Baseball Force: Newsletter Challenge (09/07/10)

I avoided that answer since it is an average, but I didn't want to throw another challenge into the mix by saying so.

Yours is clearly the expected answer. GA.

I agree with you.

Elasticity doesn't come into it because that is energy absorbed and returned.

Noise, heat and non-elastic deformation would relate to energy absorbed and not returned, and no information relating to those is supplied/available, therefore I believe your answer is the best possible with the given information.

Q.E,D,

WHERE DID THIS 75 COME FROM? ACCELERATION IS CHANGE IN VELOCITY NOT TOTAL VELOCITY.

From 40 in one direction to 35 the opposite direction is a change of 75. Some of the posts actually did the whole arithmetic: 40 - (-35) = 75. Also please ditch the all-caps.

Velocity and acceleration are both vector quantities, meaning they have both a magnitude and direction. "The opposite direction" implies that the direction of the ball after the bat hits it is precisely reversed from the incoming direction, so if we define the incoming vector as 40 m/s i, where i is the direction vector, then the outgoing velocity is -35 m/s i. 40 - (-35) = 75.

To address the instantaneous forces, we would need to know about the elasticity of the baseball and the bat. Somehow I don't think that's in the cards. Hence the simple approach.

Man, y'all can't answer that; ever'body knows baseball is measured in foots and measurin' sizes like that.

If you do the conversion it comes out to almost exactly 1 ton (imperial).

Home Run

C'mon. No one used the Impact-Momentum equation?

No ... that would be too impulsive.

150 grams!? It must have been a dirty spitball ... or a government ball.

Making an assumption of constant force through contact it would simply be delta V over the time. =75 m/1.25ms squared=60 kN

But you really need elasticities, weight of bat, location bat is hit and a few other facts to get closer tho the real forces. I think a bell graph could approximate it where when integrated it would be equal to energy change.

ooops forgat the mass .150

F=9 kN

so says the first answer

I just knew that somebody, probably somebody hiding from the daylight, would get into the technicalities that are not in the original question, that's why I hedged my bets.

600n

Force is change in momentum over time, so the average is 10kN or 1 mt.

It won't be uniform. If the shock wave is a half-sine wave, the peak is more like 1.6 mt

Newton's law of motion applied to system of constant mass:

m = 150gr = 0.15kg

dv = 45+30=75m/s

dt=1.25ms

F=0.15 x 75 / 0.00125 = 9000N

If we assume that this is an elastic collision:

F=ma => F=m(delta)v/t = [(0.150kg*(40-35)m/s)/1.25E-3 s]=600 N

However, delta v is 40 -(-35) = 75 as others have used.

Momentum is conserved, but that is assuming there are no other inputs acting on the bodies of interest. In this case, there is the force and torque being applied to the bat by the player swinging the bat, so a pure conservation of momentum analysis is invalid due to extra unknown forces being applied.

The F=ma approach applied only to the ball is still valid since the only outside force acting on the ball is the bat, which we are already considering as the force (neglecting air friction and gravity and only looking at the force in the direction of interest.)

This is a totally uninteresting question. Of more interest would be, How far does the ball travel?

And of greater interest still, one which is actually pondered amongst baseball folks is, How should the player choose the optimum weight (mass) of the bat to use which will generate the greatest force when hitting the ball.

For such questions, more information is needed. For the optimum mass, balance, and rotational inertia of the bat for minimum work in hitting a home run, see the analysis at this site: http://www.racquetresearch.com/

Not very far. Maybe most the way to second. Real hits run more like 20 kN - 40 kN. This one's a blooper or bunt.

Concur,

The ball is leaving with less energy than it arrived.

He also has to hit the ball at the center of percussion of the bat to get the maximum energy transfer.

This is going to vary from one batman to the next, and then also over time for a particular batsman and I venture suggest one possible way to determine this.

The objective is to impart the maximum energy to the ball, and the available energy is a function of the speed of the head of the bat at the point of impact and the mass of the head of the bat.

A high speed camera should be able to record the speed of the bat's head at the point of impact, and if the batsman can swing a number of bats of different masses while their speed is being recorded, then the energy can be calculated for each bat and plotted on a graph. This should be predictive of the optimum bat mass for that player at that time.

To do the problem you need a simple physics formula, force (f=ma) or force= mass x acceleration .150 (mass) x 75(acceleration)/ .00125 = 9000 Newtons (N)

600N

ie

this much force would be absorbed by the bat

force = mass*acceleration

=mass*(change in velocity / time )

=150*((40-35)/1.25)

Wrong!

The change in velocity is 75m/s! - the direction is reversed.

Direction is important, your answer would mean the ball continued in the same direction at a reduced velocity ....

I think first we need to know who's on first.

Who's on first.

What's on second.

I don't know's on third.

Why's left field.

Because is center field.

I don't give a darn is shortstop.

Tommow is pitching.

Today is catching.

Right field????????

Today we have naming of parts.

shoudn't the speed of the bat be included in the equation?

If you look at my analysis at #22 you will see that bat speed is taken into account, but it is not necessary to know the exact before and after bat speeds, nor even their difference, in order to solve for the bat force. You also don't need to know how heavy the bat is. Conservation of momentum supplies the answer if you know the ball mass, ball before and after speeds, and the dwell time, as we do here. For more fun with classical mechanics, see Racquet Research.

F = (mvf – mvi)/∆t m = 150 g = .15 kg vf = 35 m/s vi = -40 m/s ∆t = 1.25 m/s F = [(.15 kg)*(35 m/s) – (.15 kg)*(-40 m/s)] / 1.25 m/s F = 9 kg*m/s or 9,000 g*m/s

Forces are measured in newtons, kg*m/s^2. Your answer is in momentum units, and is therefore wrong. Your calculations are opaque, and with no explanation of what you are doing and why. I believe that in an exercise such as this one, which is intended to teach principles of analysis, it does not help to see even a correct dazzling display of math gibberish with no motivation. Jumping right in to the calculations, without pausing to set up the solution analytically, may be what caused the BP blowout and other engineering disasters.

Now, there is, a bunt

Rule 1.09 The ball shall weigh not less than five or more than 5 1/4 ounces. The average weight would be 5 1/8 ounces.

148.83 grams (5.25 ounces) max.

No one caught that? Must have been a bunt at PETCO park!

this punch has an total impuls of 11.25 kg*m/s, with the duration of 1.25 ms results this in an average acceleration-force of 9000N!

And

DON'T FORGET TO CHANGE YOUR BOOTS!

Force = change in momentum

= .150 kg *(40 m/s -(-35 m/s))/0.00125 s

= 90 Newtons

Recalculate, you lost two orders of magnitude!

We cannot calculate the force on the ball because the contact with the bat is not instantaneous. The best we can do is get the average force over the 1.25 ms. And even then, we cannot calculate the actual average force because we must assume that there are losses such as hysteresis, therefore not all of the work is applied to the change in momentum. We can only calculate the net work applied to the change in momentum.

To sum up, the best we can do is calculate the average force by the impact of the bat on the ball based on the change in momentum due to the net work done.

The force doesn't just vary with time; it also varies over the volume of the ball, causing it to flatten out and spring back. We don't have enough information to analyse this complex interplay of force, time and position, so I agree that calculating the net force is the best we can do.

IS THE BAT WOOD OR ALUM,

Where's this definitive answer then?

probably wouldn't be correct as op seems to have some difficulty with math

the tuesday following september 7th being september 14th

not october 5th...

If you assume a half sine shock with a pulse width of 1.25ms and a velocity change of 75m/s, I get a peak force of 14 KN (9611 g's). This is opposed to the F=MA approach which is the equivalent of assuming a rectangular pulse.

Why would one assume a half sine shock? As the bat and ball spring back from their respective compressions, they will exert mutual pressure after the maximum pressure occurs.

I guess the question I will ask is why you would assume the force was a constant. In an elastic collision, the force time history would be much more represented by a half sine wave.

Thanks,

Jim

The OP asks for "the force," without distinguishing whether peak or average. Without making further assumptions, which might be complicated, it seemed reasonable to go for the average force. Apparently everyone else also did so until just recently.

MNIce is right that the peak force will be more, but this depends on what further assumptions one makes, such as half-sine, full sine, or other force/time profile.

The answer should be 9000 Newtons. The impulse equals the average force times the time of contact. This, in turn results in the change of momentum of the ball. The momentum change is mass*change in velocity = 11.25 kg*m/s or N*s. When divided by the time of contact, the resulting average force is 9000 N.

Prof. R.

I think the batter misses the ball pretty good at an angle less than full contact.

Force=kg*m/s^2

120 J coming at the bat

93 J projected by the bat

213 J absorbed by bat

J=N*m

A bunt would give you conservation of energy, with a perfecrly elastic ball the 120 J coming into the bat would be transferred to the ball, so the ball would leave the bat at 40 m/s. However it is possible to add energy to the ball with a really good swing and perfect transfer of the energy by making good contact with the ball. Therefore conservation of energy will also still hold. Simply add the energy of the swing of the bat to the equation.

27=9000*x = 0.003 meter 3 cm = 7.62 in.

The bat is in contact with the ball for 7.62 in OR the ball absorbs the energy difference between the incomming and outgoing Energies. Is this problem a bunt?

Impulse= mv1-mv2= F*s

mv1-mv2=(75m*0.150)/s= F*s

0.150*(-40)-0.150(35)=-6-5.25=-11.25

11.25/0.00125=9000 N

I had fun trying to solve this anyways so I win even if it's the wrong answer

? 0.003 meter 3 cm = 7.62 in.?

0.003m = 3mm = 0.3cm

Your right, my error. 3mm = 1/8 inch. Makes more sense now! Thanks

Your answer is off because you used the wrong velocity for the ball after impact: -40 instead of -35. So this line of your solution

I = (0.150 kg)(40 m/s)-(0.150 kg)(-40 ms) = 12 kg∙m⁄s

gives the wrong impulse. It should be 11.25 and not 12. So the right answer is 9000 N.

The Answer given is wrong! - the speed leaving the bat is 35m/s not 40!!!!

I think your answer had a typo...the -40 ms should be -35 m/s making the impulse 11.25 kg*m/s and the force 9,000 N.

Ref Newton's 2nd law ie.rate of change of momentum (per second) = the impressed force.

Which is 40-(-35)m/sec x 0.15kgs =11.25 mkg/sec has to be divided by the impulse time which is only 1.25 millisecs. which gives 11.25 x 1000/1.25 =9000 Newtons

since 1N = 9.8 kgf. This is equivalent to being hit by a force equivalent to 900 kgs!

The initial speed of the ball is 40m/s, the speed after the impact is 35m/s; the difference speed is 75m/s, this is an overall impuls of 11.25kg*m/s equivalent to an average force of 9kN. (just a typing error?)

The given solution is calculated wrong. The answerer used 40m/s for both parts of the answer but the directions give 35m/s for the second half of the calculation. the answerer did not use the correct numbers to calculate the answer.

-just the numbers (.150*40)-(.150*-35) = 11.25 not the 12 the OP used because they used 40 in the second half of the equation even thought they specified 35 in the original question.

Using the OP's formula and the original numbers used in the question the answer is not 9600 but instead is 9000 newtons.