2 sphere system, Conservation of Energy and Momentum

[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?

 

[Vibhor]

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.

From the OP :

Now the upper sphere is pushed very slightly from it's equilibrium position

You can neglect the external work done by this very small/gentle push . It is just a nudge to get the upper ball moving .

 

[Vibhor]

Is the Mechanical Energy at the time of the upper sphere losing contact
$$\dfrac{1}{2}V_{M,earth}^2+\dfrac{1}{2}mv_{m,earth}^2+mg(R+r)(1-\cos\theta)?$$

The potential energy term is not correct . The term you have written represents the difference in potential energy .

 

[Better WOrld]

The potential energy term is not correct . The term you have written represents the difference in potential energy .

Oh Sorry, it should be $$
\dfrac{1}{2}V_{M,earth}^2+\dfrac{1}{2}mv_{m,earth}^2+mg(R+r)\cos\theta$$ right?

 

[Vibhor]

Right .

 

[Better WOrld]

Right .

And initially then, since the work done by the external force is negligible, $$mg(R+r)$$ is the total energy of the system. Thus, $$mg(R+r)=
\dfrac{1}{2}V_{M,earth}^2+\dfrac{1}{2}mv_{m,earth}^2+mg(R+r)\cos\theta$$

 

[Vibhor]

Yes .

By the way , you are missing 'M' in the first term on the right side.

 

[Better WOrld]

Yes .

By the way , you are missing 'M' in the first term on the right side.

Sorry again,
$$mg(R+r)=
\dfrac{1}{2}MV_{M,earth}^2+\dfrac{1}{2}mv_{m,earth}^2+mg(R+r)\cos\theta$$

Could you please help me apply the equation for conservation of linear momentum in the horizontal direction?

 

[Vibhor]

Could you please help me apply the equation for conservation of linear momentum in the horizontal direction?

What is troubling you in applying conservation of linear momentum?

 

[Better WOrld]

What is troubling you in applying conservation of linear momentum?

I'm trying to conserve momentum in the horizontal direction from the ground frame. However, in this frame, the velocity of the upper sphere is $v_{m,earth}$. However, I cannot understand how to find the horizontal component of $v_{m,earth}$.

 

[Better WOrld]

What is troubling you in applying conservation of linear momentum?

Sir we want to find the horizontal momentum for the upper sphere, right? So this would correspond to first finding the horizontal component of $v_{m,earth}$. Since $$v_{m,earth}=v_{m,M}+v_M$$, finding the horizontal component of $v_{m,earth}$ corresponds to finding the horizontal component of $v_M$ and $V_{m,M}$ and then adding their respective horizontal components to find the horizontal component of $v_m$. Now since $v_M$ is in the horizontal direction itself, its horizontal component is $v_M$ itself. The horizontal component of $v_{m,M}$ is $v_{m,M}\cos\theta$. Thus, the horizontal component of $v_{m,earth}$ is $$v_{m,M}\cos\theta+v_M=v_{m,M}\cos\theta-|v_M|$$ since $v_M$ is towards the negative X axis. Thus as the equation for conservation of linear momentum in the horizontal direction, I get $$0+0=-Mv_M+m(v_{m,M}\cos\theta-|v_M|)$$

 

[Vibhor]

Almost Correct .

No point in using absolute value in $|v_M|$ on right side. Writing $v_M$ is okay . The negative sign in front of $v_M$ takes care of the direction.

So , now you have three equations and three variables . Solve for $\theta$.

Note : I would again suggest you to work with the notation (free of subscripts) I mentioned in post#33 . The subscripts make things little hard to understand.

 

[Better WOrld]

Almost Correct .

No point in using absolute value in $|v_M|$ on right side. Writing $v_M$ is okay . The negative sign in front of $v_M$ takes care of the direction.

So , now you have three equations and three variables . Solve for $\theta$.

Note : I would again suggest you to work with the notation (free of subscripts) I mentioned in post#33 . The subscripts make things little hard to understand.

Just to confirm, these are the 3 equations, right?
$$\dfrac{mu^2}{R+r}=mg\cos\theta$$
$$(M-m)V=mu\cos\theta$$
$$mg(R+r)(1-\cos\theta)=\dfrac{1}{2}MV^2+\dfrac{1}{2}mv_{m,earth}^2$$
Also, $$v_{m,earth}^2=V^2+u^2-2uV\cos\theta$$
Sir, would it be possible to know the direction of $V_{m,earth}$ just by looking at the diagram and not considering the magnitude of the respective velocities? If so, please could you teach me how to do it?

 

[Better WOrld]

Just to confirm, these are the 3 equations, right?
$$\dfrac{mu^2}{R+r}=mg\cos\theta$$
$$(M-m)V=mu\cos\theta$$
$$mg(R+r)(1-\cos\theta)=\dfrac{1}{2}MV^2+\dfrac{1}{2}mv_{m,earth}^2$$
Also, $$v_{m,earth}^2=V^2+u^2-2uV\cos\theta$$
Sir, would it be possible to know the direction of $V_{m,earth}$ just by looking at the diagram and not considering the magnitude of the respective velocities? If so, please could you teach me how to do it?

Sir, could you please check the value you are getting for $\cos\theta$ with $$R=5,r=1,m=25,M=7,g=9.8$$ I'm not getting $\cos\theta=\dfrac{4}{5}$ which is the given answer.

 

[Vibhor]

$$(M-m)V=mu\cos\theta$$

Recheck this equation .

Sir, would it be possible to know the direction of $V_{m,earth}$ just by looking at the diagram and not considering the magnitude of the respective velocities? If so, please could you teach me how to do it?

Parallelogram law of vector addition .

 

[Better WOrld]

Recheck this equation .
Parallelogram law of vector addition .

Sorry, it should be $$(M+m)V=mu\cos\theta$$

If it is possible, could you please show the direction of $v_{m,Earth}$?

 

[Vibhor]

If it is possible, could you please show the direction of $v_{m,Earth}$?

The green vector represents $v_{m,Earth}$ .

 

[Better WOrld]

The green vector represents $v_{m,Earth}$ .

Thanks very much. I was about to upload the same picture!

 

[Better WOrld]

The green vector represents $v_{m,Earth}$ .

$$
mg(R+r)(1-\cos\theta)=\dfrac{1}{2}MV^2+\dfrac{1}{2}mv_{m,earth}^2$$
In this equation, how would we substitute $v_{m,earth}^2$ in terms of $u$ and $V$?
Is it $v_{m,earth}=\sqrt{u+V}=\sqrt{(u\cos\theta-V)\hat{i}+u\sin\theta\hat{j}}$

 

[Vibhor]

$$
mg(R+r)(1-\cos\theta)=\dfrac{1}{2}MV^2+\dfrac{1}{2}mv_{m,earth}^2$$
In this equation, how would we substitute $v_{m,earth}^2$ in terms of $u$ and $V$?

Refer to your post#47.

 

[Better WOrld]

$$
mg(R+r)(1-\cos\theta)=\dfrac{1}{2}MV^2+\dfrac{1}{2}mv_{m,earth}^2$$
In this equation, how would we substitute $v_{m,earth}^2$ in terms of $u$ and $V$?

Refer to your post#47.

Thanks Sir. I had just finished editing my previous post.

 

[Vibhor]

From post#47

$v_{m,earth}^2=V^2+u^2-2uV\cos\theta$

 

[Vibhor]

Is it $v_{m,earth}=\sqrt{u+V}=\sqrt{(u\cos\theta-V)\hat{i}+u\sin\theta\hat{j}}$

This doesn't make sense.

$v_{m,earth}=\sqrt{V^2+u^2-2uV\cos\theta}$

 

[Better WOrld]

This doesn't make sense.

$v_{m,earth}=\sqrt{V^2+u^2-2uV\cos\theta}$

Sorry again. I forgot to add the $^2$ symbol and remove the unit vectors. I think I do get $\cos\theta=\dfrac{4}{5}$ now.
I'm really, really grateful to you for all your help! Many, many thanks once again!

 

[Vibhor]

You are welcome .

 

[Better WOrld]

You are welcome .

Hello Vibhor. Sorry to disturb you again, but I have a few fresh doubts.

$$\dfrac{mv_{m,M}^2}{R+r}=N+mg\cos\theta$$

In this equation, is $N$ the normal force measured from the ground frame, or is it the normal force measured from the frame of the $lower sphere$ ie by keeping the lower sphere's position fixed (the way we kept the lower sphere's velocity fixed - through relative velocity)? Wouldn't the Normal force measured from these 2 frames (ground and the lower sphere) differ in direction?

I'm confused because I always learned that in circular motion, for $$F_1+F_2+...=\dfrac{mv^2}{R}$$ we consider forces $F_i$ which when drawn pass through the center of the circle about which the particle undergoes circular motion. However, when the lower sphere is moving too, I don't think it is necessary for the normal force between the upper and lower spheres (as measured from the ground frame) to pass through the center of any circle about which the particle may be undergoing circular motion.

Secondly, if we consider the motion of the upper sphere with respect to the $ground frame$, is it undergoing circular motion? Are there any specific conditions for it to undergo circular motion? I know that the upper sphere with respect to the $lower sphere$ is undergoing circular motion, but am not sure about the motion of the upper sphere with respect to the ground. In other words, over a period of time, is the green vector (in post #51) also undergoing circular motion about any point? If so, could you please show me that point?

I do hope you'll have the patience to bear with me. Many thanks in advance!