Classical mechanics problem

  • Thread starter Mechdude
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  • #1
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Homework Statement


heres a problem i working on
a block slides on a horizontal surface which has been lubricated with a heavy oil such that the block experiences a viscous drag
resistance in the opposite direction to the velocity and that varies with speed according to:
[tex]F\left( v\right) =-c\,i\,{v}^{n}[/tex]
if the initial speed is [itex] v_0 [/itex] at [itex] t=0 [/itex]
a) find [itex]v[/itex] as a function of time [itex]t[/itex]
which iv found ;
[tex]{\left( {v_o}^{1-n}+\frac{c\left( n-1\right) \,t}{m}\right) }^{\frac{1}{1-n}}[/tex]
b) For [itex]n = \frac{1}{2} [/itex] find the displacement x[itex]\vec{i} [/itex] as a function of t: i is the unit vectoer in x direction. .
which iv found
[tex] x- x_o = \frac{12\,{m}^{2}\,t\,vo+6\,c\,m\,{t}^{2}\,\sqrt{vo}+{c}^{2}\,{t}^{3}}{12\,{m}^{2}}[/tex]

c) show that for [itex] n = 1/2 [/itex] , the block will not travel further than [tex] \frac{2mv_o^{\frac{3}{2}} }{3c} [/tex]
now this is the sucker that will not work out. the answers for a and b are as given in the
assignment sheet, so they are ok.

Homework Equations





The Attempt at a Solution


give me a clue on how to do this
 

Answers and Replies

  • #2
ideasrule
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I really don't think b) can be right. For one thing, x-x0 increases without bound as t increases, which implies the block will eventually reach infinite speed.
 
  • #3
kuruman
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Use

[tex]a=v\frac{dv}{dx}[/tex]

and find v(x).
 
  • #4
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I really don't think b) can be right. For one thing, x-x0 increases without bound as t increases, which implies the block will eventually reach infinite speed.



correction
the second equation that i found has a sign problem on the second term compared to the solution provided
[tex] x- x_o = \frac{12\,{m}^{2}\,t\,vo-6\,c\,m\,{t}^{2}\,\sqrt{v_o}+{c}^{2}\,{t}^{3}}{12\,{m}^{2}}[/tex]
 
  • #5
kuruman
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OK, do it your way. Can you find the velocity as a function of time from your expression? Can you find at what time that velocity goes to zero?
 
  • #6
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OK, do it your way. Can you find the velocity as a function of time from your expression? Can you find at what time that velocity goes to zero?

so from the espression for [itex]f(v)= -cv^n [/itex]we can get using[itex] f=ma [/itex] one of newtons laws
[itex]a= dv/dt=\frac{-cv^n }{m}[/itex]
and a little separation of variables,
[tex]v(t)= {\left( {v_o}^{1-n}+\frac{c\left( n-1\right) \,t}{m}\right) }^{\frac{1}{1-n}}[/tex]
the integration was done much like that of obtaining newtons equations of motion, from 0 to t, and [itex]dv = v -v_o [/itex] but since the force is variable
the acceleration was not treated as a constant.
integrating this wrt time again we get
[tex]x-x_o= \frac{t\,\left( 12\,{m}^{2}\,v_o-6\,c\,m\,t\,\sqrt{v_o}+{c}^{2}\,{t}^{2}\right) }{12\,{m}^{2}}[/tex]
and we get this expression from v(t) at n=1/2
[tex] v(t)={\left( -\sqrt{vo}-\frac{c\,t}{2\,m}\right) }^{2}[/tex]
when the block stops moving v will be =0 hence we can solve for t at v=0
[tex]t= \pm \frac{2\,m\,\sqrt{v_o}}{c}[/tex]
inserting this into the equation for displacement x below that we found
[tex]\frac{t\,\left( 12\,{m}^{2}\,v_o-6\,c\,m\,t\,\sqrt{v_o}+{c}^{2}\,{t}^{2}\right) }{12\,{m}^{2}}[/tex]
we get after a little algebra (for the negative t, il try the positive t later)
[tex]-\frac{14\,m\,{v_o}^{\frac{3}{2}}}{3\,c}[/tex]

regards
mechdude.
 
Last edited:
  • #7
kuruman
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How do you get two times from

[tex]
0={\left( -\sqrt{vo}-\frac{c\,t}{2\,m}\right) }^{2}
[/tex]

*** Added on Edit ***

Also, negative times are unphysical. Check your signs along the derivation of the expression above.
 
Last edited:
  • #8
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Sorry, my bad, the expresion on the right is squared, so theres both negative and positive roots, but negative times are unphysical , il consider the positive root.
 
  • #9
kuruman
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Sorry, my bad, the expresion on the right is squared, so theres both negative and positive roots, but negative times are unphysical , il consider the positive root.

How do you get a positive root out of

[tex]

0={\left( -\sqrt{vo}-\frac{c\,t}{2\,m}\right) }^{2}

[/tex]

??

Check the math that got you to this point.
 
  • #10
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starting from the expression for velocity;
[tex] v(t)= {\left( \sqrt{v_o}-\frac{c\,t}{2\,m}\right) }^{2} [/tex]
and expanding it [tex] v(t)= v_o-\frac{c\,t\,\sqrt{v_o}}{m}+\frac{{c}^{2}\,{t}^{2}}{4\,{m}^{2}} [/tex]
now if we set v to zero (when it finally stops)
[tex]0 = v_o-\frac{c\,t\,\sqrt{v_o}}{m}+\frac{{c}^{2}\,{t}^{2}}{4\,{m}^{2}} [/tex]
and solve for t: using the quadratic formula:
[tex] x=-b \pm \frac{\sqrt{{b}^{2}-4\,a\,c}}{2\,a}[/tex]
[tex] t=\frac{2\,m\,\sqrt{v_o}}{c} [/tex]
if we substitute this time into the expression for displacement:
[tex] x-x_o= \frac{12\,{m}^{2}\,t\,v_o-6\,c\,m\,{t}^{2}\,\sqrt{v_o}+{c}^{2}\,{t}^{3}}{12\,{m}^{2}}[/tex]

we then get [tex] x-x_o = \frac{2\,m\,{v_o}^{\frac{3}{2}}}{3\,c} [/tex]
which is what was required.
 
  • #11
kuruman
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starting from the expression for velocity;
[tex] v(t)= {\left( \sqrt{v_o}-\frac{c\,t}{2\,m}\right) }^{2} [/tex]

That's better. Note that in the previous postings you had a minus sign in front of both terms between parentheses. Also, it is much quicker to take the square root on both sides

[tex] 0=\pm {\left( \sqrt{v_o}-\frac{c\,t}{2\,m}\right) } [/tex]
which clearly has one solution

[tex]
t=\frac{2\,m\,\sqrt{v_o}}{c}
[/tex]
 
  • #12
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That's better. Note that in the previous postings you had a minus sign in front of both terms between parentheses. Also, it is much quicker to take the square root on both sides

[tex] 0=\pm {\left( \sqrt{v_o}-\frac{c\,t}{2\,m}\right) } [/tex]
which clearly has one solution

[tex]
t=\frac{2\,m\,\sqrt{v_o}}{c}
[/tex]

thanks, i was not exactly cautious while working the problem,
regards.
 

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