2 sphere system, Conservation of Energy and Momentum

[Better WOrld]

Homework Statement

[/B]
A sphere of radius $1m$ and mass $25 Kg$ is put on another sphere of radius $5 m$ and $7 Kg$ which is placed on a smooth ground. Now the upper sphere is pushed very slightly from it's equilibrium position and it begins to fall.

Now when the line joining the centre of the 2 spheres makes an angle $\theta$ with the vertical the upper sphere loses contact with lower sphere. Then find $\theta$ in degrees.

Note that there is no Friction between the 2 spheres.

Homework Equations

$$g=9.8 ms^{-2}$$
$$m_1u_1+m_2u_2=m_1v_1+m_2v_2$$
$$U_{initial} + K_{initial}=U_{final}+K_{final}$$

The Attempt at a Solution

Unfortunately, I haven't been able to get anywhere with the solution.

I took the small sphere (of radius r) as $m$ and the large sphere (of radius R) as $M$. The velocity of the small sphere is $v_m$ and the velocity of the large sphere is $v_M$.[/B]

When the small sphere will fall off, the Normal Reaction would be 0. Using this relation,
at the time of falling off (taking radial components),

$$\dfrac{mv_m^2}{R+r}=mg\cos\theta$$

I was told that the next step would be to conserve linear momentum in the horizontal direction since there was no external force in that direction. Lastly I was to apply the Conservation of Mechanical Energy for the system sphere+sphere+Earth.

Also, I took the 0 Potential Energy reference level as the common horizontal tangent to both spheres.Unfortunately I'm unable to form the Equations for Conservation of Linear Momentum and Mechanical Energy.

I would be grateful for any help with this question. Many thanks in advance!

PS. My (incorrect?) equation for Conservation of Energy is as follows: Taking 0 Potential Energy at the top of the sphere, $$-MgR+mgr=\dfrac{1}{2}MV_M^2+\dfrac{1}{2}mv_m^2-MgR-mgR(1-\cos\theta)$$

 

[mfb]

$$\vec{F} = a^2 \cos(\theta) \cdot \{ \sin(\theta) \cos(\varphi), \sin(\theta) \sin(\varphi), \cos(\theta)^2 \}$$

The big sphere will move as well.

Also, I took the 0 Potential Energy reference level as the common horizontal tangent to both spheres.

For the initial positions?
I guess the floor would be more intuitive.

If there is friction, the spheres will rotate as well, but the problem statement is a bit unclear in several aspects.

 

[Better WOrld]

The big sphere will move as well.
For the initial positions?
I guess the floor would be more intuitive.

If there is friction, the spheres will rotate as well, but the problem statement is a bit unclear in several aspects.

Sir, I'm really sorry. I had meant to write that there is no friction. Is the problem more clear now?
Sir, why would the floor be more intuitive, please?

Also, could you please show how you arrived at $$
\vec{F} = a^2 \cos(\theta) \cdot \{ \sin(\theta) \cos(\varphi), \sin(\theta) \sin(\varphi), \cos(\theta)^2 \}?$$

 

[mfb]

why would the floor be more intuitive, please?

It is the usual way to define "zero height".

That equation in my post was a copy&paste error, sorry. Still had a formula from an unrelated thread in the clipboard. I wanted to copy this formula:

$$\dfrac{mv_m^2}{R+r}=mg\cos\theta$$

The big sphere will move as well.

 

[Better WOrld]

It is the usual way to define "zero height".

That equation in my post was a copy&paste error, sorry. Still had a formula from an unrelated thread in the clipboard. I wanted to copy this formula:

The big sphere will move as well.

Sir, I realized that it would move. This would be because of the Normal Force between the 2 spheres. However, I still cannot understand how to apply the law of Conservation of Momentum.

 

[mfb]

I realized that it would move

Then you should take this motion into account.

However, I still cannot understand how to apply the law of Conservation of Momentum.

How can you describe the horizontal momentum in terms of the velocities of the spheres?
What do you know about the horizontal position of the center of mass?

 

[Better WOrld]

Then you should take this motion into account.

How can you describe the horizontal momentum in terms of the velocities of the spheres?
What do you know about the horizontal position of the center of mass?

I'm sorry Sir, I don't know the answer to the first question. Had there been 2 separate spheres with velocities v1 and v2, the momentum (initial) of them would be $m1v1$ and $m2v2$. However, the confusion is arising since these 2 spheres are in physical contact and it thus seems that the motion of the bottom sphere will influence the motion of the top sphere. Thus, the velocity of the top sphere would not simply be $v_m$. This is what is bothering me, Sir.

Regarding the second question, I didn't understand. Wouldn't the center of mass of each sphere always be at the center of each sphere respectively?

 

[mfb]

The contact does not matter.

and it thus seems that the motion of the bottom sphere will influence the motion of the top sphere.

That is true, but it does not change for formulas for their momentum.

Thus, the velocity of the top sphere would not simply be $v_m$

You can define the velocity to be v_m. By the way, for LaTeX you can use two $ or two # here: $v_m$.

Wouldn't the center of mass of each sphere always be at the center of each sphere respectively?

The center of mass of the spheres yes, but I meant the center of mass of the whole system.

 

[Better WOrld]

The contact does not matter.
That is true, but it does not change for formulas for their momentum.
You can define the velocity to be v_m. By the way, for LaTeX you can use two $ or two # here: $v_m$.
The center of mass of the spheres yes, but I meant the center of mass of the whole system.

Sir, regarding the center of mass, in the horizontal direction, it would be at the same level (ie the $Y$ coordinate would be the same) since there is no external force acting in the horizontal direction.
Sir, please could you explain your first 2 points? I couldn't understand why the formula for momentum of the upper sphere wouldn't be affected if the velocity of the upper sphere is influenced by the lower sphere. As an example, suppose a block is placed on a surface moving towards the left with $v_1$ while the block moves on the surface with $v_2$. Wouldn't the momentum of the block in the ground frame be $m_{block}v_2# and in the frame of the surface,$m_{block}(v_2-v_1)##?
Also Sir, could you please explain what you meant by ''define the velocity'? Did you mean wrt a specific frame?

 

[mfb]

regarding the center of mass, in the horizontal direction, it would be at the same level (ie the YY coordinate would be the same) since there is no external force acting in the horizontal direction.

Right. You can write that down as formula. It helps to find a relation between velocities.

I couldn't understand why the formula for momentum of the upper sphere wouldn't be affected if the velocity of the upper sphere is influenced by the lower sphere.

The momentum of an object is always the mass multiplied with the velocity. It does not depend on which other objects it touches.
The velocity will depend on the contact, of course, but that is a different part of the problem.

Also Sir, could you please explain what you meant by ''define the velocity'? Did you mean wrt a specific frame?

What is v_m? You introduced the variable, so you have to choose what it means. If you say "v_m is the velocity of the small sphere", then it will be the velocity of the small sphere. My comment was on your question if v_m can be the velocity of the small sphere. Sure it can, if you define it to be.

 

[Better WOrld]

Right. You can write that down as formula. It helps to find a relation between velocities.

The momentum of an object is always the mass multiplied with the velocity. It does not depend on which other objects it touches.
The velocity will depend on the contact, of course, but that is a different part of the problem.

What is v_m? You introduced the variable, so you have to choose what it means. If you say "v_m is the velocity of the small sphere", then it will be the velocity of the small sphere. My comment was on your question if v_m can be the velocity of the small sphere. Sure it can, if you define it to be.

I couldn't understand why the formula for momentum of the upper sphere wouldn't be affected if the velocity of the upper sphere is influenced by the lower sphere. As an example, suppose a block is placed on a surface moving towards the left with $v_1$ while the block moves on the surface with $v_2$. Wouldn't the momentum of the block in the ground frame be $m_{block}v_2$ and in the frame of the surface, $m_{block}(v_2-v_1)$?

Sir, I'm assuming that originally, the center of both spheres lie on the Y axis. Hence, the $X$ coordinate of the center of mass will always be 0 since there are no external forces acting in the horizontal direction. Hence, $-mx_1$ where $x_1$ is the displacement of the upper sphere's center of mass, is always equal to $Mx_2$ where $x_2$ is the displacement of the lower sphere's center of mass.
$v_m$ is the velocity of the upper sphere wrt the ground frame.
Could you please tell me how to proceed next?

 

[mfb]

I couldn't understand why the formula for momentum of the upper sphere wouldn't be affected if the velocity of the upper sphere is influenced by the lower sphere.

I am traveling at 20 km/h to a town 20 kilometers away. How long do I need to go there? To find the answer, do you have to know if I run there, take a bike, a car or a train?
It simply does not matter where the velocity comes from. Momentum is always* mass multiplied by velocity.

The reference frame here is the ground. Sure, momentum does depend on the reference frame, but every other choice blows up the equations and does not help.

*ignoring relativity

 

[Better WOrld]

I am traveling at 20 km/h to a town 20 kilometers away. How long do I need to go there? To find the answer, do you have to know if I run there, take a bike, a car or a train?
It simply does not matter where the velocity comes from. Momentum is always* mass multiplied by velocity.

The reference frame here is the ground. Sure, momentum does depend on the reference frame, but every other choice blows up the equations and does not help.

*ignoring relativity

Sir, I got that. I had meant to ask how we account for the velocity of the upper sphere.

 

[mfb]

With a variable describing this. Like v_m. Better: with its horizontal component.

 

[Better WOrld]

With a variable describing this. Like v_m. Better: with its horizontal component.

So Sir, at the instant when the upper sphere loses contact, the total momentum of the system is $$-Mv_M+mv_m\cos\theta?$$ But doesn't this assume that the velocities of the respective spheres do not change with time, ie they remain the same as they initially were? How then would these velocities account for the effect of the respective spheres on each others' velocity?

 

[mfb]

The angle of the motion won't agree with the contact angle at this point.

But doesn't this assume that the velocities of the respective spheres do not change with time, ie they remain the same as they initially were?

Why? One sphere will start moving to the left, the other one to the right. The total momentum stays constant, which allows to find the ratio of these (horizontal) velocities.

 

[Better WOrld]

The angle of the motion won't agree with the contact angle at this point.
Why? One sphere will start moving to the left, the other one to the right. The total momentum stays constant, which allows to find the ratio of these (horizontal) velocities.

Sir, I couldn't understand your point regarding the angle of motion. Did you mean that the upper sphere would (at the instant of losing contact) be moving tangentially to the lower sphere? In that case, resolving the velocity into 2 components, I still get the horizontal component of $v_m$ as $v_m\cos\theta$. However, with respect to the ground, the actual value of the velocity of the upper sphere should be $v_m\cos\theta-v_M$ (hopefully this is right since the upper sphere is being carried on the lower sphere which in turn is moving with $v_M$ towards the opposite side). Thus, my (modified) equation for the total momentum of the system at the time the upper sphere loses contact is $$-Mv_M+m(v_m\cos\theta-v_M)$$ It would have been really useful if we could conserve momentum from the initial moment $t=0$ when the 2 spheres were just kept on each other and had no velocity. However, at $t+\Delta t$, an external force is applied on the upper sphere. Thus, we would have to conserve momentum from $after$ this first external force has been applied right?

But Sir, at $t+\Delta t$ at the instant the force is applied, would the 2 spheres have any velocity? In that case, could the initial momentum of the system (at $t+\Delta t$) be $0$?

 

[Better WOrld]

However, with respect to the ground, the actual value of the velocity of the upper sphere should be $v_m\cos\theta-v_M$ (hopefully this is right since the upper sphere is being carried on the lower sphere which in turn is moving with $v_M$ towards the opposite side). Thus, my (modified) equation for the total momentum of the system at the time the upper sphere loses contact is $$-Mv_M+m(v_m\cos\theta-v_M)$$ It would have been really useful if we could conserve momentum from the initial moment $t=0$ when the 2 spheres were just kept on each other and had no velocity. However, at $t+\Delta t$, an external force is applied on the upper sphere. Thus, we would have to conserve momentum from $after$ this first external force has been applied right?

But Sir, at $t+\Delta t$ at the instant the force is applied, would the 2 spheres have any velocity? In that case, could the initial momentum of the system (at $t+\Delta t$) be $0$?

Please could you ignore the part I've included in this quote? I' not sure about this; it could (and probably is) wrong.

 

[Vibhor]

Thus, my (modified) equation for the total momentum of the system at the time the upper sphere loses contact is $$-Mv_M+m(v_m\cos\theta-v_M)$$

Assuming $v_m$ represents speed of the upper sphere with respect to the lower one and $\theta$ is the angle which the line joining the the two centers make with the vertical , the above relation looks correct .

 

[Better WOrld]

Assuming $v_m$ represents speed of the upper sphere with respect to the lower one and $\theta$ is the angle which the line joining the the two centers make with the vertical , the above relation looks correct .

Sir, actually $v_M$ is the velocity of the lower sphere wrt the ground while $v_m$ is the velocity of the upper sphere wrt the ground. I know that the velocity of the lower sphere will affect the velocity of the upper sphere, but am unable to understand how to use this while writing the equation representing the conservation of linear momentu in the horizontal direction.

 

[Vibhor]

Please don't call me sir

actually $v_M$ is the velocity of the lower sphere wrt the ground while $v_m$ is the velocity of the upper sphere wrt the ground.

In that case the expression in post#17 is incorrect .

The correct expression should be $M\vec{v}_M+m\vec{v}_m$ . But note that $\vec{v}_m =\vec{v}_{m , M}+\vec{v}_M$ ,where $\vec{v}_{m , M}$ represents velocity of the upper sphere with respect to the lower one.

but am unable to understand how to use this while writing the equation representing the conservation of linear momentu in the horizontal direction.

$-Mv_M+m(v_{r,m}\cos\theta-v_M)$

 

[Better WOrld]

Please don't call me sir
In that case the expression in post#17 is incorrect .

The correct expression should be $M\vec{v}_M+m\vec{v}_m$ . But note that $\vec{v}_m =\vec{v}_{m , M}+\vec{v}_M$ ,where $\vec{v}_{m , M}$ represents velocity of the upper sphere with respect to the lower one.
$-Mv_M+m(v_{r,m}\cos\theta-v_M)$

Sorry for that (calling you Sir).
Alright, I understood that $v_m=v_{m,M}+v_M$.
But then shouldn't the horizontal component of $v_m$ be $v_m\cos\theta$?
Subsequently, shouldn't the equation for conservation of linear momentum in the horizontal direction be $-Mv_M+m(v_mcos\theta)$?
$$=-Mv_M+m(v_{m,M}+v_M)(cos\theta)$$

 

[Vibhor]

But then shouldn't the horizontal component of $v_m$ be $v_m\cos\theta$?

Do you think velocity of the upper sphere from the ground frame is making angle $\theta$ with the horizontal ?

 

[Better WOrld]

Do you think velocity of the upper sphere from the ground frame is making angle $\theta$ with the horizontal ?

Actually what I did was to draw $v_m$ as tangent to the lower sphere. Then, I drew 2 axes pointing downwards and towards the right respectively from the point of tangency, and then split $v_m$ along these axes.

 

[Vibhor]

Actually what I did was to draw $v_m$ as tangent to the lower sphere.

There lies the problem

It is $v_{m,M}$ not $v_m$which is acting at right angles to the line containing the two radii.The velocity of the upper sphere with respect to the lower sphere is in the tangential direction.

 

[Better WOrld]

There lies the problem

It is $v_{m,M}$ not $v_m$which is acting at right angles to the line containing the two radii.

Could you please hold on a moment? I'm uploading a picture of what I thought about the diagram so far.

 

[Better WOrld]

There lies the problem

It is $v_{m,M}$ not $v_m$which is acting at right angles to the line containing the two radii.The velocity of the upper sphere with respect to the lower sphere is in the tangential direction.

I've uploaded the picture. Please could you explain your point? Why is it $v_{m,M}$ acting along the tangent? How then would $v_m$ be acting? I couldn't quite understand Sir.

 

[Better WOrld]

There lies the problem

It is $v_{m,M}$ not $v_m$which is acting at right angles to the line containing the two radii.The velocity of the upper sphere with respect to the lower sphere is in the tangential direction.

Also in this attached image, a block is placed on the hypotenuse of a wedge. What would be $V_{BW}$ in this case then?
And what would be $V_B$? Would it be the hypotenuse? Or would it be the base? Please could you explain the answer? Many thanks in anticipation.

 

[Vibhor]

I have modified the picture . The red vector represents the velocity of the upper sphere with respect to the lower one and the blue represents velocity of the lower sphere .

$v_m$ is obtained by the vector addition of the two vectors .

 

[Vibhor]

Also in this attached image, a block is placed on the hypotenuse of a wedge. What would be $V_{BW}$ in this case then?
And what would be $V_B$?

$V_{BW}$ is the tilted vector along the hypotenuse . $V_B$ is the vector sum of the two arrows in your figure.

 

[Better WOrld]

$V_{BW}$ is the tilted vector along the hypotenuse . $V_B$ is the vector sum of the two arrows in your figure.

Thanks!

 

[Better WOrld]

I have modified the picture . The red vector represents the velocity of the upper sphere with respect to the lower one and the blue represents velocity of the lower sphere .

$v_m$ is obtained by the vector addition of the two vectors .

Thanks a lot! I finally understood why $v_{mM}$ is tangential to the lower sphere. Could you please show me how to proceed next with the question?

 

[Vibhor]

Could you please show me how to proceed next with the question?

Please use the following notation - 'u' for speed of the upper block with respect to the lower one and 'V' for the speed of the lower one .

For conservation of energy ,use reference level (U=0) to be a horizontal line passing through the center of the lower sphere .

So now how would you write the three equations you have written in the OP ?

 

[Better WOrld]

Please use the following notation - 'u' for speed of the upper block with respect to the lower one and 'V' for the speed of the lower one .

For conservation of energy ,use reference level (U=0) to be a horizontal line passing through the center of the lower sphere .

So now how would you write the three equations you have written in the OP ?

Sorry but I just got the first one: $$\dfrac{mu^2}{R+r}=mg\cos\theta$$
Also, how do we conserve Energy? At say $t=0$ the two spheres are at test. However, at $t+\delta t$ an external force is applied. Hence, we can conserve energy only after this external force has done work.
When considering the Potential Energy of the upper sphere at $t=0$, would we say that the Potential Energy (Gravitational) is $mgR$ or $mg(R+r)$?
It's just that even during vertical Circular Motion, I've seen the Potential Energy of the ball (radius r) moving in a circular track of radius R being taken as $mg2R$ instead of $mg(2R-r)$ (the 0 level is taken at the ground).

 

[Better WOrld]

Sorry but I just got the first one: $$\dfrac{mu^2}{R+r}=mg\cos\theta$$
Also, how do we conserve Energy? At say $t=0$ the two spheres are at test. However, at $t+\delta t$ an external force is applied. Hence, we can conserve energy only after this external force has done work.
When considering the Potential Energy of the upper sphere at $t=0$, would we say that the Potential Energy (Gravitational) is $mgR$ or $mg(R+r)$?
It's just that even during vertical Circular Motion, I've seen the Potential Energy of the ball (radius r) moving in a circular track of radius R being taken as $mg2R$ instead of $mg(2R-r)$ (the 0 level is taken at the ground).

Is the Mechanical Energy at the time of the upper sphere losing contact
$$\dfrac{1}{2}V_{M,earth}^2+\dfrac{1}{2}mv_{m,earth}^2+mg(R+r)(1-\cos\theta)?$$ What would be the initial Mechanical Energy of the 2 sphere system? Wouldn't we have to consider it after the external force has acted on the upper sphere (at $t+\delta t$), thereby causing both spheres to move?