Calculating Frictional Force for a Spinning Disk

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Homework Help Overview

The problem involves calculating the frictional force required to bring a spinning disk to a halt within a specified time frame. The disk has a mass of 1.6 kg and a diameter of 20 cm, and it is initially spinning at 240 rpm. Participants are exploring the dynamics of rotational motion, specifically focusing on the relationship between torque, moment of inertia, and angular acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to draw a free body diagram and consider the forces acting on the disk. There are attempts to calculate the moment of inertia and angular acceleration, with some participants questioning the setup of equations and the relationship between torque and frictional force.

Discussion Status

Some participants have provided guidance on using equations of motion for rotational dynamics, while others are clarifying the calculations for moment of inertia and torque. There is an ongoing exploration of how to relate torque to the frictional force, with multiple interpretations of the calculations being discussed.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There are indications of confusion regarding the correct formulas and calculations, particularly concerning the moment of inertia for a solid disk.

aligass2004
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Homework Statement



The 1.6kg, 20cm diameter disk in the figure below is spinning at 240rpm. How much friction force must the brake apply to the rim to bring the disk to a halt in 3.5s?

http://i241.photobucket.com/albums/ff4/alg5045/p13-69.gif

Homework Equations





The Attempt at a Solution



I know a free body diagram should be drawn for the disk to take all of the forces into account. I know there's the weight pulling down and the frictional force is acting to the right, but I don't know how to set up a solvable equation.
 
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Use 1.6kg, 20cm diameter disk to determine the moment of inertia.

The friction force behaves as a torque with moment r and force f.

Torque/(moment of inertia) = angular acceleration

then

use the appropriate equation of motion for rotation to determine the time to decelerate from the initial angular velocity to stop in 3.5 s.

See - http://hyperphysics.phy-astr.gsu.edu/hbase/mi
 
Last edited by a moderator:
Ok, I get that, but how do I find the frictional force?
 
The question isn't asking for time.
 
One is given time, and ask what magnitude of force is require to bring the disk to standstill (\omega = 0) in that time. Use the change in angular velocity and time to find the constant angular acceleration.

Applying an external force (friction) will cause the rotational mass to decelerate.

One must apply the appropriate equation(s) of motion, e.g.

0 = \omega_0\,+\,\alpha\,t, where \omega_0 is the initial angular velocity, and \alpha is the angular acceleration (or deceleration if negative).

With the angular acceleration (or deceleration), use the relationship between torque and moment of intertia.

Then knowing the net torque required to decelerate the disk, then find the necessary friction force applied at the appropriate moment arm (radius of disk).
 
I did the following... I=m(r^2)=.016. I then converted revolutions per minute into radians per second = 25.133. Then I found angular acceleration by using delta w/delta t = 7.181, and finally I found the torque = I(alpha) = .115. I'm still unsure about finding the friction.
 
aligass2004 said:
I did the following... I=m(r^2)=.016.
That formula is incorrect for a solid disk.

To relate torque to friction force, realize that the friction force acting with a moment arm = r creates the given torque (as Astronuc had stated).
 
Ok so instead I = .032 and T = .23. Then I use T = rF (I think) and I solved for F to get 2.3, but it wasn't right.
 
What formula are you using to calculate I? (It's still not right.) But yes, use T = rF.
 
  • #10
I'm using I = .5m(r^2) = .5(1.6)(.1^2) = .032
 
  • #11
aligass2004 said:
I'm using I = .5m(r^2) = .5(1.6)(.1^2) = .032
You're using the correct formula, but recheck your calculation.
 

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