# Thread: Dynamics Problem

1. ## Dynamics Problem

Hello, this is a series of related questions from my year 12 math class. Math is not my strongest subject so help with these would be appreciated. These are extra credit questions but i just can't seem to do much with them.

Objects falling vertically in a fluid medium are subject to a drag force due to the viscosity of the fluid. The magnitude of this drag force and hence its effect on the acceleration of the object is generally a function of the velocity of the object. Since the acceleration due to viscosity opposes gravity, a suitable model could be derived from the expression:

a = g - kv

where a is the acceleration of the object, v is its velocity, g the acceleration due to gravity, and k is a constant related to the viscosity of the fluid.

A. A ball-bearing sinking in a container of oil represents a suitable example of such motion. Using x for verticle displacement from the top of the fluid, t for time from th moment the ball is relased, and assuming v = 0, and x = 0 when t = 0, develop an expression for x in terms of g, k and v.

B. Develop an expression for v in terms of g, k and t

C. Develop an expression for x in terms of g, k and t

D. Theoretically, the ball bearing will never reach &quot;Terminal Velocity&quot;. However, it is possible to compare times and depths reached for different fluids for the ball bearing to reach say 95% of its terminal velocity. By re-arranging the expressions you have already developed, or otherwise, demonstrate the effect k has on these times and depths.

If you guys like, i could also scan the page out of my textbook and atatch it but the wording should be identicle.

Thanks in advance for anyhelp !! 2. Originally Posted by Potato Hello, this is a series of related questions from my year 12 math class. Math is not my strongest subject so help with these would be appreciated. These are extra credit questions but i just can't seem to do much with them.

Objects falling vertically in a fluid medium are subject to a drag force due to the viscosity of the fluid. The magnitude of this drag force and hence its effect on the acceleration of the object is generally a function of the velocity of the object. Since the acceleration due to viscosity opposes gravity, a suitable model could be derived from the expression:

a = g - kv

where a is the acceleration of the object, v is its velocity, g the acceleration due to gravity, and k is a constant related to the viscosity of the fluid.

A. A ball-bearing sinking in a container of oil represents a suitable example of such motion. Using x for verticle displacement from the top of the fluid, t for time from th moment the ball is relased, and assuming v = 0, and x = 0 when t = 0, develop an expression for x in terms of g, k and v.

B. Develop an expression for v in terms of g, k and t

C. Develop an expression for x in terms of g, k and t

D. Theoretically, the ball bearing will never reach &quot;Terminal Velocity&quot;. However, it is possible to compare times and depths reached for different fluids for the ball bearing to reach say 95% of its terminal velocity. By re-arranging the expressions you have already developed, or otherwise, demonstrate the effect k has on these times and depths.

If you guys like, i could also scan the page out of my textbook and atatch it but the wording should be identicle.

Thanks in advance for anyhelp !! Have you done any calculus?

CB

3. Sorry for the late reply I have been very busy with assignments.

To answer your question yes I have done calculus. I am quite good at differentiation and integration but am useless at applying in problem solving. Ihave plenty of questions like this so some basic help wouldbe appreciated.

I'm not asking you to work out the answers for me as I should be doing my own work butsome explanation on the methods and how I should approach this is more what I am after.

Thanks

4. What I probably should have said is I am okay at solving equations werethe definite integral or differential is given but am hopeless at setting up differential equations myself.

Edit:

I was fiddling with some expressions and somehow got an expression for the velocity in terms of time. I think its wrong but close to the proper answer.

My Work (i was just trying a few different things so certain aspects of my working may be incorrect, please to correct me?)

a = g - kv

(dt/dv) = (1/g-kv)

t = Integral(1/g - kv)

t = (1/k)ln(g - kv) (I'm not sure if that anti-derivitive is correct)

tk = ln(g-kv)

g-kv = e^(tk)

-kv = e^(tk) - g

v(t) = [(e^(tk)-g)]/-k

In my textbook, the answer for v in terms of t is v(t) = (g/k)(1-e^(-kt)) so i think i am close with this one, any suggestions?

If you would like the answers to the other questions from my textbook here they are , (although my teach said one was wrong) but it is method i care about not the answer.

a. x(v) = (1/k^2)[g ln{g/g-kv] - kv]

b. (Above)

c. x(t) = ( g / k ) [ { t + e^(-kt) } / k ]

5. Originally Posted by Potato Objects falling vertically in a fluid medium are subject to a drag force due to the viscosity of the fluid. The magnitude of this drag force and hence its effect on the acceleration of the object is generally a function of the velocity of the object. Since the acceleration due to viscosity opposes gravity, a suitable model could be derived from the expression:

a = g - kv

where a is the acceleration of the object, v is its velocity, g the acceleration due to gravity, and k is a constant related to the viscosity of the fluid.

a. A ball-bearing sinking in a container of oil represents a suitable example of such motion. Using x for verticle displacement from the top of the fluid, t for time from th moment the ball is relased, and assuming v = 0, and x = 0 when t = 0, develop an expression for x in terms of g, k and v.
You have:

$\displaystyle a=\dfrac{d}{dt}v = g-kv$

which we may rearrange to give:

$\displaystyle v'+kv=g$

Now this is a first order ordinary differential equation and there a number of ways of proceeding. Conceptually one of the easiest is to observe that if we multiply through by $\displaystyle e^{kt}$ the left hand side becomes the derivative of $\displaystyle e^{kt}v$:

$\displaystyle e^{kt}v'+ke^{kt}v=g\,e^{kt}$

$\displaystyle \dfrac{d}{dt}\left( e^{kt} v\right)=g\,e^{kt}$

so:

$\displaystyle \displaystyle e^{kt}v=\int_{\tau=0}^t g\,e^{k\tau}\; d\tau +v_0$

That will allow you to find $\displaystyle $$v in terms of \displaystyle$$ t,g$ and $\displaystyle $$k integrating again will give \displaystyle$$ x$

CB

6. Originally Posted by CaptainBlack You have:

$\displaystyle a=\dfrac{d}{dt}v = g-kv$

which we may rearrange to give:

$\displaystyle v'+kv=g$

Now this is a first order ordinary differential equation and there a number of ways of proceeding. Conceptually one of the easiest is to observe that if we multiply through by $\displaystyle e^{kt}$ the left hand side becomes the derivative of $\displaystyle e^{kt}v$:

$\displaystyle e^{kt}v'+ke^{kt}v=g\,e^{kt}$

$\displaystyle \dfrac{d}{dt}\left( e^{kt} v\right)=g\,e^{kt}$

so:

$\displaystyle \displaystyle e^{kt}v=\int_{\tau=0}^t g\,e^{k\tau}\; d\tau +v_0$

That will allow you to find $\displaystyle $$v in terms of \displaystyle$$ t,g$ and $\displaystyle $$k integrating again will give \displaystyle$$ x$

CB
Thanks a lot! I will try integrating that and then will try solving the other questions.

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