Rotational Motion: Stop a 10 kg Disk in 20 sec

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Homework Help Overview

The problem involves a solid disk of mass 10 kg and radius 1 m that is spinning at a rate of 20 rad/s. Participants are tasked with determining how to apply a force to stop the disk and calculating the minimum time required for it to come to a stop.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to analyze the direction of the applied force for maximum torque and calculates the moment of inertia. They question the validity of their time calculation and express uncertainty about the result.

Discussion Status

Some participants affirm the original poster's approach, indicating that the reasoning appears sound. However, there is an ongoing exploration of the implications of the applied force and the resulting time to stop the disk.

Contextual Notes

The original poster notes that the calculated stopping time of 20 seconds seems long, prompting a discussion about the magnitude of the applied force and its effect on the stopping time.

pradeepk
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Homework Statement


A solid disk of mass 10 kg and radius 1 m is spinning around its central
axis at a rate of ω = 20 rad/s. A force of magnitude 5 N is applied to the disk. Recall that the
moment of inertia of a solid disk is I=(1/2)mr^2

.
(a) Draw the disk and indicate the direction of rotation. Then draw the direction that the force
should be applied to make the disk stop as quickly as possible.
(b) What is the minimum time needed for the disk to stop?

Homework Equations


[tex]\tau[/tex]= I[tex]\alpha[/tex]

[tex]\tau[/tex]=Fr

The Attempt at a Solution


For part a, I said that the disk was moving counterclockwise, and the fastest way to stop it would be to apply a force that is perpendicular to the radius, because that would produce the largest negative torque. Is that correct?

For part b, I solved for I and got 5 kgm2 then I did this:
[tex]\tau[/tex]=I [tex]\alpha[/tex]
Fr=5[tex]\alpha[/tex]
(5N)(1m)=5[tex]\alpha[/tex]
[tex]\alpha[/tex]=1rad/2

I then solved for time with the equation:
[tex]\omega[/tex]f=[tex]\omega[/tex]i +[tex]\alpha[/tex]t
0=20rad/s - 1rad/s2(t)
t=20 seconds I don't think this is correct but I'm not sure what else to do. Thank you
 
Last edited:
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Looks okay to me!
 
hi pradeepk! :wink:

all looks fine :smile:
 
Thanks Guys!

I just thought that 20 seconds was a very long time to make it stop, but I guess 5 Newtons is a very small force
 

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