Rotating cylinder - number of revolutions before it reach 2500 rpm

In summary: I fixed it up and the answer is 1) w = -(M/k) where M is the mass of the flywheel and k is the static coefficient of friction.
  • #1
buksesele
7
0

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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  • #2
I just solved this scenario and got

[tex] \omega (t) = \omega_0 e^{\frac{-Kt}{I\omega_0}}[/tex]

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.
 
  • #3
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:
  • #4
The following should help.

Regards,

Nacer.
http://islam.moved.in/tmp/3.jpg
 
Last edited by a moderator:
  • #5
Thanks Nacer. I got the dimensions of K wrong and stuck the w0 in the exponential incorrectly.
 

1. How do you calculate the number of revolutions a rotating cylinder will make before reaching 2500 rpm?

The formula for calculating the number of revolutions is: N = (2500 rpm / 60 seconds) x T, where N is the number of revolutions and T is the time in seconds. This formula assumes constant acceleration and no external forces acting on the cylinder.

2. Can the number of revolutions be affected by the size or weight of the rotating cylinder?

Yes, the size and weight of the cylinder can affect the number of revolutions it takes to reach 2500 rpm. A larger or heavier cylinder will require more force to reach the desired speed, which may result in a higher number of revolutions.

3. Is the number of revolutions dependent on the surface the cylinder is rotating on?

Assuming the surface is smooth and there is no friction, the surface the cylinder is rotating on should not significantly affect the number of revolutions. However, if there is friction or other external forces present, it may impact the number of revolutions needed to reach 2500 rpm.

4. How does the initial speed of the rotating cylinder affect the number of revolutions?

The initial speed of the cylinder will impact the number of revolutions it takes to reach 2500 rpm. If the initial speed is already close to 2500 rpm, the number of revolutions will be less compared to a cylinder with a lower initial speed.

5. Can the number of revolutions be used to calculate the power or torque of the rotating cylinder?

No, the number of revolutions alone cannot be used to calculate the power or torque of a rotating cylinder. Other factors such as the mass and radius of the cylinder, as well as any external forces, would need to be considered in order to accurately calculate power or torque.

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