Mass of a point on an object ?

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The discussion centers on the mechanics of how a hammer delivers force during impact, emphasizing the role of torque and angular momentum. When swinging a hammer by the handle, the distance between the axis of rotation and the center of mass increases torque, allowing for a more powerful impact. The hammer head's acceleration is weak over a long distance, but it decelerates sharply upon impact, resulting in a large force due to conservation of energy. Holding the hammer head does not effectively increase its mass because the connection to the hand introduces too much "give," reducing the impact force. Understanding these principles clarifies why swinging a hammer by the handle is more effective than swinging it by the head.
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hi all. I'm currently researching angular/rotational dynamics for a hobby project I'm working on (2d physics engine). A curious thought has popped up in my mind.

How does a hammer work?

What i mean specifically is, if you hold a hammer by the handle and swing the head at something, it will hit a lot "harder" than if you hold the head and swing the handle at your target. I don't quite grasp why, though.

The total mass of the hammer is the same in both cases, and i THINK this would also apply if you swing in such a way that the velocity of the hammer is the same in both cases.

Can a point on an object (the part of the hammer that connects) be said to have mass ? Is some area around the impact point relevant in determining the mass of what hits?

A thought that came to my mind is that maybe it relates to the relationship between the impact point, the hammer's centre of mass, and the hammer's rotation point (the hammer that's holding it). Perhaps a greater distance between the latter two somehow adds more force (or power, or energy?) to the swing ?
 
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Nanako said:
the hammer's centre of mass, and the hammer's rotation point (the hammer that's holding it). Perhaps a greater distance between the latter two somehow adds more force (or power, or energy?) to the swing ?

It's torque.

Torque goes up when the axis of rotation is farther away from the center of mass.

\vec{τ} = \vec{r} \times \vec{F}

Torque is how fast the angular momentum is changing, that's the word you were looking for. Angular momentum. When the hammer hits the nail it uses the angular momentum to deliver an impulse to drive the nail.
 
ahh, that's starting to make sense =)

a few quick questions:

Can you clarify the F thingy with an arrow over it in that equation? (i'm guessing the r is axis of rotation?)

is the angular momentum of a point generally measured from it's axis of rotation, or it's centre of mass?
 
τ is torque
F is force

Nanako said:
ahh, that's starting to make sense =)
is the angular momentum of a point generally measured from it's axis of rotation, or it's centre of mass?

it's measured with the distance between the two. That's what r is.

The arrow "hats" indicate that these are vector quantities. (magnitude and direction).

Check this out...
http://en.wikipedia.org/wiki/Torque
 
wotanub said:
It's torque.

Torque goes up when the axis of rotation is farther away from the center of mass.

\vec{τ} = \vec{r} \times \vec{F}

Torque is how fast the angular momentum is changing, that's the word you were looking for. Angular momentum. When the hammer hits the nail it uses the angular momentum to deliver an impulse to drive the nail.

When a hammer hits a nail the effect depends almost entirely on the linear momentum of the hammer head. For a given impact speed of the hammer head the effect is the same whether the head is held in your hand, dropped from a height or swung by a handle.

The effect of the handle is to allow a slow (but relatively strong) motion of your hand to produce a fast (but relatively weak) acceleration of the hammer head.

The key to the force multiplication in a hammer impact is that the hammer head is accelerated weakly through a long distance and then, at impact, it decelerates strongly through a short distance. By conservation of energy, the work done on the hammer during the acceleration must be equal and opposite to the work done on the hammer during the deceleration. Since the distance covered during the deceleration is small, the force must be correspondingly large.

You can see this by hammering something with a little "give" in it. The more "give", the less effective force. Carpenters, for instance, know that hammering a nail into the middle of the flat of a board that is supported at its ends can be difficult. But if they put a two-by-four, end on, on the back side of the board under the nail, hammering becomes easy.

One might think that holding the hammer head in your hand and pounding would effectively increase the mass of the hammer head, adding the mass of your hand to the mass of the head. This does not work for the same reason. There is too much "give" between the hammer head and your hand.
 
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