Finding Angle of Sphere Falling From Table Edge

[Satvik Pandey]

There is an other method to solve the linear inhomogeneous differential equation x'-2μx=f(θ).
Take the general solution of the homogeneous part: it is x=Ce^2μθ.
Add a particular solution X_p of the inhomogeneous equation: the general solution will be x=Ce^2μθ+X_p.
In case the inhomogeneous part contains sine and cosine functions, you can seek the solution in the form Xp=Acos(θ)+Bsin(θ). Replacing X_p =Acos(θ)+Bsin(θ) into the equation, you can find the constants A and B.

I am getting

$x=-rg\left( \frac { 3sin\theta -cos\theta }{ 2 } \right) +C$

or ${ v }^{ 2 }=-rg\left( \frac { 3sin\theta -cos\theta }{ 2 } \right) +C$

Am I correct?

 

[ehild]

I do not think so. Have you tried to substitute it back into the de?

 

[Satvik Pandey]

I do not think so. Have you tried to substitute it back into the de?

What is de?

 

[ehild]

differential equation

 

[Satvik Pandey]

differential equation

No

 

[ehild]

Try.
And read my post #100 carefully.

 

[Satvik Pandey]

There were some mistakes in my solution. Now I got
$x=-\frac { rg }{ 2 } (3sin\theta +cos\theta )+C$.
I have substituted $\mu=0.5$.

 

[ehild]

What is C? Is X solution if C is not zero?

 

[Satvik Pandey]

Should I find $x$ at angle 41.82 and put it in this equation to find C?

 

[ehild]

Yes, it would be the next step if your solution were correct. But it is not. Read my post ##100.

 

[Satvik Pandey]

I was trying to solve it using IF method.:p

$\frac { dx }{ d\theta } -x=2rg(sin\theta -\mu cos\theta )$

Here $IF=e^{\int-d\theta}=e^{-\theta}$

$x{ e }^{ -\theta }=\int { { e }^{ -\theta } } \left\{ 2rg(sin\theta -\mu cos\theta ) \right\} d\theta$

$x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } \left\{ (sin\theta -\mu cos\theta ) \right\} d\theta$

$x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } \left\{ (sin\theta -\mu cos\theta ) \right\} d\theta$

$x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } sin\theta \quad d\theta -\mu \int { { e }^{ -\theta }cos\theta } d \theta$

$x{ e }^{ -\theta }=2rg\left\{ -\frac { { e }^{ -\theta }\left( sin\theta +cos\theta \right) }{ 2 } -\frac { 1 }{ 2 } \frac { { e }^{ -\theta }\left( sin\theta -cos\theta \right) }{ 2 } \right\}$

$x{ e }^{ -\theta }=-rg{ e }^{ -\theta }\left\{ \frac { \left( sin\theta +cos\theta \right) }{ 1 } +\frac { \left( sin\theta -cos\theta \right) }{ 2 } \right\}$

$x{ e }^{ -\theta }=-rg{ e }^{ -\theta }\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\}$

$x=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +C$

What am doing wrong?

 

[ehild]

I was trying to solve it using IF method.:p

$\frac { dx }{ d\theta } -x=2rg(sin\theta -\mu cos\theta )$

Here $IF=e^{\int-d\theta}=e^{-\theta}$

$x{ e }^{ -\theta }=\int { { e }^{ -\theta } } \left\{ 2rg(sin\theta -\mu cos\theta ) \right\} d\theta$

$x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } \left\{ (sin\theta -\mu cos\theta ) \right\} d\theta$

$x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } \left\{ (sin\theta -\mu cos\theta ) \right\} d\theta$

$x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } sin\theta \quad d\theta -\mu \int { { e }^{ -\theta }cos\theta } d \theta$

$x{ e }^{ -\theta }=2rg\left\{ -\frac { { e }^{ -\theta }\left( sin\theta +cos\theta \right) }{ 2 } -\frac { 1 }{ 2 } \frac { { e }^{ -\theta }\left( sin\theta -cos\theta \right) }{ 2 } \right\}$

Where is the integration constant? You must add it before multiplying by e^θ.
$x=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +C$
is not the general solution of the differential equation. You can see if you substitute it back.

 

[Satvik Pandey]

Where is the integration constant? You must add it before multiplying by e^θ.
$x=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +C$
is not the general solution of the differential equation. You can see if you substitute it back.

Oh! My silly mistake.
$x{ e }^{ -\theta }=-rg{ e }^{ -\theta }\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +C$

or $x=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +\frac { C }{ { e }^{ -\theta } }$

or $x=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +{ Ce }^{ \theta }$. Right?

 

[ehild]

It is right now.

Ce^θ is the general solution of the homogeneous equation x'-x=0. I said earlier that the general solution is equal to that + a particular solution of the original equation. You repeatedly ignored my advice to look at my post #100 or substitute back into the original equation.

 

[Satvik Pandey]

You repeatedly ignored my advice to look at my post #100 or substitute back into the original equation.

Sorry:s.
So I should now find x at angle 41.82 and put it in this equation to find C.
At $\theta=0$ velocity of the ball is also 0. Can I also use this condition to find C?

 

[ehild]

Sorry:s.
So I should now find x at angle 41.82 and put it in this equation to find C.

Yes. But take care, you should use the angle in radians in the exponent.

At $\theta=0$ velocity of the ball is also 0. Can I also use this condition 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]

No, it was a different situation, with Fs not equal to μN except the endpoint.

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?
But the value of $\mu_{s}= \mu_{k}=0.5$ so numerically Frictional force is always equal to 0.5N.

 

[ehild]

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?

In the second stage of the motion, yes.

But the value of $\mu_{s}= \mu_{k}=0.5$ so numerically Frictional force is always equal to 0.5N.

No. The static friction is a limit, not a defined value. Fs≤μ_sN while the kinetic friction is Fk=μ_kN.When the ball rolled we used the condition that the angular velocity of rotation was the same as the angular velocity of the circular motion the CoM performed.

 

[voko]

$\frac { dx }{ d\theta } -x=2rg(sin\theta -\mu cos\theta )$

In #93, the equation was $$ \frac { dx }{ d\theta } -2 \mu x + 2rg(\mu cos\theta - sin\theta) = 0. $$

 

[voko]

One thing to keep in mind is that the general solution of the equation $$ {dx \over ds} - k x = f(s) $$ is given by $$ x = e^{ks} \left[x_0 + \int\limits_0^s e^{-kt} f(t) dt\right] .$$

 

[voko]

By finding torque about the contact point

$mgsin\theta r=\frac { 7 }{ 10 } m{ r }^{ 2 }\alpha$

Is this equation correct? $\frac { 7 }{ 10 } m{ r }^{ 2 }$ is half the moment of inertia about the contact point.

 

[ehild]

It was a typo. The continuation was correct.

 

[Satvik Pandey]

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?

 

[voko]

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?

Not really. Observe that you can change variables in the original equations as follows: $$ p = {I \over 2 mr^2} \\ z = {v^2 \over r } \\ N = mg n , $$ which then gives $$ g(1 - \cos \theta_0) = pz_0 \\ g \sin \theta_0 = pz'_0 \\ g(\cos \theta - n) = z \\ 2g( \sin \theta - \mu n) = z', $$ where the prime denotes differentiation with respect to $\theta$.

$z$ and $z'$ can be eliminated from these equations, yielding a differential equation and initial conditions for $n$, without any dependence on $r$ .

 

[ehild]

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]

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!:)
I have made an equation for the motion of the CoM at the moment at which it looses contact with the cliff in #post86.
Sorry, I am little busy now a days because my tests are going to start from 24th Nov.
I will come back to this after 10 days.:p

 

[ehild]

Successful exams!

 

[Satvik Pandey]

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?

So
$v^{2}=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +{ 14.69e }^{ \theta }$

and I also have

$\frac { m{ v }^{ 2 } }{ r } =mgcos\theta$

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 $\theta$?

 

[Satvik Pandey]

Successful exams!

Thanks!:)

 

[ehild]

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?

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?

So
$v^{2}=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +{ 14.69e }^{ \theta }$

That equation should be $v^{2}=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +14.69 r e^{ \theta }$

and I also have

$\frac { m{ v }^{ 2 } }{ r } =mgcos\theta$

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. )

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 $\theta$?

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]

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 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.

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. )

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

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?

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 ${ e }^{ \theta }=(sin\theta +cos\theta )$

Taking log on both side I got

$\theta =\ln { (sin\theta +cos\theta ) }$

I don't know much about logarithms. How should I find $\theta$?

 

[ehild]

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.

When writing a report, a paper or even a test, you should explain your steps. That equation you used described the motion of the ball when it did not slip. The angle you reported was the end of that stage and the initial angle of the next stage, the ball slipping. You should have written something like that.

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

That is better...

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 ${ e }^{ \theta }=(sin\theta +cos\theta )$

You lost 3g/2 :)

Taking log on both side I got

$\theta =\ln { (sin\theta +cos\theta ) }$

I don't know much about logarithms. How should I find $\theta$?

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]

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.

Actually logarithms was earlier included in the syllabus of high school but it is not included at present.
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?

 

[ehild]

Actually logarithms was earlier included in the syllabus of high school but it is not included at present.

But you have to know it for your life... It is very basic.

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 do not think you solved the correct equation. Check.

 

[Satvik Pandey]

But you have to know it for your life... It is very basic.I do not think you solved the correct equation. Check.

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 $\theta=41.82$ $v^{2} is 3.56r$

So putting values I got

$3.56r=-rg\left\{ \frac { 3sin(41.82)+cos(41.82) }{ 2 } \right\} +{ 2.72 }^{ 0.73 }C$

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?

 

[voko]

I have long lost track of the unnecessarily over-complicated approach that Satvik has been ill-advised to undertake. Here is a simpler method. I continue from my post #124. The equations there can be simplified further by noticing that $p = {7 \over 10}$ and $\mu = 0.5$. To simplify typing, in what follows I will use the shorthands $c = \cos \theta, \ c_0 = \cos \theta_0, \ s = \sin \theta , \ s_0 = \sin \theta_0$, and I also let $z =g u$ so the simplified equations are $$ {10 \over 7}(1 - c_0) = u_0 \\ {10 \over 7}s_0 = u_0' \\ c - n = u \\ 2s - n = u'. $$ The second pair of equations are valid for all angles, including $\theta_0$, so I obtain $$ c_0 - n_0 = u_0 \\ 2s_0 - n_0 = u_0'.$$ Using the first and third pairs, I obtain $$ 17 c_0 - 4 s_0 = 10 \\ n_0 = {4 \over 7} s_0 .$$ The first of these can be converted to a quadratic equation with respect to $s_0$, and solved, yielding $$ s_0 = 0.667 \\ c_0 = 0.745 \\ \theta_0 = 0.730 \\ n_0 = 0.381. $$
Turning to the second pair, it trivially yields $$ n' - n = - 3 s, $$ whose solution, according to #120, is $$ n(\theta) = e^\theta \left[n_0 - 3 \int \limits_{\theta_0}^\theta e^{-t} \sin t dt \right] .$$ We are looking for the angle when $n = 0$, so all that remains to be done is to solve $$ 0.381 = n_0 = 3 \int \limits_{\theta_0}^\theta e^{-t} \sin t dt = 1.021 - {3 \over 2} e^{-\theta} \left(\cos \theta + \sin \theta\right).$$

 

[ehild]

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 $\theta=41.82$ $v^{2} is 3.56r$

Almost correct :). v²=3.566 r.

So putting values I got

$3.56r=-rg\left\{ \frac { 3sin(41.82)+cos(41.82) }{ 2 } \right\} +{ 2.72 }^{ 0.73 }C$

Do not substitute e with 2.72. It reduces accuracy.

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?

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. :

$3.566 r=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +{ C re }^{ \theta }.$
Here is the solution with Wolframalpha.
http://www.wolframalpha.com/input/?i=Find C if 3.566=-9.8(3sin(0.73)+cos(0.73))/2 + e^(0.73) C

 

[Satvik Pandey]

So the value of $C=8.20r$

So ${ v }^{ 2 }=-rg\left( \frac { (3sin\theta +cos\theta ) }{ 2 } \right) +8.20r{ e }^{ \theta }$

or $\frac { { v }^{ 2 } }{ r } =-g\left( \frac { (3sin\theta +cos\theta ) }{ 2 } \right) +8.20{ e }^{ \theta }$

but $\frac { { v }^{ 2 } }{ r } =gcos\theta$ (got this relation from the equation for the moment at which ball looses contact with the cliff)

or $gcos\theta =-g\left( \frac { (3sin\theta +cos\theta ) }{ 2 } \right) +8.20{ e }^{ \theta }$

Is it fine now?

According to wolframalpha the value of $\theta=0.92rad=52.71degrees$
here is solution http://www.wolframalpha.com/input/?i=solve for x if 9.8cos(x)=-9.8((3sin(x)+cos(x))/2)+8.2(2.72)^x.

 

[ehild]

Correct at last ! (sigh... )

 

[Satvik Pandey]

Correct at last ! (sigh... )

Thanks! At last got correct answer.:)
I did lots of mistakes in calculation in this thread that's why this thread became too long.