What Angle Should the Ball Strike to Initiate Pure Rolling in the Sphere?

[Satvik Pandey]

Homework Statement

A large sphere rests on a rough horizontal ground,

A ball of mass 'm' is thrown towards it along one of its radius at a speed 'u' , along an angle 'x' with the vertical

Find the angle with the vertical at which the ball should strike (find x) so that the large sphere pure rolls immediately after collision.

coefficient of friction=1

mass of large sphere = 1

radius of large sphere = sqroot(0.4)

Homework Equations

The Attempt at a Solution

During collision impulse will [/B]act in direction perpendicular to the contact surface. As the contact surface is the tangent of the sphere so the acting impulse will pass through the center of the sphere.

As the sphere do not jumps after the collision so $J_{n}+mg=Jcos(x)$. Now mg is very small as compared to these impulses so I can assume

$J_{n}=Jcos(x)$...(1)

Let $v$ be the velocity of the com of the sphere after collision so

$Jsinx-Jcosx=mv$ (value of $\mu$ is 1)...(3)

Angular impulse will only be provided by the friction force.

So $Jcos(x)R=I_{CoM}\omega$...(4)

As the sphere rolls after collision so $v=R\omega$

Solving these equations gives wrong answer.Are my equations correct?

 

[haruspex]

Homework Statement

A large sphere rests on a rough horizontal ground,

A ball of mass 'm' is thrown towards it along one of its radius at a speed 'u' , along an angle 'x' with the vertical

Find the angle with the vertical at which the ball should strike (find x) so that the large sphere pure rolls immediately after collision.

coefficient of friction=1

mass of large sphere = 1

radius of large sphere = sqroot(0.4)

Homework Equations

The Attempt at a Solution

During collision impulse will [/B]act in direction perpendicular to the contact surface. As the contact surface is the tangent of the sphere so the acting impulse will pass through the center of the sphere.

View attachment 78755
As the sphere do not jumps after the collision so $J_{n}+mg=Jcos(x)$. Now mg is very small as compared to these impulses so I can assume

$J_{n}=Jcos(x)$...(1)

Let $v$ be the velocity of the com of the sphere after collision so

$Jsinx-Jcosx=mv$ (value of $\mu$ is 1)...(3)

Angular impulse will only be provided by the friction force.

So $Jcos(x)R=I_{CoM}\omega$...(4)

As the sphere rolls after collision so $v=R\omega$

Solving these equations gives wrong answer.Are my equations correct?

Other than that you seem to have switched which mass m refers to (but the masses seem irrelevant), all looks correct to me. Your equations give me $\arcsin(\frac{7}{\sqrt{53}})$.

 

[Satvik Pandey]

Other than that you seem to have switched which mass m refers to (but the masses seem irrelevant), all looks correct to me. Your equations give me $\arcsin(\frac{7}{\sqrt{53}})$.

Yes that gives $x=74.05$.
Please clear my doubt. I have considered here the limiting value of the friction(static) but it is not given anywhere in the question to use its limiting value.

 

[haruspex]

Yes that gives $x=74.05$.
Please clear my doubt. I have considered here the limiting value of the friction(static) but it is not given anywhere in the question to use its limiting value.

Yes, good point. But there's no other way to get answer, is there?

 

[Satvik Pandey]

Yes, good point. But there's no other way to get answer, is there?

Thanks for the help!

I was wondering for what value of $x$ the ball will bounce. For that $J_{n}$ should be greater than $Jcos(x)$. But I am not finding any way to go ahead.

 

[haruspex]

Thanks for the help!

I was wondering for what value of $x$ the ball will bounce. For that $J_{n}$ should be greater than $Jcos(x)$. But I am not finding any way to go ahead.

To get bounce, there'd have to be some coefficient of restitution. We've no information on that. And it would not require x to be below any particular value.

 

[Satvik Pandey]

To get bounce, there'd have to be some coefficient of restitution. We've no information on that. And it would not require x to be below any particular value.

If $e>0$ then will it bounce for any value of J and x?

 

[haruspex]

If $e>0$ then will it bounce for any value of J and x?

As long as there is a downward component to the impulse, if e> 0 then it should bounce. But what exactly happens will also depend on c.o.r. between the two balls.

 

[Satvik Pandey]

As long as there is a downward component to the impulse, if e> 0 then it should bounce. But what exactly happens will also depend on c.o.r. between the two balls.

Thank you!:)

 

[ehild]

As long as there is a downward component to the impulse, if e> 0 then it should bounce. But what exactly happens will also depend on c.o.r. between the two balls.

If the ball arrives from above, the sphere will not bounce.

 

[haruspex]

If the ball arrives from above, the sphere will not bounce.

By what reasoning?

 

[ehild]

In the scenario Satvik has shown, the sphere will certainly bounce if the average (upward) force during the collision is greater than Mg. In this case, friction does not act during the collision.
If the force of impact is less than Mg, gravity has to be taken into account when calculating the force of friction.
If the ball arrives from above, it transfers downward vertical component to the sphere. Assuming the sphere rigid, the CM will stay in rest if the normal force from the ground cancels the impulsive force. That can happen.
In real life, the sphere can be elastic, so hitting it initiates a mechanical wave, and the wave reaching the ground, can make the sphere bounce. In this case, we have a collision among three bodies, and the outcome depends on the time between the collision with the ball and interaction with the ground, and also on the process of dissipating energy.

 

[Satvik Pandey]

In the scenario Satvik has shown, the sphere will certainly bounce if the average (upward) force during the collision is greater than Mg. In this case, friction does not act during the collision.
If the force of impact is less than Mg, gravity has to be taken into account when calculating the force of friction.

For ball to bounce after collision $J_{n}>mg+Jcos(x)$. Why did you neglect Jcos(x)?

In this case, friction does not act during the collision.
If the force of impact is less than Mg, gravity has to be taken into account when calculating the force of friction.

As long as there is interaction between the body the friction will act to oppose the motion of the sphere. $Jsin(x)$ would tend to bring the ball in motion so friction would act to oppose it. Why would friction not act in this condition?

In real life, the sphere can be elastic, so hitting it initiates a mechanical wave, and the wave reaching the ground, can make the sphere bounce. In this case, we have a collision among three bodies, and the outcome depends on the time between the collision with the ball and interaction with the ground, and also on the process of dissipating energy.

That sounds too complicated.

 

[ehild]

I understood your picture that the ball is thrown from below towards the sphere.
Make it clear please: What are J and Jn?

If the sphere bounces during the time of collision with the ball, it is in the air and is not in contact with the ground.

 

[Satvik Pandey]

I understood your picture that the ball is thrown from below towards the sphere.
Make it clear please: What are J and Jn?

If the sphere bounces during the time of collision with the ball, it is in the air and is not in contact with the ground.

J is the impulse due to collision of the ball and the sphere and Jn is the impulse which acts on the sphere by the ground.

I am bit confused. First the ball collides with the sphere and then after infinitely small time period the sphere collides with the ground. Then only impulse which is responsible for the bouncing is Jn. right? So the ball would bounce when both collisions have happened.

 

[BvU]

As the sphere do not jumps after the collision so $J_n +mg=J\cos(x)$. Now mg is very small as compared to these impulses so I can assume ...

I am surprised at this. This dimensional inconsistency returns in post #13. $mg\Delta t$ is the term I expected to see, twice.
The reaction acts at the point where the sphere touches the ground; the cause doesn't. Sum is zero, but there is an angular impulse.
And yes, it can be ignored: m doesn't lie about at the impact point for a long time :)
And no, the sphere doesn't jump: the terminology "pure rolling" excludes that.

The drawing in post #1 let's m come in from the upper right (Ehilds drawing from upper left). The result (74 degrees) doesn't indicate otherwise than that.

What happens to the vertical momentum is slightly mysterious; for me the "rough ground" indicates that it's absorbed.
Having to assume that complicates the exercise in my view: horizontal and vertical asymmetry and then still ask for pure rolling.

 

[ehild]

I think, it is too much for both of us to consider the sphere elastic. Remember, the impulse belongs to some time interval. The transfer of impulse can happen in different times at different places of the big sphere. If the impulse is J during the collision with the ball you do not know what is transferred to the ground some time interval later.
The sphere will rise up if it has upward acceleration. You need an upward resultant force. It can happen when the elastic wave arrives at the contact point at the ground and pushes the ground, and it pushes back with force higher than gravity. But that is complicated, and you have to work with rigid bodies.

Imagine a solid sphere of weight 50 N and you put a 10 N stone on the top of it. What force acts from the sphere to the ground? And what is the normal force? The force the ground exerts on the sphere? Is it not 60 N? Can it be more? Why?
When we use impulse we assume a force acting in a finite time interval, and impulse is the time integral of the force. But the acceleration is proportional to the force. What is the net force on the sphere? Will it jump when you put a stone on it?
You remove the stone. If the sphere is elastic, the weight of the stone deforms it, and removing the stone, the deformation is released The CM will rise.
In case of rigid bodies, the forces act immediately and the deformations are infinitesimally small. You can shift the forces along their line of application: If there is a force exerted on the top, or in the CM, those forces act even at the bottom. The normal force is equal and opposite to the sum of them.

 

[ehild]

What happens to the vertical momentum is slightly mysterious; for me the "rough ground" indicates that it's absorbed.
Having to assume that complicates the exercise in my view: horizontal and vertical asymmetry and then still ask for pure rolling.

The vertical momentum is transferred to the (infinite) ground, like in the case of collision with a wall. It does not matter if it is rough or not.
I prefer thinking in terms of forces. That downward impulse means downward force, which adds to mg and determines the normal force when calculating friction.
In the collision process, the CM of the sphere is accelerated by the horizontal component of the force of interaction with the ball, and its rotation is accelerated by the torque of the force of friction. At the end of the collision, the speed of the CM and the angular velocity meet the condition of pure rolling.

 

[Satvik Pandey]

Imagine a solid sphere of weight 50 N and you put a 10 N stone on the top of it. What force acts from the sphere to the ground? And what is the normal force? The force the ground exerts on the sphere? Is it not 60 N? Can it be more? Why?
When we use impulse we assume a force acting in a finite time interval, and impulse is the time integral of the force. But the acceleration is proportional to the force. What is the net force on the sphere? Will it jump when you put a stone on it?
You remove the stone. If the sphere is elastic, the weight of the stone deforms it, and removing the stone, the deformation is released The CM will rise.

Yes 60N will acts on the sphere by the ground. Net force on the sphere will be 0 because 10N is acting in he top of the sphere.

I was thinking that why ball bounces when left at some height. I came up with this--
When the ball collides with the ground it's momentum changes. Due to this drastic change in momentum a large amount of force acts on the ball (>mg) and this accelerates it upward.

But when I look at this situation. The impulse (Jn) acts on the sphere due to Jcos(x). By 3rd law the force acting on the ground by the sphere is equal to the force acting on the sphere by the ground. So the sphere can never apply a force greater than mg+Jcos(x) on the ground. So Jn can never be greater than mg+Jcos(x). So this means that sphere can never bounce if a ball is dropped on it from the above. Am I correct?

 

[ehild]

Yes 60N will acts on the sphere by the ground. Net force on the sphere will be 0 because 10N is acting in he top of the sphere.

I was thinking that why ball bounces when left at some height. I came up with this--
When the ball collides with the ground it's momentum changes. Due to this drastic change in momentum a large amount of force acts on the ball (>mg) and this accelerates it upward.

See the video. The change of momentum happens sooner - it deforms the ball, it is like a spring. The spring is compressed while the CM of the ball gets close to the ground, the compressed spring pushes the ground the ground pushes back and the ball rises and the compression releases.

But when I look at this situation. The impulse (Jn) acts on the sphere due to Jcos(x). By 3rd law the force acting on the ground by the sphere is equal to the force acting on the sphere by the ground. So the sphere can never apply a force greater than mg+Jcos(x) on the ground. So Jn can never be greater than mg+Jcos(x). So this means that sphere can never bounce if a ball is dropped on it from the above. Am I correct?

If the sphere is rigid, the force acts everywhere immediately, and the reaction force can not be greater than the force itself.
If some time elapses till the effect of the ball hitting the sphere arrives to the contact point with the ground, you do not know what force acts on the ground at a certain time instant. The reaction force is also instantaneous, it can happen that it is greater than mg at an instant.
Do not forget that the impulse refers to a time interval, during the time two bodies are in contact. It is the time integral of the force. But the acceleration is determined by the force.
In this problem, it was said that the collision ends with the sphere rolling. So it stays in contact with the ground.

Try to do some experiments with two balls - hit a ball on the ground with an other one and watch if it bounces up from the ground. Or just hit it with your fist.

 

[Satvik Pandey]

Thanks ehild. That video cleared a lot of things. So the elastic ball can bounce because of the difference in time of the effect of collision at the top and at the point at which it is contact with the ground.

 

[ehild]

Remember, the bouncing is caused by upward force. The momentum the sphere transfers to the ground can not be bigger than the momentum the sphere got from the ball. But you do not know anything about the force of interaction between the sphere and the ground.

 

[Satvik Pandey]

Remember, the bouncing is caused by upward force. The momentum the sphere transfers to the ground can not be bigger than the momentum the sphere got from the ball. But you do not know anything about the force of interaction between the sphere and the ground.

I got that. Thank you.

 

[kuruman]

Yes, good point. But there's no other way to get answer, is there?

Yes there is, if one solves the problem considering angular momentum transfer about the point of contact with the ground.

About the point of contact: The angular impulse delivered to the sphere is the horizontal component of the linear impulse multiplied by the lever arm = [J sin(x)] R [1+ cos(x)]. After the collision, the angular momentum of the rolling sphere is I_Pω where I_P = (7/5)MR² and, of course, ω = V/R. The linear momentum transfer equation J sin(x) = M V can be used to find cos(x) = 2/5. It looks too simple to be true, but I can't find a hole in the argument.

 

[haruspex]

Yes there is, if one solves the problem considering angular momentum transfer about the point of contact with the ground.

About the point of contact: The angular impulse delivered to the sphere is the horizontal component of the linear impulse multiplied by the lever arm = [J sin(x)] R [1+ cos(x)]. After the collision, the angular momentum of the rolling sphere is I_Pω where I_P = (7/5)MR² and, of course, ω = V/R. The linear momentum transfer equation J sin(x) = M V can be used to find cos(x) = 2/5. It looks too simple to be true, but I can't find a hole in the argument.

Two problems with that.
Your calculation of the moment of J about the point of contact with the ground is wrong. If you consider it as applied at point of impact, resolving it into horizontal and vertical components there, you get the expression you give from the horizontal component, but there will be an opposing moment, JR sin(x)cos(x), from the vertical component. More simply, consider the distance from the point of contact with the ground to the line of action of the impulse. You now see that the moment is JR sin(x).
In your analysis of horizontal linear momentum, you have ignored impulsive friction.