Solving a Non-Circular Path Physics Problem: Can You Do It?

In summary, the ball will go up and to the right until it reaches the circle again, and then it will go back up and to the right again.
  • #1
pchalla90
55
0
I asked my Physics teacher about this and he was stumped. He went to ask the other AP Physics teacher, and he too had no clue. So if you guys know how to do this, feel proud. You're smarter than my teachers anyway...

Imagine that there's a ball attached with a string, as shown in attachment. there is no initial tension in the string. assume that the string does not "give"; it does not stretch. one end is the ball, initially at rest; the other end of the string is at a fixed point (0,12). The larger semicircle is the path the ball will follow once it falls far enough. (I figured out the length of the string in my question using some calculus and simply drew a circle centered at (0,12) with that radius.)

We've only learned when the ball always follows a perfect circular path, but in this scenario, it starts falling completely vertically until it is caught by the string.

How do calculations with this work? How high does it go after falling and being caught by the string? what speed is it going at the instant tension becomes present in the string? what is the speed at the lowest portion of the arc?

If you need some values, assume that the ball is only a point with mass of 1 kg. It starts 12 m above the ground (x axis).

the length of the string is approximately 9.29357m.

|g|=10 m/s/s

i have studied energy including PE=mgh and KE=.5mv^2. but that's the extent of my knowledge on that topic.

Thanks in advance.
 

Attachments

  • Untitled.pdf
    86.5 KB · Views: 601
Physics news on Phys.org
  • #2
Take the case where the ball is dropped from directly under the support. It follows the normal laws of free fall until the string stops it, and now collision like physics takes over. If the string is elastic, then the ball "bounces" back upwards. If the string is not elastic, then the ball basically stops.

As the initial point of release is moved horizontally from the support, the collision effects are reduced, and the ball will have more horizontal velocity.
 
  • #3
Because of the instant acceleration, the string can't not stretch, but if it could be, KE+PE=Const applies and the ball follows the arc all the way up to the top. Conservation of Energy.
 
  • #4
so in this case the initial mechanical energy is solely potential energy, mgh=(1)(10)(12)=120J. this is partially converted into kinetic energy (partially because it doesn't go completely to the ground), but this kinetic energy's v value is vertical, meaning that the velocity isn't tangential to the new circular path at the instant the string receives tension due to the ball. how could you figure out the tangential velocity at that point? is it just the same magnitude as the instant before when it was dropping solely vertically?
 
  • #5
pchalla90 said:
How could you figure out the tangential velocity at that point?
I'm not sure about at that point, but the tangental velocity would be due to the downward pull of gravity versus the tension in the string, and the position of the ball when the string tightens up.

If the string were an elastic rubber band or spring, the resultant motion would be a bouncing of the ball up and down while it swung side to side.
 
  • #6
pchalla90 said:
how could you figure out the tangential velocity at that point? is it just the same magnitude as the instant before when it was dropping solely vertically?
No, you find it the same way you do with a pendulum: By solving for V in the kinetic energy equation.
 
  • #7
The ball will fall straight down until it reaches the circle you've drawn that represents the string's length. Assuming an ideal, massless, unstretchable, perfectly elastic string:

At the instant the ball reaches that circle, it will experience an impulse (i.e., an instantaneous change in momentum) directed along the length of the string (i.e., back toward the point (0,12)). This impulse will behave exactly like the impulse from a collision with an immovable wall; i.e., a pure reflection as though from an immovable wall tangent to the circle.

After this instant, the ball's new momentum can be used as the initial condition for simple projectile motion: it will follow a parabolic arc upwards and to the right. When it reaches the circle again, it will again reflect, and follow a new parabolic arc, etc.

In essence, you can calculate everything as though that big circle you've drawn is a perfectly rigid, spherical shell. The ball will just bounce around inside.

Note that the ball will not bounce as high as the starting point on its first bounce, because it will hit the imaginary "spherical shell" at a glancing angle; while energy will be conserved, the ball's momentum is redirected, so part of the energy is devoted to horizontal motion.

If the ball ever happens to bounce in such a way that will send it back straight up (not very likely; this is a chaotic system), then it will bounce up to its original height (i.e., 12 m). In all other cases, it will always be lower.Note that in reality, strings can stretch slightly (like springs), and that balls have non-zero spatial extent, and are therefore able to spin and turn. So if you actually conducted this experiment, both of these effects would sap energy out of the system, and redirect it into other forms. In the absence of friction, we would expect total energy to be conserved, but thermodynamics tells us that on average, the energy would be equally divided between all the various forms available: translation in x, y, and z, rotation of the ball around its three axes, and stretching of the spring. This gives 7 degrees of freedom, of which 3 are translational, so in reality, you would expect the translational motion of the ball to be as though the ball had 3/7 as much energy as in the ideal case, on average.

And of course, once you add friction to the system, the ball will eventually come to rest at the bottom of the circle.
 
Last edited:
  • #8
Ignore the string!

pchalla90 said:
there is no initial tension in the string. assume that the string does not "give"; it does not stretch. one end is the ball, initially at rest; the other end of the string is at a fixed point (0,12). The larger semicircle is the path the ball will follow once it falls far enough.

Ignore the string.

If I understand you correctly, you are saying that the ball will fall vertically until it reaches the circle, and then, without bouncing away from the circle, will follow the circle up and down, up and down …

Physically, then, the situation is the same as if, without any string, there is a semi-circular slide and the ball is assumed to move along it, without bouncing, once it hits it.

The string, and any forces in it, are irrelevant. :smile:
 
  • #9
Ben stated the situation correctly. Unless the collision with the arc (or string) is totally inelastic, the ball will bounce away from the circle, not slide along it.
 
  • #10
… let's agree on the question first …

Jeff Reid said:
If the string is not elastic, then the ball basically stops.
Well, that's what I said - and also how pchalla defined the problem:
pchalla90 said:
The larger semicircle is the path the ball will follow once it falls far enough.
Isn't it? :smile:
 
  • #11
It's what I said too. [i have some bad grammar in there, though]
 
  • #12
Ben Niehoff said:
In the absence of friction, we would expect total energy to be conserved, but thermodynamics tells us that on average, the energy would be equally divided between all the various forms available: translation in x, y, and z, rotation of the ball around its three axes, and stretching of the spring. This gives 7 degrees of freedom, of which 3 are translational, so in reality, you would expect the translational motion of the ball to be as though the ball had 3/7 as much energy as in the ideal case, on average.

Can you explain the "7 Degrees of Freedom"? I have no clue what that's about... or where you got the number "7"...

Other than that, everything else was very very clear and understandable.

thanks to everyone that replied.
 
  • #13
Ben Niehoff said:
In the absence of friction, we would expect total energy to be conserved, but thermodynamics tells us that on average, the energy would be equally divided between all the various forms available: translation in x, y, and z, rotation of the ball around its three axes, and stretching of the spring. This gives 7 degrees of freedom, of which 3 are translational, so in reality, you would expect the translational motion of the ball to be as though the ball had 3/7 as much energy as in the ideal case, on average.

pchalla90 said:
Can you explain the "7 Degrees of Freedom"? I have no clue what that's about... or where you got the number "7"...

Ben must mean:
3 of translation (that's straight movement)
3 of rotation (pitch roll and yaw)
1 of tension in the string​

But Ben's rule is totally contrary to experience: if you drop a ball (no string) on a flat horizontal surface, it does lose some energy, but it doesn't move sideways, and it doesn't rotate! It has 6 degrees of freedom (no string), but it only uses one of them. The lost energy goes into heat and noise, not into thew other 5 degrres of freedom.

Similarly, if you drop a ball on a string, vertically below the support of the string, it only uses 3 degrees: vertical movement, vertical rotation, and tension in the string.
--------​
If the ball bounces, then you can't really answer the question, since different things (tennis balls, golf balls, balls of wool …) lose different amounts of energy on bouncing!

Assuming it doesn't bounce: I said earlier "ignore the string", but I didn't say what you should do.

I assume from your original question that you know how to work out the speed, u, of the ball just before the string goes tight, and how to work out the motion along the semi-circle if you know the speed, v, along the semi-circle just after the ball joins it.

So all you don't know is how to calculate v from u.

Just use Newton's third law.

Force in any direction = change in momentum in that direction.

Can you see a direction in which there is no force?

Well, the momentum along that direction must be the same for u as for v! :smile:
 
  • #14
tiny-tim said:
Can you see a direction in which there is no force?

Unless i misunderstood you (a very good probability that I did), it's easier to answer the question "Can you see a direction in which there IS a force?"

The instant before the string catches the ball, the only force that is exerted on the ball is gravity (that i see, anyhow). After the string catches the ball, it will swing in a perfect pendulum forever (neglecting air resistance). Then the forces acting on the ball are the weight of the ball (aka gravity) and the tension in the string.

Can you please explain again?
 
  • #15
There are 2 cases here: perfectly elastic and perfectly non-elastic (and anything in between is also possible) The elastic case was described by Ben Niehoff. In the inelastic case the ball will indeed swing forever, but a lot of the energy will be lost when the string becomes tight for the first time.
If the speed of the ball at this point is v, then you can use conservation of angular momentum to figure out that the tangential speed = [tex] v sin(\theta)[/tex]
where [tex]\theta[/tex] is the angle that the rope makes with the vertical

If the length of the string is r and the ball is released a distance d horizontally from the other end of the string, the ball will fall a distance [tex] \sqrt {r^2-d^2}[/tex] it's kinetic energy will be [tex] m g \sqrt {r^2-d^2}[/tex].

after the collision only [tex]m g \frac{d^2}{r^2}\sqrt {r^2-d^2}[/tex] of this remains, so the ball will only go up a distance of [tex] \frac{d^2}{r^2}\sqrt {r^2-d^2} [/tex] above this point. This is a distance of [tex] r - \sqrt {r^2-d^2} + \frac{d^2}{r^2}\sqrt {r^2-d^2} [/tex] above the bottom of the swing

if r = 12 and d = 6 the ball will only come 4.2 cm above the bottom point, and the angle with the vertical of the pendulum will then be 49.6 degrees.
 
  • #16
oops!

pchalla90 said:
Unless i misunderstood you (a very good probability that I did), it's easier to answer the question "Can you see a direction in which there IS a force?"

The instant before the string catches the ball, the only force that is exerted on the ball is gravity (that i see, anyhow). After the string catches the ball, it will swing in a perfect pendulum forever (neglecting air resistance). Then the forces acting on the ball are the weight of the ball (aka gravity) and the tension in the string.

Can you please explain again?

oops! I forgot about gravity! sorry! :redface:

I should have said "Can you see a direction in which there is no instantaneous force (impulse)?"

When the string goes tight, the ball receives an impulse (an instantaneous force) from the string.

It instantaneously changes the momentum from vertical to perpendicular-to-the-string.

Gravity is an ordinary gradual force, not an impulse, and instantaneously has zero effect.

From Newton's second law (it's not the third law - I got that wrong:redface:), the change in momentum is proportional to the force producing it.

The only force producing this instantaneous change in momentum is along the string.

So perpendicular to the string, there's no force and so the change in momentum in that direction is zero, and we have an equation for which we don't need the value of the force! (this is the point of the exercise - to avoid calculating the force - ignore anyone who ways you should!)

And that equation is: the momentum in that direction before equals the momentum in that direction after. :smile:
 
  • #17
so the momentum before the string goes taught is equal to the momentum when the string has tension.

so MV(v)=MV(t)

(v) is vertical
(t) is tangential.

so mass cancels out, and then V(v)=V(t)

just clarifying that i understood you properly.
 
  • #18
kamerling said:
If the length of the string is r and the ball is released a distance d horizontally from the other end of the string, the ball will fall a distance [tex] \sqrt {r^2-d^2}[/tex] it's kinetic energy will be [tex] m g \sqrt {r^2-d^2}[/tex].

I don't understand any of the other math that you did kamerling, so can you explain how you got that conclusion?

also, shouldn't the Mechanical energy be [tex](1/2) mv^2 + m g (12- \sqrt {r^2-d^2} )[/tex]
 
  • #19
pchalla90 said:
so mass cancels out, and then V(v)=V(t)

The mass does indeed cancel out! :smile:

I'm not sure whether you've completely got it or not, since I'm not sure what you mean by V(v)=V(t).

Assuming that the string is at an angle [tex]\theta[/tex] to the vertical just before it goes taut, can you write it out in full, using [tex]\theta[/tex], so we can check? :smile:
 
  • #20
what i meant was that the magnitudes of the velocities are equal just before and just after the impulse (not the direction).

so immediately before the ball is caught by the string it is traveling at V @ 270 degrees
immediately after the ball is caught by the string it is traveling at V @ -theta degrees (cartesian not navigational)

i think i might have a calculation error in there somewhere

but it is traveling at V perpendicular to the circle at that point.
 
  • #21
pchalla90 said:
what i meant was that the magnitudes of the velocities are equal just before and just after the impulse (not the direction).

Ah. No.

tiny-tim said:
So perpendicular to the string, there's no force and so the change in momentum in that direction is zero, and we have an equation for which we don't need the value of the force! (this is the point of the exercise - to avoid calculating the force - ignore anyone who ways you should!)

And that equation is: the momentum in that direction before equals the momentum in that direction after. :smile:

So you need the component of velocity perpendicular to the string before and after, not the total velocity.

(Remember, you can always treat conservation of momentum separately for anyone direction at a time - in other words, the components of momentum along that direction.)
 
  • #22
so i should calculate this as if all the momentum is conserved? but it can't be that because there's an impulse from the string... unless i do what you said before to ignore the string...
 
  • #23
Always try to calculate perpendicular to an unknown force

pchalla90 said:
so i should calculate this as if all the momentum is conserved? but it can't be that because there's an impulse from the string... unless i do what you said before to ignore the string...

Hi pchalla!

You're right, there is an impulse from the string, and so momentum is not conserved.

But the bit of it perpendicular to the impulse is conserved - that's why taking components of momentum perpendicular to the impulse is such a neat trick! :smile:
 
  • #24
okay, so if i understand you properly now:
Pre-string catching the ball
mgh+.5mv^2=mgH
where h is the height where the string catches the ball and H is where the ball starts.

solve it for v

so then just before the string catches the ball, the momentum of the ball is mv.

the moment the string catches the ball, the momentum perpendicular is mvsin(theta), where theta is the angle made by the string to the vertical.

so the tangential velocity the moment the string catches the ball is mvsin(theta)/m=vsin(theta)

is that right?

T(v)=vsin(theta)
 
  • #25
Down with string theory!

pchalla90 said:
is that right?

Yay! :smile:

(… and all that string theory was irrelevant … what a surprise … :rolleyes:)
 

Related to Solving a Non-Circular Path Physics Problem: Can You Do It?

1. How do you solve a non-circular path physics problem?

Solving a non-circular path physics problem involves breaking down the motion into smaller, simpler components. This can include using trigonometry to analyze the motion in terms of horizontal and vertical components, or using vector addition to combine multiple forces acting on the object.

2. Can you provide an example of a non-circular path physics problem?

One example of a non-circular path physics problem is a ball rolling down a curved ramp. The motion of the ball can be broken down into horizontal and vertical components, and the forces acting on the ball, such as gravity and friction, can be calculated to determine the ball's acceleration and velocity at any given point.

3. What are some common challenges in solving non-circular path physics problems?

One common challenge in solving non-circular path physics problems is identifying and breaking down the different components of the motion. Another challenge can be accurately representing the forces acting on the object and understanding how they affect its motion.

4. What are some helpful tips for solving non-circular path physics problems?

To solve non-circular path physics problems, it can be helpful to draw diagrams and label all the forces acting on the object. It is also important to understand the basic principles of Newton's laws of motion and how they apply to the specific problem at hand.

5. How can solving non-circular path physics problems be applied in real life?

Solving non-circular path physics problems can be applied in various real-life scenarios, such as designing roller coasters or analyzing the motion of objects in sports, like a baseball being thrown in a curved path. It can also be used in engineering projects, such as calculating the trajectory of a rocket or the motion of a satellite in orbit.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Mechanics
Replies
21
Views
4K
  • Classical Physics
Replies
4
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
2K
Replies
1
Views
3K
  • Classical Physics
Replies
3
Views
726
  • STEM Educators and Teaching
Replies
2
Views
2K
  • Mechanics
Replies
17
Views
4K
Replies
8
Views
2K
Replies
4
Views
1K
Back
Top