# How long does it take for the disk to stop rotating?

• Istiak
In summary: Good luck with the rest of your studies!In summary, the conversation discusses an error made in a physics problem involving the moment of inertia of a solid disk. The expert summarizer points out that the error involves a factor of 2 and suggests checking the value of μ and the ½ in ½MR². The conversation concludes with the person acknowledging the mistakes and expressing gratitude for the help.
Istiak
Homework Statement
A disk of masses M and radius R is initially rotating at angular velocity \omega. While rotating, it is placed on a horizontal surface whose coefficient of friction is \mu =0.5 How long take for the disk to stop rotating?
Relevant Equations
\tau = r . F
Question :

Solution attempt :

for

Are you asking us to check your working because it does not give one of the answers in the list?

Your writing/working is difficult to read and follow. But as far as I can tell you have made a simple error: you appear to have substituted μ=1 rather than μ=0.5

Delta2
You have made a simple algebraic error. The moment of inertial of a solid disk with respect to the z-axis is ##\frac{1}{2}MR^2##. So when you divide by this to find the angular acceleration you should get $$a=\frac{4\pi\mu g\sigma R^3}{3MR^2}$$, but you have keep an additional factor of 2 (which later you get rid by replacing ##\mu=0.5##). So after simplifications you actually get $$a=\frac{2g}{3R}$$ which lead you to one of the available options.

Steve4Physics
Steve4Physics said:
Are you asking us to check your working because it does not give one of the answers in the list?

Your writing/working is difficult to read and follow. But as far as I can tell you have made a simple error: you appear to have substituted μ=1 rather than μ=0.5
2×0.5=1

Delta2 said:
You have made a simple algebraic error. The moment of inertial of a solid disk with respect to the z-axis is ##\frac{1}{2}MR^2##. So when you divide by this to find the angular acceleration you should get $$a=\frac{4\pi\mu g\sigma R^3}{3MR^2}$$, but you have keep an additional factor of 2 (which later you get rid by replacing ##\mu=0.5##). So after simplifications you actually get $$a=\frac{2g}{3R}$$ which lead you to one of the available options.

Your error is one line above from where your green arrow is pointing. You write that $$a=\frac{4 \cdot 2 \pi\mu g\sigma R^3}{3MR^2}$$ while the correct is (as i said at my earlier post) that $$a=\frac{4\pi\mu g\sigma R^3}{3MR^2}$$

Steve4Physics
Let me ask you one straight question, what do you take it to be the moment of inertia of the solid disk? $$\frac{1}{2}MR^2$$ OR $$\frac{1}{4}MR^2$$?
MoI around the z axis (perpendicular to the plane of the disk) is the first, while MoI around x or y-axis is the second...

Delta2 said:
Your error is one line above from where your green arrow is pointing. You write that $$a=\frac{4 \cdot 2 \pi\mu g\sigma R^3}{3MR^2}$$ while the correct is (as i said at my earlier post) that $$a=\frac{4\pi\mu g\sigma R^3}{3MR^2}$$
Why there won't be 4 times 2?

Istiakshovon said:
Why there won't be 4 times 2?
Because the Moment of Inertia of the solid disk (around the z-axis perpendicular to xy plane-the plane of the disk) is ##I=\frac{1}{2}MR^2## so you actually have ##2\cdot 2=4## there and NOT ##4\cdot 2=8##...

Istiakshovon said:
2×0.5=1
So I believe.

You have made an error involving a factor of 2. It might be (as @Delta2 notes) something to do with the ½ in ½MR², or as I note, something to do with the value of μ (=0.5). Maybe it's a mixture of the two - e.g. incorrectly mixing the two ½s.

For me, the main problem is that your hand-writing is very difficult to read, so it's hard to be certain exactly what you've done.

But the only 2 things you now need to do are:
1) carefully repeat your working (say from 'Iα =' onwards) to locate your mistake;
2) click 'Likes' for @Delta2 and me to acknowledge the time/effort/help we have supplied!

Delta2
Steve4Physics said:
So I believe.

You have made an error involving a factor of 2. It might be (as @Delta2 notes) something to do with the ½ in ½MR², or as I note, something to do with the value of μ (=0.5). Maybe it's a mixture of the two - e.g. incorrectly mixing the two ½s.

For me, the main problem is that your hand-writing is very difficult to read, so it's hard to be certain exactly what you've done.

But the only 2 things you now need to do are:
1) carefully repeat your working (say from 'Iα =' onwards) to locate your mistake;
2) click 'Likes' for @Delta2 and me to acknowledge the time/effort/help we have supplied!

Is it OK now?

Delta2
YES!

Istiakshovon said:
View attachment 285452

Is it OK now?
That's the same answer as I get - so I'd say yes! The important thing is that you found/understood where you made the mistake.

Istiak

## 1. How is the rotation of a disk measured?

The rotation of a disk is typically measured in revolutions per minute (RPM). This is the number of times the disk makes a full rotation in one minute.

## 2. What factors affect the rotation speed of a disk?

The rotation speed of a disk can be affected by several factors, including the size and weight of the disk, the material it is made of, and the amount of friction or resistance present.

## 3. Does the rotation speed of a disk change over time?

Yes, the rotation speed of a disk can change over time. Factors such as wear and tear, temperature changes, and changes in the environment can all affect the rotation speed of a disk.

## 4. How can the rotation of a disk be stopped?

The rotation of a disk can be stopped by applying a force in the opposite direction of the rotation, such as through braking or using a motor to reverse the direction of rotation.

## 5. Is there a specific time frame for a disk to stop rotating?

The time it takes for a disk to stop rotating can vary depending on the factors mentioned earlier. It can range from a few seconds to several minutes, depending on the size and weight of the disk and the amount of resistance present.

• Introductory Physics Homework Help
Replies
3
Views
889
• Introductory Physics Homework Help
Replies
19
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
707
• Introductory Physics Homework Help
Replies
2
Views
410
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
1K
• Introductory Physics Homework Help
Replies
30
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
799
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
321