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Inclined plane: heavy and light ball

by razal_ak47
Tags: ball, heavy, inclined, light, plane
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razal_ak47
#1
Feb24-13, 11:22 PM
P: 8
A light ball and another heavy ball is alowed to roll on an inclined plane.

which ball travel the farthest after the inclination.?

my teacher and friends argue giving difffrent points like energy,inertia...etc and reach in different conclusions.

which really will travel farther if we did it experimentaly?
:-C
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JustinRyan
#2
Feb25-13, 02:23 AM
P: 87
Without friction / air resistance they would both just keep going.......

If friction is equal for both balls then the heavier one will go further.
razal_ak47
#3
Feb25-13, 11:01 AM
P: 8
considering inertia, light ball will travel farther ,while heavy ball slows down.

hms.tech
#4
Feb25-13, 11:16 AM
P: 247
Inclined plane: heavy and light ball

Quote Quote by razal_ak47 View Post
considering inertia, light ball will travel farther ,while heavy ball slows down.
You are right in conclusion but wrong in reasoning . The heavier ball has more inertia so according to that the heavier ball must roll longer. This is clearly untrue .

The correct way of dealing with this problem is to consider energy changes involved during the different states of motion .
razal_ak47
#5
Feb25-13, 11:34 AM
P: 8
hms.tech

if we consider energy , heavier ball has greater energy but it will have to spend it for greater heat,sound,rotate...etc.... than the lighter ball.

then both would travel same distance.

one of my friends concluded like this.
deep838
#6
Feb25-13, 11:47 AM
P: 109
If there be no friction, what JustinRyan said is true... the balls keep on sliding forever with equal acceleration.
But then again, the question specifically says that the balls are rolling, and hence that possibility is eliminated. That's because a frictional force is necessary for causing the rolling motion!
If they are released from the same height, at the bottom of the inclination, they have the same velocities.
If there be friction in the ground as well, both the balls will get the same retardation! So shouldn't the balls cover the same distance?
And I don't understand how you use inertia here! I agree that the heavier ball should be harder to stop than the lighter one, but then again, greater frictional force acts on the heavier one!
Please help me understand this...
Also regarding energy changes, the heavier one has initially more energy, so what?
razal_ak47
#7
Feb25-13, 12:22 PM
P: 8
actualy which type of inertia should we take?

inertia of rest or inertia of motion?
ModusPwnd
#8
Feb25-13, 12:25 PM
P: 1,100
Quote Quote by razal_ak47 View Post
actualy which type of inertia should we take?

inertia of rest or inertia of motion?
They are the same thing. It takes only a change of frame to turn one into the other.
razal_ak47
#9
Feb25-13, 01:12 PM
P: 8
moduspwnd,

but if we take inertia of rest for a heavier ball , it slows down and stops.

and if we take inertia of motion , a heavy ball tries to go on....
ModusPwnd
#10
Feb25-13, 01:16 PM
P: 1,100
Quote Quote by razal_ak47 View Post
but if we take inertia of rest for a heavier ball , it slows down and stops.
Nope. Object in motion will stay in motion, barring any forces. Inertia means it will stay in motion, not slow down. I wouldn't try to distinguish between inertia of rest and inertia of motion. Best just to think of inertia.
JustinRyan
#11
Feb25-13, 08:46 PM
P: 87
Friction is only required to set the ball into rotation. Once the ball reaches the end of the incline its tendency will be to retain its rotational and translational motions. The only force opposing this tendency is air resistance as all other forces are normal to the motion. If the balls are of similar size and surface finish then air resistance will be equal for both balls. Since this equates to a larger deceleration for the lighter ball.....
If this is clearly untrue....enlighten me please
Menaus
#12
Feb26-13, 01:13 AM
P: 54
Quote Quote by ModusPwnd View Post
Nope. Object in motion will stay in motion, barring any forces. Inertia means it will stay in motion, not slow down. I wouldn't try to distinguish between inertia of rest and inertia of motion. Best just to think of inertia.
He means the inverse, which is also true and is described by Newton's equations--Any object at rest will stay at rest unless acted upon by a different force. So the heavier object will have a harder time starting up than the lighter.

The thing here is that because of the same law he cited, the heavier object will take longer to stop than the lighter.

If all forces are equal and opposite, then the two parts of the law will equate with each other if the balls come back to rest at another place with the same inclination. Thus, the balls will stop at the same place.

The problem is, the OP hasn't given enough specifics in order for us to answer the question accurately.
BruceW
#13
Feb27-13, 03:31 PM
HW Helper
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P: 3,464
If we assume that as the ball is going down the slope, the energy lost to dissipation is negligible, then since [itex]I=2/5MR^2[/itex] and [itex]KE=1/2MI \omega^2[/itex], both balls will be at the same speed when they reach the bottom of the slope.

So now we need to decide what kind of resistance to motion acts on the balls once they are travelling horizontally along the flat part. If it is rolling resistance, then the resistance force is proportional to the normal force (and therefore mass) of the ball. So the power lost is proportional to mass, and the initial energy of the ball in the horizontal section is also proportional to the mass. Therefore, both balls stop after the same distance in this case.

We could instead assume that most of the dissipative resistance comes from air resistance. The most basic model of air resistance says that the resistive force is proportional to speed squared and not dependent on the mass of the object. So in this case, the denser ball will go ahead of the less dense ball, since it has greater inertia than the less dense ball.
deep838
#14
Feb28-13, 01:23 AM
P: 109
Quote Quote by BruceW View Post
The most basic model of air resistance says that the resistive force is proportional to speed squared and not dependent on the mass of the object. So in this case, the denser ball will go ahead of the less dense ball, since it has greater inertia than the less dense ball.
This statement of yours is self-contradicting!
First you say that Fair resistance[itex]\propto[/itex]v2 and independant of mass!
Then you say that the denser ball keeps going further... because of inertia,ie, beacause of mass??
Previously you said (and I agree) that both the balls have same speed when they reach the bottom.
So then Fair resistance should be equal for both of them.

One more thing, if we are considering air resistance, then we should also consider it while the balls are rolling down the incline, which nobody stated before! The result will be an unnecessary complexity to this simple "innocent" question.

Why go into such troubles? Fair resistance will be negligible if the balls are small and moving with relatively small speeds!

I believe that the only retardation that will make a difference is the frictional force from the ground. But since this Fground friction is dependant on weight an the co-eff. of ground friction, its simply Fground friction=μmg.
So the retardation is simply: [itex]\frac{F}{m}[/itex]=μg.
So it all comes down to whether or not they have same μ. If they do, they will definitely stop after the same distance. Else, the one with lesser μ will go further!
Don't hesitate to discard all this if I'm worng.
JustinRyan
#15
Feb28-13, 02:53 AM
P: 87
There is nothing contradictory there. The air resistance is not dependent on mass so each ball given they are the same size and have the same velocity will experience the same resistive force. If this is equal for both balls the effect will be greater for the lighter ball as this is dependant on mass.
I think that as you say the air resistance can be minimised as can the rolling resistance by having a smooth track and ball surface. The results of the experiment will depend on these conditions and which type of resistance is more significant. As has been stated already there has not been enough information given to determine the outcome.
deep838
#16
Feb28-13, 06:11 AM
P: 109
got it!
BruceW
#17
Feb28-13, 08:06 AM
HW Helper
BruceW's Avatar
P: 3,464
Quote Quote by deep838 View Post
This statement of yours is self-contradicting!
First you say that Fair resistance[itex]\propto[/itex]v2 and independant of mass!
Then you say that the denser ball keeps going further... because of inertia,ie, beacause of mass??
This is no contradiction. If both balls are moving at some speed, then the force due to air resistance is the same on both, therefore since the denser ball has greater inertia, air resistance will have less of an effect on it. And yes, the inertia will be the inertia of a sphere, which is proportional to the mass of the ball.

Quote Quote by deep383
Previously you said (and I agree) that both the balls have same speed when they reach the bottom.
So then Fair resistance should be equal for both of them.
Yeah, I said that both balls have same speed when they reach the bottom of the incline. So then when they travel along the horizontal part, the deceleration of the denser ball will be less, so the denser ball goes further on the horizontal part. (This is if we assume air resistance is most important).

Quote Quote by deep838
One more thing, if we are considering air resistance, then we should also consider it while the balls are rolling down the incline, which nobody stated before! The result will be an unnecessary complexity to this simple "innocent" question.
I made the assumption that the incline is short enough compared to the horizontal section, so that the speeds of the two balls when they get to the bottom of the incline is approximately the same. We could instead say that air resistance affects then on the incline as well. In that case, the denser ball would simply get ahead further. And so we get the same qualitative answer.


Quote Quote by deep838
I believe that the only retardation that will make a difference is the frictional force from the ground. But since this Fground friction is dependant on weight an the co-eff. of ground friction, its simply Fground friction=μmg.
So the retardation is simply: [itex]\frac{F}{m}[/itex]=μg.
So it all comes down to whether or not they have same μ. If they do, they will definitely stop after the same distance. Else, the one with lesser μ will go further!
Don't hesitate to discard all this if I'm worng.
Yeah, well that is one possible case. But we can also imagine a situation where the air resistance is more important. As others have said, we would need more information to be able to decide if air resistance or rolling resistance is more important.

Also, I should say that the retardation due to contact with the ground is called rolling resistance (not usually called frictional force from the ground). This is because the frictional force from the ground does not necessarily cause dissipation of energy. It is only when the ball is slipping with respect to the ground, that friction causes dissipation. There are also other effects which cause dissipation due to contact with the ground, which is why the term 'rolling resistance' is used. (wikipedia have a useful page about it).
xAxis
#18
Feb28-13, 08:09 AM
P: 211
JustinRyan:"..If this is equal for both balls the effect will be greater for the lighter ball as this is dependant on mass..."
What do you mean? What effect? Isn't the effect the air resistance? If it doesn't depend on mass, then the effect will be equal. But it does, the drag is dependent on the speed and density of object, and therefore on mass.


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