Physics Help Forum Impact energy of a hammer blow along a fixed rotational axis

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 Oct 11th 2017, 08:19 AM #1 Junior Member   Join Date: Oct 2017 Posts: 3 Impact energy of a hammer blow along a fixed rotational axis Howdy, I've been trying to rectify a weird application and don't feel comfortable with my results and wanted some feedback. I have a hammer, mounted along its axis to a rotational point which allows me to raise the hammer some distance from a striking plate. When I release the hammer, I am looking to determine if there is any failure of the impact point on the steel plate. The requirement is that *any* hammer tested would impact the steel plate at 0.7 J of equivalent energy. What I need another set of eyes on is my model: VariablesMass of rod/connector = M Mass of hammer = m Length of rod/connector = L Height from plate = H Angle formed of rod/connector and plate = Theta EquationsH=L*SIN(Theta) Energy = (M+m)*g*H Am I right to just set Energy equal to the 0.7 J and solve for H? Do I need to factor in the center of gravity since the hammer adds mass at the end, thus shifting the center? Driving myself CRAZY!!! Thanks in advance for any guidance.
 Oct 11th 2017, 12:03 PM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,009 Conventional testing machines have a vertical pivot arm and the hammer is swung to impact the target sideways, for good reasons. Yours sounds as if it is mounted with the pivot arm horizontal for the hammer to swing vertically. Perhaps because this is a fatigue tester with the arm lifted by cam? Have a look (on Google) at the Izod test and the Charpy test. topsquark likes this.
 Oct 11th 2017, 12:16 PM #3 Senior Member     Join Date: Aug 2008 Posts: 113 you need to calculate the hammer's center of mass relative to the pivot, which I assume is the end of the rod opposite the hammer head. Assuming the rod is of uniform mass ... $\bar{x} = \dfrac{M \cdot \frac{L}{2} + m \cdot L}{M+m} = \dfrac{L(M+2m)}{2(M+m)}$ height of the center of mass above the plate is $H_c = \bar{x} \cdot \sin{\theta}$ ... $H_c = \dfrac{L(M+2m)}{2(M+m)} \cdot \sin{\theta}$ gravitational potential energy ... $(M+m)g \cdot H_c = 0.7 \, J$ $(M+m)g \cdot \dfrac{L(M+2m)}{2(M+m)} \cdot \sin{\theta} = 0.7 \, J$ $g \cdot \dfrac{L(M+2m)}{2} \cdot \sin{\theta} = 0.7 \, J$ solve for $\theta$ ... topsquark likes this.
 Oct 12th 2017, 03:06 PM #4 Junior Member   Join Date: Oct 2017 Posts: 3 Thanks for the breakdown. So basically its θ=arcsin (1.4/(g*L*(M+2m))) Dumb question is that θ is in degrees right? Thanks again for the help thus far.
 Oct 12th 2017, 03:12 PM #5 Senior Member     Join Date: Jun 2016 Location: England Posts: 786 Depends on if you have your calculator set to degrees or radians __________________ ~\o/~
Oct 13th 2017, 01:08 AM   #6
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 Originally Posted by mongo2001 Thanks for the breakdown. So basically its θ=arcsin (1.4/(g*L*(M+2m))) Dumb question is that θ is in degrees right? Thanks again for the help thus far.

Did you note skeeter's caveat about the length L?

This is equivalent to the condition that the pivot is at the centre of percussion.

Otherwise there is a pivot reaction to take into account.

https://en.wikipedia.org/wiki/Center_of_percussion

 Oct 13th 2017, 10:51 AM #7 Junior Member   Join Date: Oct 2017 Posts: 3 Greetings, The hammer is to be attached at the end of the rod. Think of it as just a longer handle with it pivoting perpendicular to the table surface. This is different than an Izod or Charpy test since there is no specimen with a notch. The component under test is the hammer which has been thermally soaked at high/low temp before hitting a steel plate. Based on this, I think that the breakdown is correct. Just not trusting my plug and chug values and wanted to confirm.

 Tags axis, blow, energy, fixed, hammer, impact, rotational

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