Calculating Angular Velocity of 0.600 m Flywheel

AI Thread Summary
The discussion revolves around calculating the angular velocity of a flywheel with a diameter of 0.600 m, subjected to a steady pull of 40.0 N on a rope. The user initially calculated the linear velocity from the unwound rope, but confusion arose between average and final velocity, leading to incorrect angular acceleration results. Key points include the need to use the radius for calculations and the relationship between angular and tangential acceleration. The final angular velocity is determined to be twice the average velocity, and the user is guided on how to calculate kinetic energy once mass is determined. The conversation emphasizes understanding the relationships between linear and angular motion to solve the problem correctly.
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

\theta = \frac{\alpha t^2}{2}
 
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

\theta = \frac{\alpha t^2}{2}

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
 
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  • #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 hav 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
 
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  • #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 hav 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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