# Rotating cylinder - number of revolutions before it reach 2500 rpm

## 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 :)

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.

siddharth
Homework Helper
Gold Member
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).

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.

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The following should help.

Regards,

Nacer.
http://islam.moved.in/tmp/3.jpg [Broken]

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Thanks Nacer. I got the dimensions of K wrong and stuck the w0 in the exponential incorrectly.