How can a ball rise higher than originally dropped?

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In summary, the conversation discusses the effects of friction on a ball being thrown on the floor. It is noted that the presence of friction does not increase the normal force, but rather affects the bounce angle. This can be achieved by either reducing horizontal velocity or increasing vertical velocity. The question is raised whether a friction-ball can jump higher than a frictionless ball, and it is argued that this is possible due to conservation of angular and linear momentum.
  • #1
FallenApple
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So here is a case where a ball is thrown on the floor with no friction.

NoFric.png
Then is a case with friction
WithFriction.png
Now what I don't understand is how is the presence of friction going to make the normal force larger than usual. The friction is dependent on the normal force, not the other way around. And the normal force is dependent on the vertical impulse, which should the same as the previous case, since the CM of the ball has the same final vertical velocity before impact.
 
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  • #2
Well there is angular momentum, the ball when it hits a surface with friction will convert angular momentum into deformation so that would account for an additional boost.
 
  • #3
FallenApple said:
Now what I don't understand is how is the presence of friction going to make the normal force larger than usual. The friction is dependent on the normal force, not the other way around. And the normal force is dependent on the vertical impulse, which should the same as the previous case, since the CM of the ball has the same final vertical velocity before impact.
You are correct. The normal force would not be larger than normal. But I think that you are misinterpreting the drawing. It is not intended to show that the bounce is reaches a higher maximum height than normal. It is intended to show that the bounce angle is higher than normal.

You can achieve a higher bounce angle in either of two ways: By reducing the horizontal velocity or by increasing the vertical velocity. The text speaks of the former and not the latter.
 
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  • #4
jbriggs444 said:
You are correct. The normal force would not be larger than normal. But I think that you are misinterpreting the drawing. It is not intended to show that the bounce is reaches a higher maximum height than normal. It is intended to show that the bounce angle is higher than normal.

You can achieve a higher bounce angle in either of two ways: By reducing the horizontal velocity or by increasing the vertical velocity. The text speaks of the former and not the latter.

Oh ok. That makes sense. The height must be the same, so the reduced distance(due to friction) traveled in the x direction would cause the hypotenuse to be angled higher.
 
  • #5
jbriggs444 said:
You are correct. The normal force would not be larger than normal. But I think that you are misinterpreting the drawing. It is not intended to show that the bounce is reaches a higher maximum height than normal. It is intended to show that the bounce angle is higher than normal.

You can achieve a higher bounce angle in either of two ways: By reducing the horizontal velocity or by increasing the vertical velocity. The text speaks of the former and not the latter.

The question remains if the friction-ball can actually jump higher (achieve higher vertical velocity) than the frictionless ball. Without initial rotation, the friction-ball loses linear KE to rotational KE, so the final linear speed is lower than for the frictionless ball. Can the vertical velocity still be higher for the friction-ball, due to the steeper angle? Consider the extreme case of a very flat initial angle.
 
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  • #6
A.T. said:
Can the vertical velocity still be higher for the friction-ball, due to the steeper angle?
I believe so - no conservation law is violated if you include the Earth so that the system is properly closed with regard to conservation of angular and linear momentum.
 

Related to How can a ball rise higher than originally dropped?

1. How does the height of the drop affect the height of the bounce?

The height of the bounce is directly related to the height of the drop. The higher the ball is dropped from, the higher it will bounce back up. This is due to the law of conservation of energy, which states that energy cannot be created or destroyed but can only be transferred. When the ball is dropped from a higher height, it gains more potential energy, which is then converted into kinetic energy during the bounce, resulting in a higher bounce.

2. Does the material of the ball affect its bounce height?

Yes, the material of the ball does affect its bounce height. Different materials have different elasticity, which is the ability to deform and regain their original shape. A ball made of a more elastic material, such as rubber, will have a higher bounce than a ball made of a less elastic material, such as clay.

3. Can air resistance affect the height of the bounce?

Yes, air resistance can affect the height of the bounce. When a ball is dropped, it experiences air resistance as it falls, which slows it down. This means that it has less kinetic energy when it reaches the ground, resulting in a lower bounce. However, the effect of air resistance is usually minimal and can be ignored for most practical purposes.

4. How does the angle of the drop affect the bounce height?

The angle of the drop does not have a significant effect on the bounce height. As long as the ball is dropped from the same height, it will bounce to the same height regardless of the angle of the drop. However, if the ball is thrown at an angle, it will travel a longer horizontal distance before bouncing back, resulting in a lower bounce height.

5. Is the gravitational pull of the Earth a factor in the bounce height?

Yes, the gravitational pull of the Earth is a major factor in the bounce height. The acceleration due to gravity, which is 9.8 m/s² on Earth, determines how fast the ball will fall and how much kinetic energy it will have upon impact. The higher the acceleration due to gravity, the higher the bounce height will be.

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