- #106
ehild
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Try.
And read my post #100 carefully.
And read my post #100 carefully.
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Where is the integration constant? You must add it before multiplying by eθ.Satvik Pandey said:I was trying to solve it using IF method.:p
##\frac { dx }{ d\theta } -x=2rg(sin\theta -\mu cos\theta )##
Here ##IF=e^{\int-d\theta}=e^{-\theta}##
##x{ e }^{ -\theta }=\int { { e }^{ -\theta } } \left\{ 2rg(sin\theta -\mu cos\theta ) \right\} d\theta ##
##x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } \left\{ (sin\theta -\mu cos\theta ) \right\} d\theta ##
##x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } \left\{ (sin\theta -\mu cos\theta ) \right\} d\theta ##
##x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } sin\theta \quad d\theta -\mu \int { { e }^{ -\theta }cos\theta } d \theta##
##x{ e }^{ -\theta }=2rg\left\{ -\frac { { e }^{ -\theta }\left( sin\theta +cos\theta \right) }{ 2 } -\frac { 1 }{ 2 } \frac { { e }^{ -\theta }\left( sin\theta -cos\theta \right) }{ 2 } \right\} ##
ehild said:Where is the integration constant? You must add it before multiplying by eθ.
##x=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +C##
is not the general solution of the differential equation. You can see if you substitute it back.
ehild said:You repeatedly ignored my advice to look at my post #100 or substitute back into the original equation.
Satvik Pandey said:
Sorry:s.
So I should now find x at angle 41.82 and put it in this equation to find C.
No, it was a different situation, with Fs not equal to μN except the endpoint. You get v at 41.82° from that solution.Satvik Pandey said:At ## \theta=0## velocity of the ball is also 0. Can I also use this condition to find C?
ehild said:No, it was a different situation, with Fs not equal to μN except the endpoint.
In the second stage of the motion, yes.Satvik Pandey said:Fs is static friction. right?
Do you want to say that till angle 41.82 static friction acts on the ball and after that kinetic friction acts on the the ball and in differential equation we have assumed ## \theta## to be the angle at which the ball is slipping and rotating together that's why we can't use the initial condition at which v and ## \theta## are equal to zero.right?
Satvik Pandey said:But the value of ## \mu_{s}= \mu_{k}=0.5## so numerically Frictional force is always equal to 0.5N.
Satvik Pandey said:##\frac { dx }{ d\theta } -x=2rg(sin\theta -\mu cos\theta )##
Satvik Pandey said:By finding torque about the contact point
## mgsin\theta r=\frac { 7 }{ 10 } m{ r }^{ 2 }\alpha ##
Satvik Pandey said:I think I need the value to ##r##(radius of that sphere) to find the angle at which sphere looses contact with the cliff. right?
No. Finish your solution. No need to start a new one . Find C from the angle and speed of the CoM at the end of the first stage when the slipping started. Then you know how v2/r changes with the angle during the second stage. You had an equation where both v2/r and N were involved. Eliminate v2/r. The ball looses contact with the cliff if N=0.Satvik Pandey said:I think I need the value to ##r##(radius of that sphere) to find the angle at which sphere looses contact with the cliff. right?
ehild said:No. Finish your solution. No need to start a new one . Find C from the angle and speed of the CoM at the end of the first stage when the slipping started. Then you know how v2/r changes with the angle during the second stage. You had an equation where both v2/r and N were involved. Eliminate v2/r. The ball looses contact with the cliff if N=0.
Thanks!:)ehild said:Successful exams!
Satvik Pandey said:I am replying exactly after 10 days.:D
So 41.82 deg.=0.73 radians
Now from equation
##g(1-cos\theta )=\frac { 7 }{ 10 } \frac { { v }^{ 2 } }{ r } ##
I got value of ##v^{2}## at angle 41.82 is 3.56##r##.
Putting that in solution of that differential equation I got
##C=14.6859r##. Is it right?
Satvik Pandey said:So
##v^{2}=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +{ 14.69e }^{ \theta }##
From where do you have that equation? What does it mean? (Again, I know, but other people reading your solution might be confused, what it means. )Satvik Pandey said:and I also have
##\frac { m{ v }^{ 2 } }{ r } =mgcos\theta ##
Satvik Pandey said:I need to find ##\theta## from these equations. So I need to eliminate ##r##
Do you want me to eliminate ##r## using these--
##mgcos\theta -N=m\frac { { v }^{ 2 } }{ r } ##
and ##\\ mgsin\theta -\mu N=ma##
But these equations are for the moment at which the ball's CoM is doing circular motion about the contact point. So even if the ball is slipping we can use these equations. right?
Should I find another equations from these equations between ##r##,##v## and ##\theta##?
ehild said:Be more specific. What was the equation you got v2? (I remember that it was the equation for the no-slip case at the end when slipping started, but other might not know.) What do you mean with "Putting that in solution of that differential equation " - which differential equation do you mean?
I have mentioned that equation is #post 86.ehild said:From where do you have that equation? What does it mean? (Again, I know, but other people reading your solution might be confused, what it means. )
ehild said:You had two equations at the start of the problem concerning the slipping stage of the motion. You have solved them. There are no more equations and you do not need more.
You see that r cancels if you correct your mistake concerning C. And you need to find theta, where the ball detaches - what does it mean for N?
Satvik Pandey said:I got this equation by conservation of energy. I have mentioned it in #post . And I am talking about differential equation which is mentioned in #post 133.
That is better...Satvik Pandey said:I have mentioned that equation is #post 86.
The moment at which ball looses contact with the cliff the normal force acting on it will be 0. So at that moment the component of weight(mg) is responsible for providing the centripetal force.
So the component of mg = centripetal force
Satvik Pandey said:I tried this
Now
##\frac { { v }^{ 2 } }{ r } =-g(\frac { 3sin\theta +cos\theta }{ 2 } )+14.69{ { e }^{ \theta } }^{ \\ }##
and ##\frac { { v }^{ 2 } }{ r } =gcos\theta ##
So
##\frac { { v }^{ 2 } }{ r } =-g(\frac { 3sin\theta +cos\theta }{ 2 } )+14.69{ { e }^{ \theta } }^{ \\ }##
or ##gcos\theta +\frac { gcos\theta }{ 2 } +\frac { 3gsin\theta }{ 2 } =14.69{ e }^{ \theta }##
or ##\frac { 3g }{ 2 } (sin\theta +cos\theta )=14.69{ e }^{ \theta }\\ ##
or ##{ e }^{ \theta }=(sin\theta +cos\theta )##
Satvik Pandey said:Taking log on both side I got
##\theta =\ln { (sin\theta +cos\theta ) } ##
I don't know much about logarithms. How should I find ##\theta##?
Actually logarithms was earlier included in the syllabus of high school but it is not included at present.ehild said:You must have learned about logarithm.
You can not solve the equation in closed form. Apply some numerical method.
But check C first. I do not think it is correct.
Satvik Pandey said:Actually logarithms was earlier included in the syllabus of high school but it is not included at present.
Satvik Pandey said:I checked C.
##C=4.677r##.
Please See this http://www.wolframalpha.com/input/?i=Find C if 3.56r=-9.8r(3sin(0.73)-cos(0.73))0.5 + 2.72^(0.73) C.
Is it correct now?
I hate long calculations.##\frac { 7 }{ 10r } { v }^{ 2 }=g(1-cos\theta )##ehild said:But you have to know it for your life... It is very basic.I do not think you solved the correct equation. Check.
Almost correct :). v2=3.566 r.Satvik Pandey said:I hate long calculations.##\frac { 7 }{ 10r } { v }^{ 2 }=g(1-cos\theta )##
I got this from equation energy conservation-
Putting g=9.8 ##\theta=41.82## I got ##v^{2}=3.56r##
Now the the solution diffrential equation was
##x=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +{ Ce }^{ \theta }.##
where ##x=v^{2}##
So at ##\theta=41.82## ##v^{2} is 3.56r##
Do not substitute e with 2.72. It reduces accuracy.Satvik Pandey said:So putting values I got
##3.56r=-rg\left\{ \frac { 3sin(41.82)+cos(41.82) }{ 2 } \right\} +{ 2.72 }^{ 0.73 }C##
You typed in 3sin(41.82)-cos(41.82) instead of 3sin(41.82)+cos(41.82). Do not mix radians and degrees in the same equation. Use 0.73 for theta everywhere. And you get a number times r for C. :Satvik Pandey said:According to wolfram alpha ##C=4.67726##.
Please see this http://www.wolframalpha.com/input/?i=Find C if 3.56r=-9.8r(3sin(41.82)-cos(41.82))/2 + 2.72^(0.73) C&a=*C.r-_*Variable.dflt-&a=*C.C-_*Variable.dflt-&a=TrigRD_D
Where am I going wrong?
ehild said:Correct at last ! (sigh... )