How long the ball remains in contact with the ground when dropped

In summary: This is not the right way to solve the problem .In summary, the conversation discusses a problem involving a ball dropped from a height and the estimation of its contact time with the ground. The ball is modeled as an ideal spring with a spring constant of 100Nm, and the gravitational acceleration is taken as 9.8m/s^2. The conversation also includes a discussion on the equation for the period of oscillation and the proof for it. It is concluded that the ball undergoes half an oscillation while in contact with the ground, but this approximation is not entirely accurate due to the ball's deformation when in contact with the ground. An alternative method for solving the problem is proposed, but it is not considered to be the correct
  • #1
Satvik Pandey
591
12

Homework Statement


A ball of mass 1kg is dropped from a height 1m . Estimate how long the ball is in contact with the ground in seconds ?

Details and assumption

Simplify the question by modeling the ball as an ideal spring of spring constant 100Nm .

Take gravitational acceleration if necessary as 9.8m/s2

Homework Equations


Work done by spring = 1/2kx^2.
Change in kinetic energy = Work done.


The Attempt at a Solution


Let the spring stops(ball) after it is bring displaced from Xm from its mean position.
Using the work energy theorem-
mgX-[itex]\frac{kx^{2}}{2}[/itex]=0-m/2*2gh
where m is the mass of spring.I got square of velocity before impact as 2gh(h is the height mentioned above).
So I got the displacement of spring from its mean position to stop.
I tried to use equations of motion but acceleration is not constant.So I tried to use calculus.
I got Acceleration as a function of x =g-kx/m .Could you please help me in solving this question further.
 
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  • #2
The answer is 0.341 seconds.It can be wrong.I am not sure about it.
 
  • #3
Satvik Pandey said:
The answer is 0.341 seconds.It can be wrong.I am not sure about it.

Is not it 0.314 s?

ehild
 
  • #4
Here's a hint: remember that the period of oscillation of a spring is [itex]2\pi \sqrt{\frac{m}{k}}[/itex] where [itex] k [/itex] is the spring constant and [itex] m [/itex] is the mass. See if you can use that fact to solve this problem.
 
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  • #5
ehild said:
Is not it 0.314 s?

ehild

Yes you are right sir.It was my mistake.
 
  • #6
ZetaOfThree said:
Here's a hint: remember that the period of oscillation of a spring is [itex]2\pi \sqrt{\frac{m}{k}}[/itex] where [itex] k [/itex] is the spring constant and [itex] m [/itex] is the mass. See if you can use that fact to solve this problem.
On putting values in this equation I got t=0.628 seconds.But this value is twice of the answer(0.314) above.
 
  • #7
Satvik Pandey said:
On putting values in this equation I got t=0.628 seconds.But this value is twice of the answer(0.314) above.

Think about the motion of the ball when it's in contact with the ground. It doesn't oscillate through a full period, does it?
 
  • #8
ZetaOfThree said:
Think about the motion of the ball when it's in contact with the ground. It doesn't oscillate through a full period, does it?
When the spring touches the ground it is then compressed but after some displacement from the mean position it is not possible to compress it more then spring relax itself.Is this not the full oscillation?
 
  • #9
Satvik Pandey said:
When the spring touches the ground it is then compressed but after some displacement from the mean position it is not possible to compress it more then spring relax itself.Is this not the full oscillation?

Not. Imagine a sine function. It starts from zero - goes to maximum- decreases to zero - turns to negative - reaches minimum-goes to zero. That is a full period. The ball can only be compressed, the ground can not stretch it. There is no "negative compression".

ehild
 
  • #10
ehild said:
Not. Imagine a sine function. It starts from zero - goes to maximum- decreases to zero - turns to negative - reaches minimum-goes to zero. That is a full period. The ball can only be compressed, the ground can not stretch it. There is no "negative compression".

ehild
Before colliding with the the ground spring was displaced 0cm from its mean position.It is then compressed and it get displaced by maximum displacement in negative direction.It then relax itself and reaches the mean position again.So this oscillation was from zero to minimum and again to zero.So this is only half oscillation.So the required time is half of the total time period.I understood this.Is this right?
 
  • #11
Yes, it is right.

ehild
 
  • #12
What is the proof of T=2∏√(m/k).
I have not studied Harmonic motion and oscillation yet.It would be nice if somebody could send me link from where I could study about this.Link of a video will be appreciated.
 
  • #13
You can find lot of videos and pages browsing "simple harmonic motion".

The equation for the time period T=2∏√(m/k) is equivalent to ω=√(k/m), as the angular frequency is ω=2π/T. The derivation of the equation for ω can be found here, for example.

Or see http://en.wikipedia.org/wiki/Simple_harmonic_motion

ehild
 
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  • #14
Thank you all of you for helping me.
 
  • #15
Funny things happen to a ball dropped on the ground. See . So the approximation that it is in contact with the ground for half period is very rude. The ball really vibrates like a spring when it is in the air again, but during the time it is in contact with the ground it gets very much deformed and it does not look like performing SHM.

ehild
 
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  • #16
ehild said:
Funny things happen to a ball dropped on the ground. See . So the approximation that it is in contact with the ground for half period is very rude. The ball really vibrates like a spring when it is in the air again, but during the time it is in contact with the ground it gets very much deformed and it does not look like performing SHM.

ehild


Thank you
 
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  • #17
Hello ehild

Somehow I am not convinced that the ball undergoes half the oscillation , for the reason that ball just touching the floor is not the equilibrium position . I tried a little differently .Not sure if I have got it right .

Let ##v_0## be the speed of the ball when it comes in contact with the ground . ##v_0 = \sqrt{2gh}## . x-axis points vertically downwards with x=0 is the location of CM when the ball first comes in contact with the floor.

Applying conservation of energy

$$\frac{1}{2}m{v_0}^2 = -mgx + \frac{1}{2}mv^2+ \frac{1}{2}kx^2$$ . Here 'x' is the displacement of the CM .

When the ball is fully compressed , v=0 and let x_0 be the maximum displacement of the CM .

$$\frac{1}{2}m{v_0}^2 = -mgx_0 + \frac{1}{2}kx_0^2$$

$$ x_0 = \frac{mg}{k}+ \sqrt{\frac{{v_0}^2mk}{k^2}+\frac{m^2g^2}{k^2}}$$

$$ x_0 - \frac{mg}{k} = \sqrt{\frac{{v_0}^2mk}{k^2}+\frac{m^2g^2}{k^2}}$$

From the first equation , $$ v = (\sqrt{\frac{k}{m}})\sqrt{\frac{{v_0}^2mk}{k^2}+\frac{m^2g^2}{k^2}-(x-\frac{mg}{k})^2}$$

$$ v = (\sqrt{\frac{k}{m}})\sqrt{(x_0-\frac{mg}{k})^2-(x-\frac{mg}{k})^2}$$

Now writing v=dx/dt and integrating with x(0→x_0) and t(0→T/2) doesn't give the time period as ##\frac{\pi}{10}## .
 
  • #18
Tanya Sharma said:
$$\frac{1}{2}m{v_0}^2 = -mgx_0 + \frac{1}{2}kx_0^2$$
.

This means my first step was right.I did it in the question (post#1).I learned it in doing 'Mechanics problem -- 2 masses & spring on a surface' from you and partially by sankalpmittal on this forum.
 
  • #19
Tanya Sharma said:
Hello ehild

Somehow I am not convinced that the ball undergoes half the oscillation , for the reason that ball just touching the floor is not the equilibrium position . I tried a little differently .Not sure if I have got it right .

Let ##v_0## be the speed of the ball when it comes in contact with the ground . ##v_0 = \sqrt{2gh}## . x-axis points vertically downwards with x=0 is the location of CM when the ball first comes in contact with the floor.

Applying conservation of energy

$$\frac{1}{2}m{v_0}^2 = -mgx + \frac{1}{2}mv^2+ \frac{1}{2}kx^2$$ . Here 'x' is the displacement of the CM . ......(1)

When the ball is fully compressed , v=0 and let x_0 be the maximum displacement of the CM .

$$\frac{1}{2}m{v_0}^2 = -mgx_0 + \frac{1}{2}kx_0^2$$

$$ x_0 = \frac{mg}{k}+ \sqrt{\frac{{v_0}^2mk}{k^2}+\frac{m^2g^2}{k^2}}$$

$$ x_0 - \frac{mg}{k} = \sqrt{\frac{{v_0}^2mk}{k^2}+\frac{m^2g^2}{k^2}}$$...(2)

From the first equation , $$ v = (\sqrt{\frac{k}{m}})\sqrt{\frac{{v_0}^2mk}{k^2}+\frac{m^2g^2}{k^2}-(x-\frac{mg}{k})^2}$$...(3)

$$ v = (\sqrt{\frac{k}{m}})\sqrt{(x_0-\frac{mg}{k})^2-(x-\frac{mg}{k})^2}$$

Now writing v=dx/dt and integrating with x(0→x_0) and t(0→T/2) doesn't give the time period as ##\frac{\pi}{10}## .
I understood it till till eq(2).How did you get eq(3).Have I mentioned the first equation correctly?Is this the equation you were talking about.How did you eliminate x[itex]_{0}[/itex] from eq(2) and get it in terms of v.I am really confused .
 
  • #20
Tanya Sharma said:
Hello ehild

Somehow I am not convinced that the ball undergoes half the oscillation , for the reason that ball just touching the floor is not the equilibrium position . I tried a little differently .Not sure if I have got it right .

Let ##v_0## be the speed of the ball when it comes in contact with the ground . ##v_0 = \sqrt{2gh}## . x-axis points vertically downwards with x=0 is the location of CM when the ball first comes in contact with the floor.

Applying conservation of energy

$$\frac{1}{2}m{v_0}^2 = -mgx + \frac{1}{2}mv^2+ \frac{1}{2}kx^2$$ . Here 'x' is the displacement of the CM .
And what is v?

Remember the problem says that the ball is like a spring. It is not a rigid body. Its parts move with different velocities with respect to the CM. The kinetic energy has contribution from that internal motion.

If the problem was " a small ball of mass m falls onto the top of a vertical spring from high h, how long is the ball in contact with the spring" then it could be solved by this method.

ehild
 
  • #21
ehild said:
And what is v?

:shy:

ehild said:
Remember the problem says that the ball is like a spring. It is not a rigid body. Its parts move with different velocities with respect to the CM. The kinetic energy has contribution from that internal motion.

Right.

Still I am not convinced that ball undergoes half the oscillation.

This may sound stupid but - do you believe if the ball is placed gently on the floor it completely maintains its shape ?
 
  • #22
Tanya Sharma said:
:shy:



Right.

Still I am not convinced that ball undergoes half the oscillation.

This may sound stupid but - do you believe if the ball is placed gently on the floor it completely maintains its shape ?

I am even convinced that the ball does not perform oscillation (SHM) during the impact. It can do after leaving the ground as shown in the video (post #13). And the time it is in contact with the ground is certainly not exactly half of the oscillation period, but can not be calculated with simple methods.

The ball, placed on the ground changes its shape, but not uniformly. If you model it with a spring of the same mass, the upper parts are less compressed than the bottom parts, as they have to balance more weight. See this: . The upper part of the ball is almost unchanged, while the bottom part is heavily distorted.

I think, it is a badly worded problem again. The usual problems sound as I have written before: a point-like rigid ball of mass m falls onto the top of a vertical spring. How long is it in contact with the spring? That problem is solvable, and the time is not half of the oscillation period -longer than that. It would be useful to solve that problem with the given data.

ehild
 
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  • #23
ehild said:
The usual problems sound as I have written before: a point-like rigid ball of mass m falls onto the top of a vertical spring. How long is it in contact with the spring? That problem is solvable, and the time is not half of the oscillation period -longer than that. It would be useful to solve that problem with the given data.

ehild

I agree :smile:
 
  • #24
I try to make sense to the problem: Assuming the ball is of radius R and mass 1kg; dropped from high H=1 m so as its bottom is at high H above the ground; and the vertical diameter deforms (decreases) by x during the impact, otherwise the ball stays homogeneous, only the shape changes; and the deformation performs SHM about the equilibrium deformation with angular frequency ω=√(k/m)=10s-1: x=Asin(ωt+θ)+Xe.
Xe=mg/k = 0.098 m.

See figure.

Assume conservation of energy. The CM is at hight H+R above the ground. At full compression, the CM is at hight h and it is in rest. The vertical diameter is shortened by X, h=R-0.5X.
Write up the equation for conservation of energy and find X. Well, the ball is not small if it can be deformed by that much!

x performs SHM about the equilibrium compression: x=Xe+Asin(ωt+θ).
At t =0 the ball just touches the ground, x=0: Asin(θ) + Xe=0.
At the deepest position, the compression is maximum, ωt+θ=pi/2. x=X, so A=X-Xe. Knowing A, you can find θ and then t.
The time at full compression is T=(pi/2-θ)/ω . The same time is needed to go back to the original state.


ehild
 

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  • #25
ehild said:
Xe=mg/k = 0.098 m.

Why have you not considered normal force from ground ?
 
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  • #26
How would you do it?

ehild
 
  • #27
Interesting .

Do you mean normal from ground is equal to kX_e ?
 
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  • #28
Tanya Sharma said:
Interesting .

Do you mean normal from ground is equal to kX_e ?

I do not mean anything. :devil: I just would like to know your opinion. I want to learn how to include normal force. :-p

ehild
 
  • #29
Hmmm... I can see you are enjoying pulling my leg :cool:.

But I think normal equal to kx makes sense as kx is the force with which the spring pushes the ground .So you were right in your assessment :smile:
 
  • #30
Tanya Sharma said:
Hmmm... I can see you are enjoying pulling my leg :cool:.

Of course :-p
Tanya Sharma said:
But I think normal equal to kx makes sense as kx is the force with which the spring pushes the ground .So you were right in your assessment :smile:

Add to it that the ball (or spring) is in equilibrium. The force of gravity and the normal force balance each other. mg-N=0.

The interaction between the ground and spring is the normal force from the ground and the spring force by the spring. They are of equal magnitude and of opposite direction. So N=kXe.

ehild
 

Related to How long the ball remains in contact with the ground when dropped

1. How does the height of the drop affect the time the ball remains in contact with the ground?

The height of the drop does not affect the time the ball remains in contact with the ground. This is because the time it takes for an object to fall to the ground is determined by the force of gravity and the object's initial velocity, not the height of the drop.

2. Does the weight of the ball affect the time it remains in contact with the ground when dropped?

No, the weight of the ball does not affect the time it remains in contact with the ground. The time it takes for an object to fall to the ground is determined by the force of gravity and the object's initial velocity, not its weight.

3. What is the relationship between the time the ball remains in contact with the ground and the surface it is dropped on?

The surface the ball is dropped on does not affect the time it remains in contact with the ground. As long as the surface is flat and solid, the time will remain the same. However, if the surface is soft or uneven, it may affect the bounce of the ball after it leaves the ground.

4. Can the shape of the ball affect the time it remains in contact with the ground when dropped?

The shape of the ball does not affect the time it remains in contact with the ground. As long as the ball is round and symmetrical, the time will remain the same. However, if the ball is oblong or has uneven weight distribution, it may affect the bounce of the ball after it leaves the ground.

5. Is there a maximum or minimum time the ball can remain in contact with the ground when dropped?

Yes, there is a minimum time the ball can remain in contact with the ground when dropped. This is determined by the force of gravity and the object's initial velocity. However, there is no maximum time as the ball can continue bouncing indefinitely if there is no external force acting on it.

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