Solve for Velocity of Two Falling Balls | Challenging Physics Problem

[troy5538]

Two heavy balls of equal mass M are attached to a long but light metallic rod standing on the floor. The rod with the balls falls to the floor. Find the velocity of each ball at the moment when the rod hits the ground. Neglect mass of the rod and the friction between the balls and the floor.

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O This ball is already touching the floor. Assume that the system falls to the right.
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[learningphysics]

Try to use conservation of energy and also conservation of momentum ideas...

 

[troy5538]

my solution so far is that as the system falls to the right, the ball touching the floor will start to move to the left a distance equal to the distance the top ball falls; therefore the top ball moves down with velocity sqrt(2gh) and the velocity of the bottom ball is sqrt(2gh)/4, but I am not sure about this at all since it must be more complicated than this

 

[Doc Al]

What can you say about the motion of the center of mass of this object?

 

[learningphysics]

Just as the higher mass is about to hit the ground, what is the horizontal velocity of each mass?

There's no force in the horizontal direction, so momentum is conserved horizontally.

 

[troy5538]

well I am not exactly sure what you guys mean since the only force acting on this system is gravity which accelerates the higher mass; however, the final velocity of the higher mass is directed downwards whereas the velocity of the lower mass is to the left. i would greatly appreciate it if you guys could please give a more detailed explanation of a possible solution. thanks.

 

[learningphysics]

well I am not exactly sure what you guys mean since the only force acting on this system is gravity which accelerates the higher mass; however, the final velocity of the higher mass is directed downwards whereas the velocity of the lower mass is to the left. i would greatly appreciate it if you guys could please give a more detailed explanation of a possible solution. thanks.

The main idea I used is that momentum is conserved in the horizontal direction... Let's just look at the horizontal direction. initially the momentum of the system in the horizontal direction is 0.

so it remains 0 throughout the fall... ie the horizontal momentum of the right mass and the horizontal momentum of the left mass (which is its entire momentum since it doesn't move vertically) are equal and opposite.

so at the very bottom when the rod is horizontal... the right mass has a horizontal velocity of v. The left mass has a horizontal velocity of -v. What is v? Hint think of the rod being rigid.

 

[troy5538]

ok i understand what you are saying, but i am still a bit confused with the fact that the upper ball will have its velocity vector pointing downward while the lower ball will have its velocity vector pointing to the left and so the velocities will be perpendicular

 

[learningphysics]

ok i understand what you are saying, but i am still a bit confused with the fact that the upper ball will have its velocity vector pointing downward while the lower ball will have its velocity vector pointing to the left and so the velocities will be perpendicular

Are you saying they're perpendicular when the upper ball reaches the bottom? That's true, but they aren't perpendicular throughout the fall... only at the bottom. So at the bottom the lower ball has its velocity vector to the left, the upper ball has its velocity vector downward... what must the lower ball's velocity be? use conservation of momentum in the horizontal direction.

 

[learningphysics]

The idea is that the horizontal velocity of the 2 balls is 0 at the bottom... net linear momentum horizontally is 0...

But we have a rigid rod... we can't have the balls moving in the direction of the rod...

So all we have is the vertical velocity of the ball that has been falling...

 

[troy5538]

ok so if i understand you correctly, then the velocity is just sqrt(2gh) since we just have vertical velocity?

 

[learningphysics]

ok so if i understand you correctly, then the velocity is just sqrt(2gh) since we just have vertical velocity?

yes. the falling ball has a downward velocity of sqrt(2gh). The other ball has 0 velocity.

 

[troy5538]

ok. just one last question, why doesn't the other ball move? (i mean there is no friction on the floor and the upper ball is applying a force on it)

 

[learningphysics]

ok. just one last question, why doesn't the other ball move? (i mean there is no friction on the floor and the upper ball is applying a force on it)

that ball moves... it gains velocity to the left initially... reaches some maximum velocity... then slows down back to zero...

 

[troy5538]

isn't acceleration that rises initially, reaches some max, and then slows down back to zero, but velocity is always increases up to the moment it hits the ground

 

[learningphysics]

isn't acceleration that rises initially, reaches some max, and then slows down back to zero, but velocity is always increases up to the moment it hits the ground

can you explain your reasoning for the acceleration ? not saying you're wrong there... I just didn't think about that...

The net kinetic energy is increasing... since potential energy is being converted to kinetic energy... but the velocity of the ball on the ground increases then decreases back to zero... the velocity of the other ball keeps increasing all the way down.

 

[troy5538]

well the upper ball has mass M and thus gravity pulls down with Mg; however this force changes with the angle as the upper balls falls down (for example, initially Mg is directed straight down, but as the ball falls to the right, Mg starts to play a role in the x component of the tension force of the rod until it hits the floor where Mg has no effect on the other ball)

 

[learningphysics]

well the upper ball has mass M and thus gravity pulls down with Mg; however this force changes with the angle as the upper balls falls down (for example, initially Mg is directed straight down, but as the ball falls to the right, Mg starts to play a role in the x component of the tension force of the rod until it hits the floor where Mg has no effect on the other ball)

The acceleration of the upper ball towards the right starts at zero... becomes positive... comes back to zero... and becomes negative... before it hits the bottom.

reason being that the velocity towards the right, goes up to some max... and comes back down to zero...

so the force of the rod I think goes from compressive to tensile...

 

[troy5538]

how exactly does the acceleration ever become negative because Vmax is the moment the ball hits the ground and the acceleration on the upper ball is always just gravity, the acceleration of the lower ball changes due to the angle.

 

[learningphysics]

how exactly does the acceleration ever become negative because Vmax is the moment the ball hits the ground and the acceleration on the upper ball is always just gravity, the acceleration of the lower ball changes due to the angle.

gravity isn't the only force on the upper ball... the rod exerts a force too... the Mg downward force is always acting... but the force the rod exerts is changing... initially it is straight upward... at the bottom, there's no force exerted by the rod...

acceleration of the lower ball to the left = acceleration of the upper ball to the right...

do you agree that the lower ball accelerates to the left... then decelerates (since it has to get back to 0m/s) ?

 

[troy5538]

ok i guess that makes sense...so just to clarify, the velocity of the upper ball is sqrt(2gh) and the velocity for the lower ball is 0 at the moment the system hits the floor?

 

[learningphysics]

ok i guess that makes sense...so just to clarify, the velocity of the upper ball is sqrt(2gh) and the velocity for the lower ball is 0 at the moment the system hits the floor?

Yes. I'm pretty sure that's the answer.

upper ball: sqrt(2gh) downwards

lower ball: 0

 

[troy5538]

ok, thanks a lot for all your help

 

[wags finger]

Congratulations on violating the Stanford Honor Code

Well troy5538, great job sullying the reputation of our University, and brazenly violating one of the core pillars of a Stanford education.

I quote from the Fall 2007, Physics 22, Extra Credit Problem Set:

"Two heavy balls of equal mass M are attached to a long but light metallic rod standing on the floor. The rod with the balls falls to the floor. Find the velocity of each ball at the moment when the rod hits the ground. Neglect mass of the rod and the friction between the balls and the floor."

It wouldn't be a clear violation of the Honor Code except that it goes on to say:

"Unlike the usual homework, you should work on this problem alone, without any help of others. If you solve the problem correctly, it will give you as much points as one regular homework."

Maybe you should spend some time reviewing "honorcode.stanford.edu" particularly the bit about "Unpermitted collaboration."

I don't want to report you to Judicial Affairs, who would then ask the site admins for your ISP address, and that would probably earn you the punishment for a first offense which as the website says, "includes a one-quarter suspension from the University and 40 hours of community service."

Here's what you need to do immediately:
1) Do not turn in the problem for extra credit.
2) Email Prof. Linde and tell him exactly what you did and apologize. When you send that email Bcc wags.finger@gmail.com
3) Post a copy of that email here but redact your email address and name.