Physics Help Forum Falling in a viscous liquid problem

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Jul 27th 2015, 01:12 PM #1 Member   Join Date: Jan 2015 Posts: 96 Falling in a viscous liquid problem Hi Everyone , In have a tricky problem that has me stumped at present .. it reds as follows - A ROCK WITH MASS m = 3KG FALLS FROM REST IN A VISCOUS MEDIUM. THE ROCK IS ACTED ON BY A NET CONSTANT DOWNWARD FORCE OF 18 NEWTONS AND BY A FLUID RESISTANCE FORCE f = kv , WHERE v IS SPEED IN m/s AND k = 2.2 N.s/m. NOW I HAVE MANAGED TO SOLVE ALMOST ALL OF THE PROBLEMS ASKED - INCLUDING .. a) INITIAL ACCELETRATION .. b) SPEED WHEN a = 3m/s/s .. c) SPEED AT 0.1 INITIAL ACCELERATION .. d) TERMINAL SPEED AND e) DISTANCE, SPEED AND ACCELERATION 2.0 SECONDS AFTER START OF MOTION. HOWEVER I CANNOT SEEM TO SEE HOW TO GO ABOUT FINDING .. THE TIME TAKEN TO REACH 0.9 OF THE TERMINAL SPEED ?? COULD ANYONE GIVE ME AN EXPLANATION OF HOW YOU GO ABOUT SOLVING THIS FINAL PART OF THE PROBLEM ? REGARDS JACKTHEHAT
 Jul 27th 2015, 01:40 PM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 This is really a university problem and involves solving differential equations. Attached Thumbnails
 Jul 27th 2015, 02:39 PM #3 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,347 I assume you have the equation for v(t), with all constants established from boundary conditions, right? If you know the terminal velocity, and set v(t) to 90% of that, why can't you simply solve for t? Perhaps if you show us the equation you have for v(t) we can help better.
 Jul 28th 2015, 11:41 AM #4 Member   Join Date: Jan 2015 Posts: 96 Hi ChipB, Sorry to take so long to get back to you but I am just back home from work in the last hour and have just got back on to the forum now. Well to answer your question .. the equation I have for velocity changing with time for this problem is ... v(t) = Vt [t - m/k (1 - e to the power -(k/m)t ) where Vt = terminal velocity; t = time; m = mass of body; k = fluid resistance force of the medium; and e is the exponent. In case my equation above is still a bit confusing .. the exponent e is raised to the power whose value is equal to .. -(k/m)t. I hope this is clear. regards, JacktheHat.
Jul 28th 2015, 12:13 PM   #5
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Join Date: Jun 2010
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 Originally Posted by jackthehat Hi ChipB, the equation I have for velocity changing with time for this problem is ... v(t) = Vt [t - m/k (1 - e to the power -(k/m)t )
You're missing a closing bracket - I assume you mean this:

v(t) = V_t [t - m/k (1 - e ^( -kt/m)]

Note that I used "V_t" for terminal velocity (as "Vt" looks too much like velocity times time) and the convention of using "^" to mean "raised to the power of."

I see your difficulty solving for t using this equation - however, this equation is not correct. A quick sanity check: V_t is the limit of the velocity as t goes to infinity, but from this equation you would have:

V_t = v(infinity) = V_t[infinity-m/k(1-e^(-k infinity/m)] (note - your equation is missing the closing bracket, but I assume this what you meant).

which equals Vt[infinity-m/k] = infinity.

How did you get the t-m/k factor inside the brackets? Clearly the value of whatever is inside the brackets should approach 1 as t goes to infinity.

Last edited by ChipB; Jul 28th 2015 at 12:17 PM.

 Jul 28th 2015, 12:53 PM #6 Member   Join Date: Jan 2015 Posts: 96 Hi again ChipB, I am afraid that you have lost me in your explanation .. as far as I gathered there are a finite time, distance, velocity and acceleration in this problem ,,, your inclusion in your equations of infinity has totally confused me. Firstly you, were right i did miss out the closing square bracket in my equation .. secondly my equation as it stands was gleaned from the physics book I am studying from ..."University Physics with Modern Physics" by Young and Freedman (international edition) ... and the chapter I am working on derives the equations for a, v, and distance .. free-body diagram and force equations .. and I used these equations when dealing with these quantities as they vary with time ... the one I used in this problem is in the book as.. v = v_t[t - m/k(1-e power -(k/m)t)] .. this is derived from integrating formula for acceleration from Newton's 2nd Law ... ?? Surely there is an easier way to work out the time ?
 Jul 28th 2015, 01:30 PM #7 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,347 Here's the issue. Starting with your equation: v = v_t[t - m/k(1-e power -(k/m)t)] multiply the V_t term through the brackets: v = V_1 t - V_t m/k(1-e^-(kt/m)] The left hand side is simply velovity. But note that the first term of the right hand side has units of velocity times time. So the equation is clearly incorrect. As for why I introduced infinity: as the object falls it goes faster and faster, but its rate of acceleration decreases, with the net result that after some period of time acceleration reaches zero and hence velocity stops increasing. That is when it has reached terminal velocity. How long does it take for acceleration to reach zero? It's the length of time it takes e^(-kt/m) to reach 0, which means t = infinity. I suggest you start with basic differential equation for velocity. From F=ma = m dv/dt, where F = mg-kv, you have mg-kv -= m dv/dt Divide through by m and rearrange: dv/dt + (k/m)v =mg The general solution to this is v(t) = A-Be^(Ct) where A, B and C are constants, which you can solve from initial conditions. I had assumed you had gotten this far, since you already had determined the terminal velocity. How did you do that?
 Jul 28th 2015, 02:28 PM #8 Member   Join Date: Jan 2015 Posts: 96 Hi again, Yes I get your last post that is nearer to the way I solved the other parts of the problem .. the reason I got lost i think is .. I assumed that since the book seemed to be advocating using the equation I quoted earlier for problems involving finding .. a, v, distance when it involved these terms varying with time that I should go down that route to solving this part of the problem. I guess I didn't understand the equation as much as I thought. I solved the terminal velocity part of the question by using the force equation ... Net F = F (downward) + (-k.v_t) = 0 and .. F (downward) we are told is .. 18 Newtons; k = 2.2 N.s/m; v_t = terminal velocity .. Thus 18N + (-2.2 . v_t) = 0 Rearranging ... v_t = 18/2.2 => v_t = 8.18 m/s I suppose the lesson is top keep things simple if possible. 8.18 m/s is the answer at the back of the book ! Thanks for your help. Regards, JacktheHat
 Jul 29th 2015, 09:34 AM #9 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,347 Yes, I agree that 8.18 m/s is equal to V_t. Have you been able to make progress with the differential equation I provided earlier, in order to solve the problem of determining t for v= 0.9 V_t? One correction I need to make to it: when I set that up I assumed (wrongly) that the downward force is gravity, and hence F_down=mg. But I see the problem actually specifies the downward force as 18 N, which is not the same as mg. It seems weird to me that they did that, but OK - ignoring gravity and using F_down in place of mg yields: m dv/dt +kv = F_net From this can you derive the equation for velocity as a function of time? Last edited by ChipB; Jul 29th 2015 at 12:16 PM.
 Jul 29th 2015, 11:36 AM #10 Member   Join Date: Jan 2015 Posts: 96 Hi ChipB, What I did was to go back to the derived book equations .. which when I re-read the particular section is actually not quite what I had put down in my original e-mails to you .. I mixed up the derived equation for distance travelled with the equation for velocity .. (sorry about that) .. the correct equation should be .. v(t) = v_t (1 - e to the power -(k/m)t ) .. My problem with this originally would have still been the same by the way .. that the term for time t forms part of exponent's power .. and how to get a value for t from that equation .. but now I am thinking (after going away and refreshing myself with some basic maths .. as I left full-time education 40 years ago !) that if you take the log (logarithm function) of each side of the above equation you can take t down from the power as I believe ... log(e to the power x) = x ... that's right isn't it ? my only problem now (I think) is that the equation would then become .. log (v(t)) = v_t(1 - (-k/m)t) but how do I then get rid of the log function on the left hand side of the new equation ? Any insight on this would be appreciated. Regards, JacktheHat

 Tags falling, liquid, problem, viscous