Methods of Solving Dynamics Problems for Non-Engineers

By Joel Jackson

This website will show an average college student two ways to
solve complex Dynamics problems. The first way we are going to solve this
problems is through basic methods acquired from our dynamics class. We will
explain these methods in a way that a regular college student could understand
them. This process is complicated, so my second method of showing someone to
solve complex problems is through a program called * Working Model*.
This program allows a person to set up a simulation of a problem on the computer
and change its conditions, and then the program will compute a solution to your
problem.

There are two problems that I will solve for demonstrational
purposes. One will be a problem that was previously assign to me is class. That
problem will solve using normal dynamics method and by using * Working
Model.* The second problem will
just be solved by

Problem 13.174

A 1-kg block is moving with a velocity of magnitude of 2 m/s
as it hits the 0.5-kg sphere A, which is at rest and hanging from a cord
attached to *O.* Knowing that the coefficient of kinetic friction is 0.6
between the block and the horizontal surface and the *e* = 0.8 between the
block and the sphere, determine after impact (*a*) the maximum height *h*
reached by the sphere, (*b*) the distance *x *traveled by the block.

(Beer, Johnston Vector Mechanics for Engineers- Dynamics)

How To Solve:

__Dynamics Method__:

First consider our given conditions:

Mass of the Block = m_{b=} 1.0 kg Initial Velocity of
Block = v_{b} = 2.0 m/s

Mass of the Ball = m_{a} = 0.5 kg Coefficient of Kinetic Friction = *u _{k}*=
0.6

Coefficient of Restitution = *e *= 0.8 Gravity = *g *= 9.81
m/s^{2}

Now you must find the forces that are acting on the block.
There is a force due to gravity coming straight down from the block this denote
*mg. *Another force that is action on the block is the normal force (N)
that is pointing up at a 90-degree angle to the block. The final force is the
frictional force that is opposing the movement of the ground. Next we must draw
an inertia diagram that shows the inertia force vectors that cause the system
to be at dynamic equilibrium.

Now all the forces that we drew in the free body diagram must be put together in a sum of X-axis components and Y-axis components.

∑ F_{x = }F_{f } = m_{b}a

Given Equation for friction-

F_{f} = *u _{b}* *N

∑ F_{y }= N-m_{b} g = 0

N = m_{b} g

So now you must put both equation together to
solve for the acceleration ( a )

*u _{b}* * m

a = *u _{b}** g

We can solve for a because we have *u _{b }*and
g stated as givens so substitute them in

a = 0.6*9.81

a = 5.88 m/s^{2}

^{ }

We have found the acceleration equation we must put that aside, for we will use it later.

Now we should proceed to solve part “a” of problem 13.174

To
find the ratio of velocities we can use a helpful equation for the coefficient
of restitution (*e*)

*e* = (v_{b2} –v_{a1})/ (v_{a}
–v_{b})

The v_{a
}and v_{b }are the velocities of the block and ball originally,
which are given. The v_{a2 }and v_{b2 }are the final velocities
of the block and ball. We know that the v_{a }is zero because it starts
off a rest.

Substitute in the given values:

v_{a2} –v_{b2 = }0.8 ( 0- 2 m/s)

v_{b2} –v_{a2 = }-1.6 m/s

v_{b2 = }v_{a2 }– 1.6

Another helpful equation that will be used is the Conservation of Momentum Equation:

m_{a}v_{a} + m_{b}v_{b}
= m_{a}v_{a2} + m_{b}v_{b2}

Again
we know v_{a }and v_{b }are the velocities of the block and
ball originally, which are given. The v_{a2 }and v_{b2 }are the
final velocities of the block and ball. We know that the v_{a }is zero
because it starts off a rest.

0 + 1kg * 2 m/s = 0.5 kg*v_{a2} + 1 kg *v_{b2}

v_{b2} = 2- 0.5*v_{a2}

Substitute in v_{b2}

1.5 v_{a2
}= 3.6

v_{a2}=
2.4 m/s

v_{bs }=0.8
m/s

Now since we have found the final velocity of
the ball we can find the maximum height by using the equation that max
potential (where the height is max) equal maximum kinetic energy.

mgh = 0.5*mv^{2}

^{ }

h = v^{2}/
2g

h = 0.2936 m

Finally to solve part “b” of the problem you
must use the formula for uniform accelerated rectilinear motion:

v_{f}^{2
}= v_{o}^{2} + 2a( x_{f} –x_{o})

The Final velocity of the block is zero
because it comes to rest and the acceleration found previously in our work.

a = 5.88 m/s^{2}

^{ }

0 = 0.8^{2}
+ 2(5.88) *(Δ x)

Δ x = 0.0544 m

__Working
Model Method__:

To solve this problem on working model, I first constructed the model that is similar to the one given to the book. I gave the objects the same initial conditions that were given in the book. Well when you run this system it seems like it is going fine, but there are not any results to your questions. This is where you must play with working model a little bit. I created measuring devices that calculate the change of distance in the x-direction that the block traveled and the distance the ball traveled in the y-direction. To find the max height of the ball just find the original point of ball and subtract it from the point when the ball is at its highest point. To find the distance traveled by the block, you have to find original point on the x-axis subtracted by the point on the x-axis where the ball came to rest.

My results on Working Model:

X_{o}= 6.15 m X_{f
= }6.076 m

ΔX = 0.076 m

Y_{o}=
-3.55 m Y_{f = }-3.28 m

ΔY= 0.27 m

To try this for yourself go to the following link:

__Conclusion
of Part 1__:

There
was a small difference in the answer I got in part by doing the problem by hand
and * Working Model*. I think this is because I use a different gravity
force then the program use in their calculations. We used g = 9.8 m/s

__Second
Problem:__

The second problem is a more real world applications and could help people understand the principals of dynamic in everyday life. I chose the following problem because this situation could benefit many people and has affected me on a personnel level. If you have ever been to a Nebraska Cornhusker football game you have noticed that there is one man of all the 78,000 in capacity that everyone has their eye one. This noble person is the Fairbury Viener Slinger Man. He is the guy who shoots Fairbury hotdogs in to the crowd of hungry fans. How does this guy know how fast to shoot these hotdogs to allow them to reach the fans in the stands?

__Problem: __

Lets say that the Viener Slinger Man comes up
to me and says, “ Joel, I am really having a problem shooting my hotdogs 13.5
meters high, when I am 34 meters away. Last game I missed I guy who was approximately
that far away. This caused the crowd to become distraught and indirectly caused
the Huskers to lose to Colorado.” I would say, “ Excuse me Mr. Slinger, but
have you consider that the air resistance and wind could play a factor?” He
would tell me no because we were in an idea world. So I would then decided to
help him.

I would then take him to room 308 in the Walter Scott
Engineering Building and give him a first class lesson of projectile motion on
the * Working Model*. First I would set up a simulation much like
the one he experiences everyday. I would set a target that is 13.5 meters away
with a height of 34 meters. I didn’t know, but he told me that the hotdogs that
he fires weigh approximately a quarter pound. I did the math and calculated that
to be about 0.1136 kg. With all these
conditions set he play with the program until he found the right initial
velocity that would be need to hit his target.

To Try:

1) Adjust the x and y velocities on the velocity panel

2) Your x any y velocities should be the same because the Viener Slinger Man likes to shoot his Fairbury hotdog at a 45-degree angle

3) You can turn the tracking on or off to follow the projectile motion

__Conclusion__:

I found out by testing that the projectile would hit its
target if it were shoot an initial velocity of 17.0 m/s to 17.9 m/s. This is if
the angle it is shoot at is 45-degrees, which is the preferred angle of the
Viener Slinger. As you can see the Dynamics problem are very difficult and time
consuming, but if you have * Working Model* it allows interactively
solve these problems.