Trouble with PE exam sample problem 2

[bshadel]

Hi folks,

New to the forum, I’m getting ready to take the mechanical PE exam this Friday. I’ve been studying for a better part of the year and hope for some positive results.

I’ve come across a sample problem where I don’t agree with the solution. Perhaps someone can set me straight. By the way, this problem is from “Six-Minute Solutions for Mechanical PE Exam”.

Problem:
Water balloons are launched from a catapult on the rooftop of a building 50 feet off the ground. Balloons must achieve at least 100 yards (horizontal). Catapult spring constant is 0.5 lbf/in and can be extended no more than 5 inches. Weight of each balloon is 4.5 lbf, and each one takes approximately 1 second to launch. Under these conditions, how many springs in parallel will it take to launch 100 yards?

My solution:
I solved for an initial velocity that would send the balloon 100 yards. I then used energy methods to solve for the total spring constant: 1/2kx^2 = 1/2mv^2. Unfortunately my answer was way off.

Book solution:
The acceleration of the balloon was found from the 1 second launch to the terminal velocity in 5 inches. Knowing that, they said: (mass of balloon x acceleration) = (spring constant x 5 inches).

While the units work out, I don’t agree with this simply because the spring force is not constant and varies with the deflection. This is not the only problem like this…there is another that utilizes this type of solution. Am I missing something???

Any assistance is greatly appreciated.

Thanks,
Bryan.

 

[tlee123]

I found zekeman’s idea to try to satisfy all the criteria of the question including the 1 second launch time to be very interesting. I think he also got the correct answer of a lever ratio of 260, but there were problems. The differential equation he used is wrong which you can tell by setting t=0 into his solution to the differential equation.

He used x=5/12(1-cos(sqrt(k/m)t)). Setting t=0 we get x=0 which isn’t right. X should be 5/12 ft at the start. X=0 means the spring is not exerting a force (F=kx).

Here is my solution of the massless, frictionless rack and pinion based catapult.

We know from the energy based analysis that 3888 springs are required which gives a spring rate k=23328 lb/ft. We also know that these springs give up their energy to the water balloon thru a 5/12 ft stroke and will give a launch velocity of 170 ft/sec (what we want). Any other solution that gives an answer of fewer springs violates the Conservation of Energy, so I’d be very careful to promote those or your reputation may get tattered.

Before the balloon is released, let’s say that x is the motion at the end of the springs and y is the tangential motion of the mass (the water balloon). We’ll say that the radius of the pinion (where the springs push) has length r1 and the long arm welded to the pinion that holds the mass has length r2. So:

y = (r2/r1)x and
d[sup]2[/sup]y/dt[sup]2[/sup] = (r2/r1) d[sup]2[/sup]x/dt[sup]2[/sup]

If we do a free body diagram and take the sum of the moments around the center of the pinion we get:

Fspring(r1) – m(r2)d[sup]2[/sup]y/dt[sup]2[/sup] = 0

Substitute d[sup]2[/sup]y/dt[sup]2[/sup] = (r2/r1) d[sup]2[/sup]x/dt[sup]2[/sup]
into above equation gives:

Fspring = m(r2[sup]2[/sup]/r1[sup]2[/sup])( d[sup]2[/sup]x/dt[sup]2[/sup])

From F=ma, we can see that the “effective mass” from the point of view of the springs is:

m[sub]effective[/sub]= m(r2[sup]2[/sup]/r1[sup]2[/sup])

Now the solution to the differential equation is:

x = x0(cos(k/m[sub]effective[/sub])[sup]1/2[/sup]t)

When t=1 second we want x=0. We already know from the energy analysis that the velocity of the mass will be correct when the springs have moved 5/12 ft (x=0).

For the cosign term in the above equation to be zero, (k/m[sub]effective[/sub])[sup]1/2[/sup] must be equal to pi/2.

Substitute m[sub]effective[/sub]= m(r2[sup]2[/sup]/r1[sup]2[/sup]) and solve for r2/r1 we get

r2/r1=(4k/(pi[sup]2[/sup]m))[sup]1/2[/sup]=260

 

[zekeman]

Tlee Quote:
"He used x=5/12(1-cos(sqrt(k/m)t)). Setting t=0 we get x=0 which isn’t right. X should be 5/12 ft at the start. X=0 means the spring is not exerting a force (F=kx)."
--------------------------------------------------------------
To set the record straight the differential equation I used was NOT incorrect but
mx"=5/12k-kx which references x to the start position zero where the spring is compressed at time t=0 and is quite arbitrary (just picture a spring compressed and the starting position of the system is 0). Saying that x=0 does NOT imply that the force is zero. This ODE shows that at the start there is a force of 5/12k .
The solution to this differential equation is as I showed
x=5/12(1-cos(sqrt(k/m)t)) which totally satisfies the ODE
And by the way your solution effectively says that x=-5/12 at time t=0 which is just as arbitrary but conveniently becomes a valid solution of the homogeneous equation which is correct for your proposed starting conditions in which you have x referenced to the the equilibrium position of the spring
mx"=-kx
x=-5/12cos(sqrt(k/mt))
If you look carefully you will see that the difference in the both solutions is 5/12 foot, which has only to do with the reference positions we each took.
So, in essence the both solutions are correct and not surprisingly lead to the same result.

 

[tlee123]

OK, I agree with you that you weren't wrong. Thanks for the explanation.