Calculating Angular Velocity of 0.600 m Flywheel

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Homework Help Overview

The discussion revolves around calculating the angular velocity and acceleration of a flywheel with a diameter of 0.600 m, which is subjected to a steady pull of 40.0 N via a rope. The flywheel starts from rest, and the problem involves unwinding 5.00 m of rope in 2.00 s.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between linear and angular velocity, with some attempting to derive angular acceleration from given distances and times. Questions arise regarding the distinction between average and final velocities, as well as the correct application of kinematic equations.

Discussion Status

There is ongoing exploration of various methods to calculate angular acceleration and velocity, with participants sharing their calculations and questioning assumptions. Some guidance has been offered regarding the use of average versus final velocities and the implications for angular motion.

Contextual Notes

Participants note potential confusion regarding the use of diameter versus radius in calculations, as well as the need for mass to determine kinetic energy. The discussion reflects uncertainty about the correct application of formulas and the interpretation of results.

masterthephysics
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Question:
A flywheel 0.600 m in diameter pivots on a horizontal axis. A rope is wrapped around the outside of the flywheel, and a steady pull of 40.0 N is exerted on the rope. The flywheel starts from rest, and 5.00 m of rope are unwound in 2.00 s.

Okay i have tried finding the angular velocity by first finding the linear velocity=5/2 then putting it into v=omega*r and making omega the subject. Then after finding omega I substituted it into the equation of (omega - 0)/2 but its wrong :(

Where have I gone wrong? Can someone help me out? thanks in advance
 
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masterthephysics said:
Question:
A flywheel 0.600 m in diameter pivots on a horizontal axis. A rope is wrapped around the outside of the flywheel, and a steady pull of 40.0 N is exerted on the rope. The flywheel starts from rest, and 5.00 m of rope are unwound in 2.00 s.

Okay i have tried finding the angular velocity by first finding the linear velocity=5/2 then putting it into v=omega*r and making omega the subject. Then after finding omega I substituted it into the equation of (omega - 0)/2 but its wrong :(

Where have I gone wrong? Can someone help me out? thanks in advance

You have not actually stated the question you are trying to answer. Something about acceleration I bet. In any case, you have probably confused average velocity with final velocity. The information given in the problem gives you the average velocity. From that you have to decuce final velocity and probably use that to find acceleration.
 
What is the angular acceleration of the flywheel?
 
So.. what did you actually calculate? What was your answer?
 
i got 4.17 as angular accerleration but it was wrong
 
Acceleration = Velocity / Time.. you found the velocity, you have the time..
 
would that give me the tangential acceleration?
the i can work out a_tan = r*alpha?
alpha = angular acceleration
 
Alternatively if oyu can find the angle swept during the 2s, you can find the acceleration using the relationship

[tex]\theta = \frac{\alpha t^2}{2}[/tex]
 
masterthephysics said:
i got 4.17 as angular accerleration but it was wrong

Either you have confused the average and final velocity, or used the given diameter instead of the radius in your calculation.
 
  • #10
so to work out theta i would use s=r*theta with s=5 and r=0.3?
 
  • #11
masterthephysics said:
so to work out theta i would use s=r*theta with s=5 and r=0.3?

That would be correct for using

[tex]\theta = \frac{\alpha t^2}{2}[/tex]

You can also do it using your original velocity calculation, but you need to understand that is the average velocity, which is 1/2 the final velocity.
 
  • #12
alright - i tried using the equation above with theta but I still get the wrong answer for the acceleration: i did -
5/0.3 = (alpha(4))/2 and got 8.33
 
  • #13
masterthephysics said:
alright - i tried using the equation above with theta but I still get the wrong answer for the acceleration: i did -
5/0.3 = (alpha(4))/2 and got 8.33

That looks right.. what are they giving as the correct answer?

By velocity it would be average velocity = 5m/2 sec = 2.5m/sec. So omega = v/r = 5m/2sec/.3m = 8.33/sec. The final angular velocity would be twice that and the time to reach that angular velocity is 2 seconds

alpha = 2*8.33/sec/2sec = 8.33/sec^2
 
Last edited:
  • #14
thanks! i just realized that i did something wrong haha
 
  • #15
how would u work out the final kinetic energy here?
i tried using k_e=.5*I*(omega)^2

but I is an unknown, as to find I, u use I=kmr2, so I=(1/2)*m*(.03)^2 (as this is a solid cylinder)
and i have no idea how to work out m.

wait can u use f=ma? so 40 n = m* a, where a =r*alpha, or=2.499

so 40=m*2.499 and m=16.01kg?

___NVm got the answer
 
Last edited:
  • #16
You would need the mass to find kinetic energy.
 
  • #17
la673 said:
how would u work out the final kinetic energy here?
i tried using k_e=.5*I*(omega)^2

but I is an unknown, as to find I, u use I=kmr2, so I=(1/2)*m*(.03)^2 (as this is a solid cylinder)
and i have no idea how to work out m.

wait can u use f=ma? so 40 n = m* a, where a =r*alpha, or=2.499

so 40=m*2.499 and m=16.01kg?

___NVm got the answer

If you have found the angular deceleration correctly, you now have everything needed to figure out I, and you need to do it.
 

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