# A disk on a slippy surface (Rotation and Translation)

• srecko97
In summary, the conversation revolves around a problem involving a disk with mass and a hanging weight on a frictionless surface. The question is how far the center of mass of the disk moves in 10 seconds and how many turns the disk makes in the same time period. The conversation covers various attempts at solving the problem, including relating the acceleration of the hanging weight to the acceleration of the disk and the angular acceleration of the disk. Ultimately, the correct solution is found and the discussion ends with a question about the rotation of the disk and different arguments for which axis it rotates about.
srecko97

## Homework Statement

There is a disk with mass 50 grams put on a slippy surface (no friction!). Its radius is 5 cm. Mass of the weight is 20 grams. At the beginning there is no motion. How far does the center of mass of the disk move in 10 seconds? (answer: 91 m) How many turns does the disk in 10 seconds?

F=ma
F*r = J*alpha

## The Attempt at a Solution

The only force on the disk is the force of the rope if there is no friction. -->

Is the acceleration of the center of the disk the same as the acceleration of the hanging weight?

srecko97
Yeah, right, they are not. But I do not know how to solve it in other way

acceleration of the disk = force of the rope / mass of the disk ... Is that correct?

srecko97 said:
acceleration of the disk = force of the rope / mass of the disk ... Is that correct?
Yes. ("Acceleration of the disk" means the acceleration of the center of the disk.)

srecko97
Yes, I meant that

You will need to relate the acceleration of the hanging weight to the acceleration of the center of the disk and the angular acceleration of the disk.

srecko97
angular acceleration (alpha) = (2 * force_rope )/ (mass * radius)

TSny
yeah,yeah, i think i got it! circimferential acceleration + acceleration of center of mass = acceleration of weight , right?

Ok, thanks, I got the right result! Your hints are very useful! Can you give me your phonenumber? ... just joking! have a nice day, thanks!

srecko97 said:
yeah,yeah, i think i got it! circimferential acceleration + acceleration of center of mass = acceleration of weight , right?

srecko97 said:
Ok, thanks, I got the right result! Your hints are very useful! Can you give me your phonenumber? ... just joking! have a nice day, thanks!
Good work!

srecko97
I have been struggling with as a recreational, not homework problem.

Does the disk rotate? If so, about which axis does it disk rotate? And why?

I can make three arguments but cannot see which is correct.

1. The disk rotates about its centre

2. The disk rotates about the point opposite to where the rope leaves the disk. I prefer this argument. It is smilar to the wheel on a car which, when viewed from a frame attached to the ground (as we have here), rotates about its point of contact with the ground. If there was friction then this is the intuitive answer.

3. The disk does not rotate because there is only one force acting on it. I can draw a box enclosing the disk and the rope is then applying the only external force to the system.

I tried simplifying the problem to a stick on a table with friction which will then rotate about the far end of the stick and is how I came to argument 2. But why would it rotate if there was no friction - why isn't it just pulled along without rotation? Replacing the rope with a little rocket didn't help.

Any guidance will be welcomed. It's two years old so any answer shouldn't be a spoiler.

Frodo said:
I have been struggling with as a recreational, not homework problem.

Does the disk rotate? If so, about which axis does it disk rotate? And why?

I can make three arguments but cannot see which is correct.

1. The disk rotates about its centre

2. The disk rotates about the point opposite to where the rope leaves the disk. I prefer this argument. It is smilar to the wheel on a car which, when viewed from a frame attached to the ground (as we have here), rotates about its point of contact with the ground. If there was friction then this is the intuitive answer.

3. The disk does not rotate because there is only one force acting on it. I can draw a box enclosing the disk and the rope is then applying the only external force to the system.

I tried simplifying the problem to a stick on a table with friction which will then rotate about the far end of the stick and is how I came to argument 2. But why would it rotate if there was no friction - why isn't it just pulled along without rotation? Replacing the rope with a little rocket didn't help.

Any guidance will be welcomed. It's two years old so any answer shouldn't be a spoiler.
There's actually a current thread on this problem, I believe:

## 1. How does a disk move on a slippy surface?

A disk on a slippy surface will experience both rotation and translation. The rotation is caused by the torque applied to the disk, while the translation is caused by the force of friction between the disk and the surface.

## 2. What factors affect the rotational motion of a disk on a slippy surface?

The rotational motion of a disk on a slippy surface is affected by the torque applied to the disk, the moment of inertia of the disk, and any external forces acting on the disk.

## 3. Can a disk on a slippy surface come to rest?

Yes, a disk on a slippy surface can eventually come to rest due to the forces of friction and air resistance acting on the disk. However, in a perfectly idealized scenario, the disk may continue to rotate indefinitely due to the conservation of angular momentum.

## 4. How does the shape and size of the disk affect its motion on a slippy surface?

The shape and size of the disk can affect its rotational motion on a slippy surface due to changes in its moment of inertia. A disk with a larger moment of inertia will resist changes in its rotational motion more than a disk with a smaller moment of inertia.

## 5. What other factors, besides friction, can affect the translation of a disk on a slippy surface?

Besides friction, the translation of a disk on a slippy surface can be affected by external forces such as gravity, air resistance, and any additional applied forces. The mass and shape of the disk can also play a role in its translation on a slippy surface.

• Introductory Physics Homework Help
Replies
11
Views
224
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
24
Views
2K
• Classical Physics
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
852
• Introductory Physics Homework Help
Replies
18
Views
2K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
30
Views
3K
• Introductory Physics Homework Help
Replies
1
Views
886
• Introductory Physics Homework Help
Replies
18
Views
2K