Solid ball dropped on ground again

In summary, the conversation discusses a problem involving a ball of radius R spinning at an angular speed ω and then being lowered onto a horizontal surface with a coefficient of kinetic friction μ. The goal is to find the linear speed of the ball and the distance it travels before achieving pure rolling motion. The discussion touches on the use of conservation of angular momentum, torque, and energy to solve the problem, with the conclusion that a combination of Newton's 2nd law and kinematics is the most suitable method.
  • #1
vladimir69
130
0

Homework Statement


I got struck with this problem again in the review questions and still can't do it, I think this problem shows a major gap in my knowledge of the subject.

A solid ball of radius R is set spinning with angular speed [tex]\omega[/tex] about a horizontal axis. The ball is then lowered vertically with negligible speed until it just touches a horizontal surface and is released. If the coefficient of kinetic friction between the ball and the surface is [tex]\mu[/tex], find the linear speed of the ball once it achieves pure rolling motion and the distance it travels before its motion is pure rolling.


Homework Equations


[tex]W=\Delta K[/tex]

[tex]I_0\omega_0=I\omega[/tex]

[tex]I=\frac{2}{5}MR^2[/tex]

[tex]v=R\omega[/tex]

The Attempt at a Solution


Now i don't think i can use conservation of angular momentum because the external force of friction is acting on the ball, which brings me to another question...how can one use conservation of mechanical energy for a ball rolling down an incline? ie mgh=1/2 mv^2 + 1/2I[itex]\omega ^2[/itex]

To solve this problem i would probably say
[tex]-\frac{1}{2}I\omega^2+\frac{1}{2}I\omega_f ^2 + \frac{1}{2}MR^2\omega_f ^2=F_fs[/tex]
but i don't think i can use [tex]F_f = \mu Mg[/tex]
and i am not given how far the ball slips before its in pure rolling motion (actually its asked int he question to find that) so i don't think i have enough information to solve this problem.
 
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  • #2
Forget energy conservation. Just fall into momentum conservation.

The torque about the point of the ball which touches the gorund at a given instant is zero. So you can apply momentum conservation. For the case initial momentum is just due to the rolling but the final momentum is rolling plus translational. And in the final momentum term do not forget to replace v by wr.
 
  • #3
i am a bit confused
in this problem
https://www.physicsforums.com/showthread.php?t=301129
we had a cylinder rolling down a hill and it had friction working on it
other times we don't seem to pop in friction to the equations
whats the deal with that?
 
  • #4
Friction is inevitable. It always acts.

But the torque bu the friction about the point of contact is

Torque=(Friction)(Distance)

The distance is zero
Hence torque is zero
 
  • #5
vladimir69 said:
Now i don't think i can use conservation of angular momentum because the external force of friction is acting on the ball,
Certainly the angular momentum about the center of mass is not conserved, but you can apply conservation of angular momentum about a fixed point on the surface. In that case the torque due the external force is zero.

which brings me to another question...how can one use conservation of mechanical energy for a ball rolling down an incline? ie mgh=1/2 mv^2 + 1/2I[itex]\omega ^2[/itex]
In that kind of problem the friction is static friction and does no work. So energy is conserved.

To solve this problem i would probably say
[tex]-\frac{1}{2}I\omega^2+\frac{1}{2}I\omega_f ^2 + \frac{1}{2}MR^2\omega_f ^2=F_fs[/tex]
The problem with this is that the ball both rolls and slips, so defining "s" is tricky. (It's not just the displacement of the ball's center.)
but i don't think i can use [tex]F_f = \mu Mg[/tex]
Why not?
and i am not given how far the ball slips before its in pure rolling motion (actually its asked int he question to find that) so i don't think i have enough information to solve this problem.
Either use conservation of angular momentum or basic dynamics/kinematics. There's a force on the ball: Apply Newton's 2nd law for rotation and translation, then solve for the point where the condition for rolling without slipping is met.
 
  • #6
doc al, you have enlightened me somewhat but i have been thinking about your post for about 20 mins now and not sure about "conservation of angular momentum about a fixed point on the surface" and also the last bit about applying Newtons 2nd law for rotation and translation.
i tried to write a single equation down but nothing springs to mind

if the question said static friction instead of kinetic friction would the conservation of energy equation i used apply?
 
  • #7
vladimir69 said:
doc al, you have enlightened me somewhat but i have been thinking about your post for about 20 mins now and not sure about "conservation of angular momentum about a fixed point on the surface"
If you take some point along the surface as your origin, then angular momentum about that point will be conserved. (The friction force exerts no torque about that point, since its line of action will pass through that origin.) But you don't need to use this method, if it's not clear to you.

and also the last bit about applying Newtons 2nd law for rotation and translation.
i tried to write a single equation down but nothing springs to mind
You need two equations: One for rotation; one for translation. Just apply Newton's 2nd law to find the accelerations. Then use a bit of kinematics. The center of the ball speeds up from v = 0 to some final speed vf, while the angular speed slows down from the initial ω to some final angular speed ωf. The final translational and rotational speeds must satisfy the condition for rolling without slipping. (Hint: Solve for the time it takes to reach that final speed.)

if the question said static friction instead of kinetic friction would the conservation of energy equation i used apply?
If the friction were static, no work would be done and energy would be conserved.
 
  • #8
thanks for the help doc al, greatly appreciated
i will have to do some thinking about this one and a bit of reviewing
 

1. What happens when a solid ball is dropped on the ground?

When a solid ball is dropped on the ground, it will fall due to the force of gravity. As it falls, it gains speed until it reaches its maximum velocity. Upon hitting the ground, the ball will experience a sudden change in momentum, causing it to bounce back up.

2. Does the height from which the ball is dropped affect its bounce?

Yes, the height from which the ball is dropped does affect its bounce. The higher the drop, the greater the ball's initial potential energy and the higher it will bounce. This is because the ball has more potential energy to convert into kinetic energy as it falls.

3. Why does the ball bounce back up after hitting the ground?

The ball bounces back up after hitting the ground due to the conservation of energy. When the ball hits the ground, some of its kinetic energy is transferred into the ground, but the remaining energy is converted back into potential energy, causing it to bounce back up.

4. How does the surface on which the ball is dropped affect its bounce?

The surface on which the ball is dropped can affect its bounce. A harder surface, such as concrete, will cause the ball to bounce higher compared to a softer surface, such as grass. This is because the harder surface absorbs less of the ball's energy, resulting in a greater bounce.

5. Will the ball bounce the same way every time it is dropped on the ground?

No, the ball will not bounce the same way every time it is dropped on the ground. Factors such as the height of the drop, the surface it is dropped on, and the angle at which it hits the ground can all affect the ball's bounce. Additionally, the ball may lose some energy with each bounce, causing it to bounce less and less each time.

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