# Calculating Force on a Glockenspiel Bar Using a Dropped Ball

1. Sep 25, 2012

### preston8

Hello,

I'm looking to calculate how much force is needed to strike the bars on my glockenspiel (musical instrument).

My theory is that I can use a wooden ball of a known mass, and drop it at varying heights until the proper sound is achieved. I can then use equations that model free-falling objects (I am neglecting air resistance) to determine the amount of force the ball exerts on the bar. Is this the best way to find the force?

Here's the problem: this is an inelastic collision which means that the kinetic energy is not conserved on impact. I am not sure how to account for this loss.

Mass of ball: 2.2g
Starting height: 0.1397m (5.5 in)
End height: 0m
Initial Velocity: 0m/s
Final Velocity: ?
Elapsed time: ?

I tried "messing around" with some equations as shown below, but didn't come up with anything substantial:

Free Fall:
yf=yi + (Vy)iΔt -1/2g(Δt)2
0 = 0.1397m - 0.5*9.8*Δt2

Δt=0.169s
(Vy)f = (-9.8m/s2 * 0.1397m) = 1.37m/s

This isn't really helping me out here. Could someone set me on the right path?

2. Sep 25, 2012

### Simon Bridge

Some of the energy goes into making the sound, yes.

Once you know the height to drop the ball, you can replace it with a small hammer on a pivot. How far the hammer rebounds tells you the energy that was lost in the collision ... hopefully into sounding the key.

Note: you'll get a different result with different materials for balls.

The key to this sort of thing is to start with the end in mind: what do you want to know for? This tells you what you need to calculate. The balls won't strike the bar with a single force at an instant - the force will be distributed over time. So it is unclear what "force" you want to calculate.

3. Sep 25, 2012

### preston8

Hello Simon,

The force I am trying to determine will be used to select a solenoid. The solenoid will be used to hit a mallet arm near the fulcrum point. To determine which solenoid to purchase, I need to know the force.

I'm going to create the pivoting hammer. Would you suggest using a video camera to gauge the arc length that the hammer travels?

I'll post pictures of the results so that others may learn from this experience.

Preston

4. Sep 26, 2012

### Simon Bridge

1. are the solenoids you are contemplating rated in units of force then?
2. you don't need the arc-length of the hammer swing - just the initial height... it's still conservation of energy to get the energy of the impact. However - the camera would help you determine how high the hammer bounces back.

Since you want the solenoid to play the thing, you'll just be using the rubber ball (or whatever) that will end up on the end of the solenoid shaft for your test?

5. Sep 26, 2012

### preston8

1) Most solenoid data sheets give the maximum force. The force does vary slightly depending on the location within the stroke length.

2) I'll be using a wooden ball. The final version will use a rigid plastic ball.

6. Sep 26, 2012

### Simon Bridge

1. OK - maximum force it is then. If you know the change in momentum and the impact time, you can get the peak force required (modelling as a quadratic).

7. Sep 26, 2012

### preston8

I've created a video demonstrating my test.

1) The bar absorbs a majority of the energy. It only rebounds about 1/4 inch. The mallet was dropped at 89mm above the bar.

2) When you play a glockenspiel, you need to pull back the mallet arm immediately after the strike. If you let the mallet sit on the bar, it will prevent the bar from vibrating.

Ball mass: 2.2g
Arm mass: Appx 1.1g
Distance from mallet head to axis: 100mm

How would I be able to determine the impact time?

Last edited by a moderator: Sep 25, 2014
8. Sep 26, 2012

### Simon Bridge

That would be the amount of time the head is in contact with the bar. I've normally used a pressure sensor - so it records force as well as time.

It sounds like about three strikes in 1/2 second, suggesting 1/4 second between strikes ... so 1/10 second impact time would be generous.

The analysis is complicated by the arm mass being significant.
If the ball starts at height $h_1$ and rebounds to height $h_2$ then:

$m_bg(h_1-h_2)+\frac{1}{2}m_ag(h_1-h_2) = \Delta E$

You may decide to neglect the rebound - since all but about 6% of the energy goes into the strike. Probably the rebound is due to the springiness of the padding since you don't get the same % loss with each bounce.

A very simple calculation would just go: $$m_b\sqrt{2gh} = \Delta p = \int_0^{\Delta t}F(t)dt$$

Where $F(t)=\frac{4}{\Delta t^2}t(\Delta t - t)F_{peak}$

This would give an under-estimate of the force ... you could try working out a range of heights that give you a good sound so we have some idea of the tolerances involved.

The need to pull the ball away after a strike just means you need to reverse the direction of the current to the solenoid at the right time... or add a small spring to the mechanism and just deliver a pulse to the solenoid.

9. Sep 27, 2012

### preston8

Thanks Simon,

Help me understand a few things here:

1) What is the difference between F(t) and Fpeak
2) What is the difference between Δt and t

I plan on utilizing a solenoid with a built-in spring return, by the way.

Preston

10. Sep 27, 2012

### Simon Bridge

1. when you strike something, the force of the strike varies with time - starts zero, goes to a maximum, and then drops to zero again. F(t) is how the force varies with time, and Fpeak is the maximum force it goes to.
2. t is the time axis for the motion ... the number of microseconds or whatever since the clock started. I chose to start the clock (t=0) when the ball first touches the bar. The ball remains in contact with the bar from t=0 to t=Δt...

I worked out the integral:

$$F_{peak} = \sqrt\frac{9gh}{2}\left ( \frac{m_{eff}}{\Delta t} \right )$$

You will probably want to put $m_{eff}=m_b+\frac{1}{2}m_a$ (mass of ball plus half the mass of the arm - "half" because the center of mass of the arm is about half way along it) ... that should give you a slight over-estimate. If you put $m_{eff}=m_b$ as per the simple example above, you'll get an under-estimate.

The trick will be to estimate the contact time Δt.
The less time the ball spent in contact with the bar, the higher the peak force that was delivered. However, I suspect you don't need to be very accurate - otherwise I'll have to model how bouncy the bar support is ... and use damped harmonic motion on the way the hammer moves.

Basically the model can get as exact as you want it to be ... the trick is to get it as exact as you need it to be.

Disclaimer: I'm going off your notes here - I can help you make the decision but I cannot make the decision for you. I will not take responsibility if you buy a less than ideal solenoid: price of "free" science work.

Last edited: Sep 27, 2012
11. Sep 27, 2012

### preston8

Simon,

Here's the value I have reached:
$\sqrt{9*9.8*0.089m/2}$* (0.0022kg + 0.5 * 0.0011kg)

Result: 0.0054N

That's about 1/1000 of a pound. Does that make sense?

Would an estimate be easier to arrive at if we assumed that we were simply dropping the ball (no arm attached) from a set height and assumed that all of the energy was absorbed by the bar?

It sounds good from 90-140mm above the bar.

12. Sep 27, 2012

I would of made it really simple by just treating it as a dropped ball. F = mdv/dt. You know the mass, you can calculate the final velocity easily from that and the drop height, and the contact time from a video of the test. You could slow the video down and calculate it if you know the frame rate of the camera. Although, I like Simon's estimate, that would be about right for an ineslastic object. Simon's is pretty much the same though.

13. Sep 27, 2012

### Simon Bridge

That's pretty much what I did.
The point of attaching the ball to an arm was to get an estimate of how much gravitational PE got turned into sound ... but we saw that pretty much all of it did.

Note: I don't see a value for Δt in there: you are effectively assuming an impact time of one second. If we say Δt=0.1s the Fpeak=0.054N ... if Δt=0.01s then it's 0.54N ... you can get an idea what is reasonable by tapping it against your hand and comparing with hefting the masses. The wooden hammer does not hit very hard does it?

You could also try hitting a sensitive electronic balance with a hammer - though that usually doesn't work well since it takes the balance a while to react.

What we are doing with the Fpeak is working out what sort of impact profile the solenoid needs to reproduce the impact energy of the dropped ball. So you should really use the Δt for the solenoid there.

14. Sep 28, 2012

### preston8

I analyzed the audio from the video and I believe that the impact time is considerably shorter than 1/10 second. As you can see in the waveform, the metallic "ting" sound lasts only for about 0.018 seconds. I believe that the initial ting sound is the impact, and the rest of the waveform is simply the reverberation of the bar.

I re-calculated and got some numbers that seem reasonable.

Δt = 0.018s
102mm above bar: 0.322N (32.3g)
130mm above bar: 0.366N (36.6g)
165mm above bar 0.412N (41.2g)

Last edited by a moderator: May 6, 2017
15. Sep 28, 2012

### Simon Bridge

Oh that's good!
Shows what can happen with collaboration.

If the solenoids are rated in lbs, then 0.008-0.01lbs would seem fair.

16. Sep 28, 2012

### preston8

Hey thanks.

I'd like to clarify the poundage calculations. I believe 1N = 0.2248 pound

102mm above bar: 0.322N (32.3g) = 0.072lb
130mm above bar: 0.366N (36.6g) = 0.083lb
165mm above bar 0.412N (41.2g) = 0.093lb

17. Sep 29, 2012

### preston8

I'd like to keep this thread alive with another question. I've looked over my physics book to review torques and equilibrium and believe I have setup the following problem correctly.

Basically, I want to find out approximately how much force is translated to the tip of the mallet.

Would Simon or someone else mind looking this over?

Thanks

http://ubuilds.com/myPhotos/ScannedImage-3_Small.jpg [Broken]

Last edited by a moderator: May 6, 2017
18. Sep 30, 2012

### Simon Bridge

Presumably you have a spring holding the mallet head above the bar ready to strike?

You could hinge the mallet at the end of the shaft and strike it from above 15mm in ;) (Also lets you vary the lever arm to fine-tune the effect.)

If you want to use the lever to amplify the force, then you need to get the mallet head on the short part of the lever.

But see: http://en.wikipedia.org/wiki/Hammer

19. Sep 30, 2012

### preston8

The mallet head needs to be moving at an adequate speed during impact. To get the speed, I need to mount the solenoid very close to the fulcrum. I understand that I will need to get a solenoid that is powerful enough to deliver 0.5N of force. This may require a solenoid with force equivalent to 1.4N or greater.

I'd like to ask you where you believe I should place the W.

Also, you're right in that I will have a spring that holds the mallet above the bar. The spring will be built into the solenoid. To "hold" onto the mallet, the solenoid's shaft will have a ball on the end that fits into a groove on the back of the mallet.

20. Sep 30, 2012

### Simon Bridge

See the wiki link about how a hammer/lever setup amplifies the speed.
The calculation is very similar to the one I did using the energy of the impact.