moment of inertia about this axis is determined solely by the head of the hammer, which is treated as a point mass.

The moment of inertia about axis O is the lowest for the head.

Axis M (closer to the center of mass than N):


The distance between axis M and the center of mass (O) is shorter than for axis N.

The moment of inertia about axis M is lower than that for axes P and N.

Axis L (closest to the center of mass):


The axis through point L is the axis of rotation for the handle, which is treated as a thin rod.

The moment of inertia about this axis depends on the mass distribution of the handle but is generally lower than for the head and other axes.

So, in summary, the ranking of the rotational inertia (I) of the sledgehammer about the given axes, from highest to lowest, is as follows:


Axis P

Axis N

Axis O

Axis M

Axis L


answered by: anonymous
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The sledgehammer shown consists of a head (with a high mass) and a handle (with a much, much lower mass). Points P, O, L, N, and M represent possible axes of rotation that are perpendicular to the screen and evenly spaced. Point O is at the exact middle o
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