# Sphere rolling problem

1. Nov 25, 2013

### Saitama

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
I have some difficulty understanding the problem statement. The problem says the paper moves horizontally but then in the very next line, it states the velocity is perpendicular to the initial velocity of sphere.

I can't visualize the motion of sphere. How to form the equations?

Any help is appreciated. Thanks!

#### Attached Files:

• ###### sphere rolling.png
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2. Nov 25, 2013

### TSny

Let the table be the x-y plane and let the initial velocity of the ball be along the x axis. The paper is then suddenly pulled in the y direction.

3. Nov 26, 2013

### Saitama

Hi TSny!

I drew a sketch, is it correct?

How to form the equations?

Thanks!

#### Attached Files:

• ###### sphere on paper.png
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4. Nov 26, 2013

### voko

There are two horizontal directions. The ball is rolling along one of those, the paper is moved along the other

5. Nov 26, 2013

### TSny

The paper slides along the table in a direction perpendicular to the initial velocity of the ball. In your diagram, you would pull the paper either towards you (out of the screen) or away from you (into the screen).

6. Nov 26, 2013

### Saitama

Okay, I understand the situation, thanks both of you but how to make the equations?

7. Nov 26, 2013

### Saitama

I thought some more upon the problem. Please look at the attachment.

The green colour denotes the paper and the purple for sphere.

The $v_i$ remains unchanged as there is no force in that direction. The friction acts in the direction of paper's velocity.

Now I am confused at one point. I can easily find the velocity of sphere in the direction of paper's motion as $f=\mu_k mg$. Also, there will be angular velocity generated in a direction different to the initial one, so the final angular velocity is obtained by the resultant of two angular velocities. But then where do I need the coefficient of static friction ($\mu_s$)?

Am I heading in the correct direction?

#### Attached Files:

• ###### top view.png
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8. Nov 26, 2013

### voko

I agree with the outline of the solution. I do not think that static friction could play a role in this problem.

9. Nov 26, 2013

### TSny

Maybe you won't need $\mu_s$. You'll have to work out the problem to see.

I think you are heading in the right direction.

10. Nov 26, 2013

### Saitama

Thanks a lot both of you.

Velocity after two direction in the motion of paper is $\mu_k gt=0.8g$.
Hence, resultant velocity is $\sqrt{v_i^2+(0.8g)^2}$

Similarly, I found the final (resultant) angular velocity to be
$$\sqrt{\left(\frac{v_i}{R}\right)^2+\left(\frac{2g}{R}\right)^2}$$

Should I use conservation of angular momentum now?

11. Nov 26, 2013

### TSny

You can't assume that the friction force will last for 2.0 seconds. Can you see the condition that will determine when the friction force stops?

12. Nov 26, 2013

### Saitama

When the pure rolling starts. Then what is the implication of 2 seconds in the problem?

13. Nov 26, 2013

### voko

You should realize that as the paper stops, the sphere's velocity w.r.t. the paper changes abruptly again.

14. Nov 26, 2013

### Saitama

I am not sure but do you mean that the frictional force reverses its direction after two seconds?

15. Nov 26, 2013

### TSny

You will need to find the time the friction must act to produce rolling without slipping on the paper and make sure it's less than 2 seconds.

16. Nov 26, 2013

### haruspex

You can break the problem into two parts: what is the velocity as it leaves the paper, then what is the final velocity. The second part is a standard question I'm sure you've seen before; it turns out to be independent of the coefficient of friction there (as long as it's not zero).

Conservation of momentum (linear and angular) is definitely the way to go. You can calculate how long before the sphere is rolling on the paper. While still sliding on the paper what is the torque? If it slides for time Ts, what angular velocity will it then have perpendicular to the movement of the paper (i.e perpendicular to its original axis of rotation)? What linear velocity (in the ground frame) in the direction of movement of the paper? What velocity in that direction relative to the paper? If it now starts rolling, what is the relationship between the angular and linear velocities?
Note that you can do all this in regard to the one spatial dimension, so no need to get tangled up with Pythagoras.

Last edited: Nov 26, 2013
17. Nov 27, 2013

### Saitama

Here's what I think:

The velocity in the direction of motion of paper at any time t is $\mu_k gt$.

The angular velocity is $\mu_k mgR/(2/5 mR^2) t=(5/2)\mu_k gt/R$.

When pure rolling starts, we have the condition that $v+r\omega=1$
$$\mu_k gt+\frac{5}{2}\mu_k gt=1 \Rightarrow t=0.07288 \,\, \text{seconds}$$

This is less than 2 seconds so the pure rolling starts before the paper stops.

For the time calculated above, the velocity is $(2/7) m/s$ and angular velocity is $(5/(7R)) rad/s$.

When the paper stops, we apply conservation of angular momentum,
$$mvr-I\omega=mv'R+I\omega \Rightarrow \frac{2}{7}mR-\frac{2}{5}mR^2\frac{5}{7R}=mv'R+\frac{2}{5}mR^2\frac{v'}{R}$$
Solving for v' gives zero.

18. Nov 27, 2013

### nil1996

i am too trying to solve this
can't Rω = 1cm/s be the condition for pure rolling?
where ω is the angular velocity due to the friction force perpendicular to the intial angular velocity.
by doing that i get t=0.103 s

19. Nov 27, 2013

### TSny

Good.

What is the answer to the original question based on this result?

20. Nov 27, 2013

### Saitama

The answer is 1 which is correct. Interesting.

Can this be explained through the use of words? I mean, the final speed is unaffected by the motion of paper, is there a interpretation for this interesting result?

Thanks a lot everyone.