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- Thread starter iliepins
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E = 1/2 * I * w^2

As an approximation, consider the hammer as consisting of two parts. Approximate the metal head with mass M1 at radius r=R1, and e.g. a rod of some mass M2 going from r=0 to r=R1 (assuming homogeneous mass distribution along its length).

The moment of inertial about r=0 of the head in this approximation is I_h = M1*R1^2 since we are approximating it as a point.

The moment of inertial about r=0 of a radial rod of mass M2 of length R1 is (from looking it up on wikipedia) I_r = 1/3 * M2 * R1^2 (please double-check this...)

Thus, the moment of inertial about r=0 of the entire hammer in this approx is I = I_h + I_r.

Then the kinetic energy if the hammer is E = 1/2 * I * w^2 where w is the angular velocity (in radians per second) of the rotational motion upon impact.

It is of course possible to find a more accurate moment of inertia of the hammer by not assuming the head to be located at one point.

Of course, if the head is much more massive than the handle, then your can just approximate everything by incoming linear motion of the head at v=5m/s.

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tiny-tim

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hi iliepins! welcome to pf!

Looking for quidance on calculating the mass at an impaction point of a swinging hammer. The hammer is set up to be swung pneumatically just at the edge of the handle in the horizontal plane. I understand the calculation E=1/2mv^2.

why do you want the

the effect is measured by the

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