Rotating cylinder, how long will it take before the angular velocity is halved

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SUMMARY

The discussion centers on calculating the time it takes for a rotating homogeneous cylinder, specifically a flywheel with a mass of 40.0 kg and a radius of 0.50 meters, to halve its angular velocity after the motor is switched off. The moment of inertia is calculated using the formula I = (m*r^2)/2, and the frictional torque is defined as M = -kw, with k being 1.2 * 10^2 Nm/s. The initial attempt at solving the equation resulted in an incorrect time value of T = -416, indicating a need for a clearer derivation of the equation of motion.

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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 are caused by friction.
How long will the cylinder rotate before the angular velocity is halved?


Homework Equations



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


The Attempt at a Solution



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

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

w = 1/2*w0

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

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

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

(1/2*w0) = -((((m*r^2)/2)*((1/2*w0)/T))/k)

When solving this equation the answer is: T = -416
BUT sadly we are quite sure this answer is incorrect, any help will be appreciated.
 
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This exercice would have been more funny if only the absorbed power before switch-off had been given, and not the friction factor "k".

Anyway, you need to solve the equation of motion.
You should write the equation of motion, to make your derivation clearer,
and it could help you too.
 
Aha. Please see my post in your earlier thread.
 

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