Stopping distance of a ball bearing down a slope

[TheBigDino]

Hello,

 

I am just wondering, why does the stopping distance of a ball bearing start to decrease after the angle of the slope exceed 45o

I really don't understand why....

Thanks.

[JaKiri]

Can you clarify the question?

 

I suspect it's because of the inelastic collision with the ground, but can't be sure that's the answer.

[5614]

why does the stopping distance of a ball bearing start to decrease after the angle of the slope exceed 45o

 

surely it'd be because the ball is gaining acceleration from the force of gravity as it goes down the slope, once it stops going down it stops having the acceleration from gravity (and the hill) and therefore its not accelerating as fast so will stop in a shorter distance.

[swansont]

Velocity is a vector. Once the ball reaches the bottom, there is no way to increase the x component, which gets smaller as the angle is increased. As JaKiri surmised, it's the action that the ball will bounce (y component of the velocity)

 

Consider the limiting case of 90 degrees. The stopping distance is 0. But who knows how long it will keep bouncing vertically.

[JaKiri]

surely it'd be because the ball is gaining acceleration from the force of gravity as it goes down the slope, once it stops going down it stops having the acceleration from gravity (and the hill) and therefore its not accelerating as fast so will stop in a shorter distance.

 

What?

[Panic]

 

at 45 degree, the horizontal acceleration = vertical accelleration which is 1/2 of the gravitational accelerations. but increasing the angle past 45 degree will generate a large initial impact with table. Eventually if the angle of the ramp is 90 degree the horizontal acceleration wx Is decreased to 0 ... since it's horizontal acceleration Wx that will puss it across the surface of the table.

 

at 90 degree the ball would act as a free falling object and bound inplace ... total horizontal displacement of 0... again due to 0 horizontal accelerations.

 

Wx = sin(angle)

 

Wy = Wcos(angle)

[Mart]

the horizontal acceleration = vertical accelleration which is 1/2 of

 

There's no horizontal accleration.

I think you mean the acceleration parallel to the incline and the acceleration at right angles to the incline

[JaKiri]

There's no horizontal accleration.

 

If there's no horizontal acceleration, how does it move down the slope?

[TheBigDino]

Can you clarify the question?

 

I suspect it's because of the inelastic collision with the ground' date=' but can't be sure that's the answer.[/quote']

 

Ok, the situation is that a ball bearing is being rolled down the slope from being at rest (The ball is let free and is not given any other force [like me pushing it or something] to make it move).

 

Thankyou very much for all your answers, but I weally don't get what you mean (Mart).

I think you mean the acceleration parallel to the incline and the acceleration at right angles to the incline

[swansont]

There's no horizontal accleration.

I think you mean the acceleration parallel to the incline and the acceleration at right angles to the incline

 

Of course there is. The normal force will have a horizontal component.

[5614]

What?

 

OMG apologies, i misread the original question, i thought he said 'why did the stopping distance decrease once the ball left the slope which was at 45 degrees' when he really said something totaly different!

 

well, i would answer it but there isnt really much that i can add on top of what everyone else has said.

[[Tycho?]]

There's no horizontal accleration.

I think you mean the acceleration parallel to the incline and the acceleration at right angles to the incline

 

If there is no horizontal acceleration how does it move horizontally? Gravity may be acting vertically on it, but the result of it being on a slope means that its acceleration will have both an x and y component.

[Ophiolite]

I'm glad we got that resolved......

[bloodhound]

Consider the limiting case of 90 degrees. The stopping distance is 0. But who knows how long it will keep bouncing vertically.

 

if u try to sum the total distance the ball bounces, its a geometric series. and converges for all e<1.

[ecoli]

'']If there is no horizontal acceleration how does it move horizontally? Gravity may be acting vertically on it, but the result of it being on a slope means that its acceleration will have both an x and y component.

 

Because there is still velocity in the x direction. Velocity remains constant in the x...but will accelerate in the y, due to gravity.

[Mart]

If there's no horizontal acceleration, how does it move down the slope?

 

It has a horizontal velocity not a horizontal acceleration.

 

Think of a gun held horizontally (parallel to the ground). Fire the gun. The bullet is accelerated horizontally but the force that creates the acceleration ceases after a very short time. The bullet is now subject only to the force of gravity (as long as we neglect air friction). Therefore only vertical acceleration. The bullet eventually hits the ground. Strange as it may seem a bullet dropped from your hand (where you're holding the gun) will take the same time to hit the ground as the fired bullet. Impossible I hear you say!

[JaKiri]

It has a horizontal velocity not a horizontal acceleration.

 

I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.

 

It's stationary when it begins.

 

Therefore only vertical acceleration.

 

We were talking about it on the slope. You said that it didn't have a horizontal component of acceleration on the slope.

 

There's no horizontal accleration.

I think you mean the acceleration parallel to the incline and the acceleration at right angles to the incline

 

 

Resolve parallel to the ground, please. (The angle is there because I've just pasted it from another thread)

 

Impossible I hear you say!

 

Similarly, I hear you say that's impossible for the earth to orbit the sun, or for there to be indivisible particles. Put words into my mouth all you like, I have an exemplary understanding of basic mechanics.

[The Rebel]

I think the gun is confusing the idea a bit. The gun doesn't really accelerate a bullet in the same terms as this topic is on. A gun works on collisions and the transfer of collision energies. A ball bearing on a slope works on potential energy (GPE) accelerating towards Kinetic Energy.

 

Correct me if I'm wrong as I struggled visualing these angles during mechanics, however, lets consider the ramp above.

 

Gravity acts straight down, and as a result there is a reaction force perpendicular to the incline. The formula being R = g cos (angle).

 

If we consider the horizontal component of this reaction, we find that there is an acceleration (we're assuming the friction is zero) against the surface. F = R sin (angle)

 

Therefore F = R sin (angle) = g cos (angle) sin (angle), which gives a parabolic when plotted between 0 and 90. Showing how the force is at a maximum when the ramp is at 45 degrees and reduces in either direction.

 

As the ball leaves the incline, the reaction is equal to gravity, i.e. there is no horizontal component, it is down to friction to slow the ball down. However the initial velocity as the ball reaches the end of the incline, that determines the stopping distance would already have been decided on by sqrt(2*acceleration*distance)

[Mart]

Resolve parallel to the ground' date=' please. (The angle is there because I've just pasted it from another thread)

[/quote']

 

 

Sorry about putting words in your mouth. Force of habit

I've resolved the forces in your diagram parallel to the ground. I found that Fcos20 = Rcos70. Is that what you got?

 

And I did it without the F and found that you were right there is a horizontal force and therefore there is a horizontal acceleration. A big THANKS.

[TheBigDino]

I have another question, do you think that by rolling a ball bearing down a slope (without ANyyyyy friction whatsoever), it would make a quite a large difference? (Like 500cm??)

[The Rebel]

I have another question, do you think that by rolling a ball bearing down a slope (without ANyyyyy friction whatsoever), it would make a quite a large difference? (Like 500cm??)

 

It depends on the co-efficient of friction.

 

The equations I used above, F = R sin (angle), assumed no friction. If there was friction the formula would be F = R sin (angle) - umg sin (angle), where u is the co-efficient of friction.

 

So, we can see that if the co-efficient is high (rough), the ball experiences less acceleration by gravity. Therefore the ball will have less velocity when it reaches the end of the incline, therefore it will go a shorter distance before stopping.