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

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.
  • #1
Istiak
158
12
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 :

1625401961807.png


Solution attempt :

1625401977600.png
for
 
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  • #2
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
 
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  • #3
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.
 
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  • #4
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
 
  • #5
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.
1625411680313.png
 
  • #6
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 $$a=\frac{4\pi\mu g\sigma R^3}{3MR^2}$$
 
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  • #7
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...
 
  • #8
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?
 
  • #9
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##...
 
  • #10
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!
 
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  • #11
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!
1625413468631.png


Is it OK now?
 
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  • #13
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.
 
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Related to How long does it take for the disk to stop rotating?

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.

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