Rotating cylinder - number of revolutions before it reach 2500 rpm

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Homework Help Overview

The problem involves a homogeneous cylinder functioning as a flywheel, which is initially rotating at 5000 rpm. The cylinder is subjected to a moment of force due to friction when the motor is turned off, and the task is to determine how many revolutions it makes before coming to a stop.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the moment of inertia and the effects of friction on the cylinder's motion. There are attempts to derive equations related to angular velocity and frictional forces. Some participants question the completeness of the problem statement and the assumptions regarding friction.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some have attempted to derive equations but express uncertainty about the implications of their findings. There is no clear consensus, and participants are seeking clarification on the problem's parameters.

Contextual Notes

Some participants note potential missing information regarding the orientation of the flywheel and the nature of the friction acting on it. There is also mention of confusion regarding the dimensions of constants used in calculations.

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Homework Statement



A flywheel has shape as a homogeneous cylinder with mass m = 40,0 kg and radius r = 0,50 meters. The cylinder is rotating round the axis of symmetry, and is runned by a motor with constant angular velocity (w0). When the motor is switched off, the cylinder is affected by a moment of force that is caused by friction.

When the motor is being swiched off, the rotational frequency of the cylinder was 5000 rpm. How many revolutions does the cylinder make before it stops?


Homework Equations



The friction: M = -kw
w = omega
k = (1,2 * 10^2) Nm/s
w0 = 5000 rpm


The Attempt at a Solution



The moment of inertia: I = (m*r^2)/2 (cylinder)

M = -kw => 1) w = -(M/k)

w = 1/2*5000

alpha = w/T = (1/2*w0)/T

1) (1/2*w0) = -(M/k)

(1/2*w0) = -((I*alpha)/k)

Any help will be appreciated :)
 
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I just solved this scenario and got

\omega (t) = \omega_0 e^{\frac{-Kt}{I\omega_0}}

which means it's never going to stop. Because there is no element of static friction, the friction is tending to zero as omega.

I must have made an error, maybe someone can correct me.
 
When the motor is switched off, the cylinder is affected by a moment of force that is caused by friction.

Are you sure that's how the problem was worded? Some information seems to be missing, like how the flywheel was oriented, and where the friction acts (ie, axle, point in contact with the ground if vertical, or whole surface if horizontal).

Mentz114 said:
which means it's never going to stop. Because there is no element of static friction, the friction is tending to zero as omega.

IMO, since the flywheel will be moving, there's always going to be dynamic friction till it stops.
 
Last edited:
The following should help.

Regards,

Nacer.
http://islam.moved.in/tmp/3.jpg
 
Last edited by a moderator:
Thanks Nacer. I got the dimensions of K wrong and stuck the w0 in the exponential incorrectly.
 

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