Rotational motion -- Energy stored in a flywheel

In summary, the conversation discusses a cylindrical shape pulley with given measurements and a constant torque causing it to stop. The conversation then moves on to calculating the angular velocity, finding the moment of inertia for a cylinder, and determining the energy stored in the rotating cylinder. Finally, the conversation concludes that to stop the cylinder, 1065 J of work must be done.
  • #1
189
1

Homework Statement


Cylindrical shape pulley ( m = 6 kg, R = 0.18 m) is rotating at a frequency f = 10 s-1. Due to constant torque it stops. Calculate the work done by the breaking force.

Homework Equations


w = 2πf

The Attempt at a Solution


[/B]
I can calculate the angular velocity:
w = 2πf = 62.8 rad/s

I don't know what to do now.
 
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  • #2
(I added to your thread title for you...)

How much energy is stored in that spinning flywheel? What is its moment of inertia?
 
  • #3
I don't know, it is not given. I think I'm suppose to find some kind of ratio.
 
  • #4
Calculate it. Look up the formula for the moment of inertia for a cylinder in your textbook or use Google or Wikipedia...
 
  • #5
Here is moment of inertia for a cylinder:

I = 1/2 MR2 = 1/2 * 6 kg * 0.18 m = 0.54 kg/m
 
  • #6
kaspis245 said:

Homework Statement


Cylindrical shape pulley ( m = 6 kg, R = 0.18 m) is rotating at a frequency f = 10 s-1. Due to constant torque it stops. Calculate the work done by the breaking force.

Homework Equations


w = 2πf

The Attempt at a Solution


[/B]
I can calculate the angular velocity:
w = 2πf = 62.8 rad/s

I don't know what to do now.

kaspis245 said:
I don't know, it is not given. I think I'm suppose to find some kind of ratio.

kaspis245 said:
Here is moment of inertia for a cylinder:

I = 1/2 MR2 = 1/2 * 6 kg * 0.18 m = 0.54 kg/m

And what is the energy stored in that rotating cylinder? Once you have the energy stored in the rotating cylinder, how much work would it take to slow it down and stop it?
 
  • #7
Energy stored in a rotating cylinder:

Erotational = ½Iω2 = 1065 J

So in order to stop this cylinder one must do 1065 J work?
 

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