Collision of a particle with a thin rod

[Satvik Pandey]

I got $V=\frac { 2mvcos\theta }{ 4m+M\\ } \quad$. Using equation $v cos\theta=V+\frac{l\omega}{2}+v_{0} cos\alpha$ and

$6V=l\omega$ and $MV=m(v_{0}cos\alpha+vcos\theta)$

and I got $\omega =\frac { 6 }{ l } (\frac { 2mvcos\theta }{ 4m+M } )$

and $\alpha =arcsin(\frac { vsin\theta }{ { v }_{ 0 } } )$

So $cos\alpha =\frac { \sqrt { { v }_{ 0 }-{ v }^{ 2 }{ sin }^{ 2 }\theta } }{ { v }_{ 0 } }$

And from energy conservation equation I got

${ v }_{ 0 }=\sqrt { { v }^{ 2 }-\frac { 4M{ V }^{ 2 } }{ m } }$

or ${ v }_{ 0 }=\sqrt { { v }^{ 2 }-\frac { 4M }{ m } { \left( \frac { 2mvcos\theta }{ 4m+M\\ } \right) }^{ 2 } }$

Now angle ACB is 90. So $BC=lsin\phi$

Now bu applying cosine rule in triangle OBC I got

${ (lsin\phi ) }^{ 2 }=\left( { \frac { l }{ 2 } } \right) ^{ 2 }+\left( { \frac { l }{ 2 } } \right) ^{ 2 }-2\frac { { l }^{ 2 } }{ 4 } cos(\beta )$

I have an expression of $\phi$ in terms of $v_{0}$ $V$ and $\alpha$. Am I right till here?

 

[Satvik Pandey]

Is there supposed to be an explicit expression for the solution to this problem? I'm getting a transcendental equation for θ that I can only solve numerically. Also, I get solutions only for a certain range of the ratio of the two masses.]

I forgot to mention an information in the question.

It is given that $\frac { M }{ m } =\frac { 12 }{ \pi } +2$.
Here $M$ and $m$ are the masses of rod and the particle respectively.

This might help.

 

[ehild]

I forgot to mention an information in the question.

It is given that $\frac { M }{ m } =\frac { 12 }{ \pi } +2$.
Here $M$ and $m$ are the masses of rod and the particle respectively.

This might help.

It does! Why did not say earlier ?

Have you noticed that β=2Φ?

 

[Satvik Pandey]

It does! Why did not say earlier ?

Have you noticed that β=2Φ?

Yeah! β=2Φ. I think I should put it in last equation in #post31. I think I now have sufficient number of equation to find $\theta$. right?. Are my calculations in post #31 correct?

 

[ehild]

Yeah! β=2Φ. I think I should put it in last equation in #post31. I think I now have sufficient number of equation to find $\theta$. right?. Are my calculations in post #31 correct?

No, it will give an identity. Use that the time for the particle to go from A to C is the same as β/ω. You know the speed of the particle in terms of known data.
Let us see what you get.

 

[TSny]

I forgot to mention an information in the question.

It is given that $\frac { M }{ m } =\frac { 12 }{ \pi } +2$.
Here $M$ and $m$ are the masses of rod and the particle respectively.

This might help.

That information helps a lot. With it you should find that $\phi$ works out nicely and you can then find a simple expression for $\theta$.

 

[Satvik Pandey]

That information helps a lot. With it you should find that $\phi$ works out nicely and you can then find a simple expression for $\theta$.

Sorry to all of you. I didn't mention that earlier.

No, it will give an identity. Use that the time for the particle to go from A to C is the same as β/ω. You know the speed of the particle in terms of known data.
Let us see what you get.

I got

$\frac { lcos\phi }{ { v }_{ 0 } } =\frac { 2\phi }{ \omega }$

or $\frac { lcos\phi }{ \sqrt { { v }^{ 2 }-\frac { 4M{ V }^{ 2 } }{ m } } } =\frac { 2\phi }{ \frac { 6V }{ l } }$

or $9{ V }^{ 2 }{ cos }^{ 2 }\phi ={ \phi }^{ 2 }\left( { v }^{ 2 }-\frac { 4M{ V }^{ 2 } }{ m } \right)$.

 

[Satvik Pandey]

Sorry to all of you. I didn't mention that earlier.
I got

$\frac { lcos\phi }{ { v }_{ 0 } } =\frac { 2\phi }{ \omega }$

or $\frac { lcos\phi }{ \sqrt { { v }^{ 2 }-\frac { 4M{ V }^{ 2 } }{ m } } } =\frac { 2\phi }{ \frac { 6V }{ l } }$

or $9{ V }^{ 2 }{ cos }^{ 2 }\phi ={ \phi }^{ 2 }\left( { v }^{ 2 }-\frac { 4M{ V }^{ 2 } }{ m } \right)$.

I think I made mistake. I have to consider the velocity of the particle wrt CoM of the rod. It is too late here(now the time is about 11pm). I will try it tomorrow. Good Night.

 

[ehild]

I got $V=\frac { 2mvcos\theta }{ 4m+M\\ } \quad$. Using equation $v cos\theta=V+\frac{l\omega}{2}+v_{0} cos\alpha$ and

$6V=l\omega$ and $MV=m(v_{0}cos\alpha+vcos\theta)$

and I got $\omega =\frac { 6 }{ l } (\frac { 2mvcos\theta }{ 4m+M } )$

and $\alpha =arcsin(\frac { vsin\theta }{ { v }_{ 0 } } )$...

or ${ v }_{ 0 }=\sqrt { { v }^{ 2 }-\frac { 4M }{ m } { \left( \frac { 2mvcos\theta }{ 4m+M\\ } \right) }^{ 2 } }$

All formulas are correct, but you need the components of vo, vo_y =vocosα and vo_x=vo sin α = vsinθ. No need to calculate vo.

 

[Satvik Pandey]

All formulas are correct, but you need the components of vo, vo_y =vocosα and vo_x=vo sin α = vsinθ. No need to calculate vo.

Considering speed in horizontal I got

$\frac { \frac { l }{ 2 } -\frac { l }{ 2 } cos\left\{ -\left( 180-2\phi \right) \right\} }{ Vsin\theta } =\frac { 2\phi }{ \omega }$
(angle is in clockwise direction so it is -ve)

or $\frac { \frac { l }{ 2 } \left( { 2{ sin }^{ 2 }\phi \quad } \right) }{ Vsin\theta } =\frac { 2\phi }{ \omega }$

or $\frac { { sin }^{ 2 }\phi }{ sin\theta } =\frac { \left( 4m+M \right) \phi }{ 6mcos\theta }$

Now considering speed in vertical direction I got

$\frac { \frac { l }{ 2 } -\frac { l }{ 2 } cos\left\{ -\left( 180-2\phi \right) \right\} }{ Vsin\theta }$ =$\frac { 2\phi }{ \omega }$....(1)

Now

${ v }_{ 0 }cos\alpha +V=\sqrt { { v }^{ 2 }-{ \frac { 4MV }{ m } }^{ 2 }-{ (vsin\theta ) }^{ 2 } } +V$

or $\sqrt { { v }^{ 2 }{ cos }^{ 2 }\theta -{ \frac { 4M }{ m } \left( { \frac { 2mvcos\theta }{ 4m+M\\ } \quad } \right) ^{ 2 } } } +\frac { 2mvcos\theta }{ 4m+M\\ } \quad$

or $\sqrt { \frac { { v }^{ 2 }{ cos }^{ 2 }\theta (16{ m }^{ 2 }+{ M }^{ 2 }-8mM) }{ m{ (4m+M) }^{ 2 } } } +\frac { 2mvcos\theta }{ 4m+M } \quad =\frac { vcos\theta (4m-M)+2mvcos\theta }{ m{ (4m+M) } }$

I have considered the value of expression in numerator inside square root in LHS to be the expansion $(4m-M)^{2}$ because I think 4m is greater than M.

Putting this value and the value of $\omega$ in eq(1) I got

$\frac {sin(180-2\phi)}{ 2(4m-M)+2m) } =\frac { 2\phi }{ 12m }$

or $\phi =\frac { 3msin2\phi }{ 2m-M } \\$.

 

[ehild]

You made it utterly complicated. Go back to the beginning of #Post21, and express vocos(α) in terms of v and θ. You do not need alpha. . I do not have the nerves to check your last derivation
And it is high time to use the given M=(12/pi +2)m. 4m is not greater than M!
Anyway, it might be correct. Correct the sign and plug in the value of M. You get a very simple equation, easy to solve by inspection.

 

[Satvik Pandey]

You made it utterly complicated. Go back to the beginning of #Post21, and express vocos(α) in terms of v and θ. You do not need alpha. . I do not have the nerves to check your last derivation
And it is high time to use the given M=(12/pi +2)m. 4m is not greater than M!
Anyway, it might be correct. Correct the sign and plug in the value of M. You get a very simple equation, easy to solve by inspection.

I agree that I made that complicated. There are so so many equations. I just forgot about the equation at start of #post21.
I just made guess that 4m is greater than M. I should have checked that.

On altering the signs I got $\frac{\pi sin(2\phi)}{4}=\phi$

So $\phi=\frac{\pi}{4}$.

So putting that in this equation --

$\frac { { sin }^{ 2 }\phi }{ sin\theta } =\frac { \left( 4m+M \right) \phi }{ 6mcos\theta }$

I got $\theta ={ tan }^{ -1 }\left( \frac { 2 }{ \pi +2 } \right)$.

Is it right?oo)

 

[ehild]

I got $\theta ={ tan }^{ -1 }\left( \frac { 2 }{ \pi +2 } \right)$.

Is it right?oo)

It is the same I got. You are great !

 

[Satvik Pandey]

It is the same I got.

If you got that then it must be correct.:D

You are great !

That compliment belongs to you!:)

Thanks for the help ehild and TSny. :) You guys are awesome!

 

[ehild]

I have learned something from you Satvik: The velocity of separation. I never use it in case of collision with an extended object, but it proved to be correct.

 

[Satvik Pandey]

I have learned something from you Satvik: The velocity of departure. I never use it in case of collision with an extended object, but it proved to be correct.

Thank you ehild. :).
I think you use other concepts instead of that for framing equations. So you never required that.

 

[Satvik Pandey]

I have a question regarding impulse.

Suppose at a certain point of time ball has velocity $vcos\theta$ in horizontal direction and $v\sin\theta$ in vertical direction (downwards) and angular velocity $\omega$ in clockwise direction. Suppose if the coefficient if restitution of the ground to be $e$ and coefficient of friction to be $\mu$. I am wondering what will be there horizontal and vertical components of the velocity of the ball after collision.

In common normal direction I can use $v_{separation}=ev_{approach}$.

Let $V_{x}$ and $V_{y}$ be the final velocity in X and Y direction respectively.

So $V_{y}=evsin\theta$...(1)

So $J=m(ev sin\theta+v sin\theta)$...(2)

So in horizontal direction impulse is $\mu m(ev sin\theta+v sin\theta)=J_{x}$....(3)

So $J_{x}=V_{x}-vcos\theta$.....(4)

Now impulsive torque is $RJ_{x}$

So $RJ_{x}=I(\omega_{f}-\omega)$....(5)

From equation 3 and 4 we can find $V_{x}$. But it doesn't depends on $\omega$. This sounds strange to me. I think I have done a mistake. Am I correct?

 

[TSny]

So $V_{y}=evsin\theta$...(1)

So $J=m(ev sin\theta+v sin\theta)$...(2)

So in horizontal direction impulse is $\mu m(ev sin\theta+v sin\theta)=J_{x}$....(3)

So $J_{x}=V_{x}-vcos\theta$.....(4)

Now impulsive torque is $RJ_{x}$

So $RJ_{x}=I(\omega_{f}-\omega)$....(5)

From equation 3 and 4 we can find $V_{x}$. But it doesn't depends on $\omega$. This sounds strange to me. I think I have done a mistake. Am I correct?

I believe your conclusion is correct as long as you assume that there is pure slipping throughout the contact of the sphere with the surface so that J_x = -μJ_y. Here μ is the coefficient of kinetic friction.

However, it is generally not true that you will have pure slipping where J_x = -μJ_y. For example, a rubber ball hitting a rough surface will tend to "grab" the surface and you cannot model it as pure slipping. Generally, the final value of v_x will depend on the initial spin of the ball. I think it can get very complicated.

 

[Satvik Pandey]

I believe your conclusion is correct as long as you assume that there is pure slipping throughout the contact of the sphere with the surface so that J_x = -μJ_y. Here μ is the coefficient of kinetic friction.

However, it is generally not true that you will have pure slipping where J_x = -μJ_y. For example, a rubber ball hitting a rough surface will tend to "grab" the surface and you cannot model it as pure slipping. Generally, the final value of v_x will depend on the initial spin of the ball. I think it can get very complicated.

So ,it is not essential that the ball slides on the surface it can even roll on it.

I think we can conserve angular momentum about the contact point even if ball rolls or slips. right?

$Rvsin\theta+I\omega=Rv_{0}sin\alpha+I\omega_{0}$
Considering clockwise as +ve.

Suppose if the coefficient of kinetic and static friction are equal then friction acting on the ball in either cases will be $\mu J$. right?

But I think I am going to end up as same equation as in last post. I don't see any way to deal with this situation.Thanks for the help TSny.:)

 

[ehild]

Imagine that you kick a ball towards a wall perpendicularly. The ball rolls before hitting the wall. There is friction between the wall and the ball. What happens during the impact? What forces are exerted on the ball? How will it move after the impact?

An oblique collision can be quite complicated, as the relative velocity between the surfaces in contact is not horizontal if the ball rolls.
And both the ball and the wall are elastic to some degree. Interesting things can happen during the impact!
See:

This article looks interesting, I have not read it yet.
http://www.physics.usyd.edu.au/~cross/Gripslip.pdf

 

[Satvik Pandey]

Imagine that you kick a ball towards a wall perpendicularly. The ball rolls before hitting the wall. There is friction between the wall and the ball. What happens during the impact? What forces are exerted on the ball? How will it move after the impact?

As ball collides the velocity of it's CoM changes it's direction but the angular velocity doesn't. So the ball tends to move left after the collision however $\omega$ remains in the same direction. So the lower end of the ball slips on the ground so $\mu_{k}mg$ acts at the bottom of the ball to the right. As the vertical wall is rough so vertical component of impulse acting on it will be $\mu_{k}J$ if the ball slides on the wall during collision. I think it will bounce after collision.

This article looks interesting, I have not read it yet.
http://www.physics.usyd.edu.au/~cross/Gripslip.pdf

Thanks for the video and article. The video is amazing. I will surely read that article.:)

 

[ehild]

View attachment 78041
As ball collides the velocity of it's CoM changes it's direction but the angular velocity doesn't.

What impulsive torque acts on the ball during impact? Does it not change the angular velocity? How does it accelerate the CoM of the ball?

 

[Satvik Pandey]

What impulsive torque acts on the ball during impact? Does it not change the angular velocity? How does it accelerate the CoM of the ball?

The impulsive torque will be $\mu_{k}JR=J_{y}$ in anti clockwise direction. I have neglected the friction force at the bottom. I think it will be too small as compared to $\mu_{k}JR$. Can we do that? I have written a wrong statement. The angular velocity changes but I assume the direction of final angular velocity same as the initial.

So $\omega_{f}=\omega-\frac{\mu_{k}JR}{I}$

Let $V_{y}$ be the final velocity of the ball in Y-direction.

So $V_{y}=\frac{\mu_{k}J}{m}$

 

[ehild]

Will the ball bounce up a bit during the impact? I think it will. We need to do some experiments :)

 

[Satvik Pandey]

Will the ball bounce up a bit during the impact? I think it will. We need to do some experiments :)

But that doesn't effect the final velocity in Y direction (it will remain same as $\frac{\mu_{k}J}{m}$).right?

 

[ehild]

If the ball both rotates and slips during the impact, you do not know the direction of the relative velocity of the surface in contact with the wall. The force of friction opposes that relative velocity, but you do not know the direction.
And think of the tennis ball in the video. Nothing is sure what happens during an impact.

 

[Satvik Pandey]

I have a confusion regarding pure rolling motion.

In figure the ball's CoM has $v$ velocity and angular velocity is $\omega$. For pure rolling motion $v=r\omega$ must be true. Here there is +ve angular acceleration due to friction force but -ve acceleration of ball's CoM due to that force. In $dt$ time the angular acceleration will increase and velocity of ball's CoM will decrease. So the equation $v=r\omega$ will not hold for long. Where am I wrong in the logic?

 

[ehild]

I have a confusion regarding pure rolling motion.
View attachment 78179
In figure the ball's CoM has $v$ velocity and angular velocity is $\omega$. For pure rolling motion $v=r\omega$ must be true. Here there is +ve angular acceleration due to friction force but -ve acceleration of ball's CoM due to that force.

In case of pure rolling with constant velocity the friction is static, and the force of friction is zero.:)

 

[Satvik Pandey]

In case of pure rolling with constant velocity the friction is static, and the force of friction is zero.:)

Thank you! I got it. That was a silly question.