Calculating Kinetic Energy of a Rotating Grinding Wheel

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Homework Help Overview

The discussion revolves around calculating the kinetic energy of a rotating grinding wheel, with a focus on the relationship between kinetic energy and gravitational potential energy. The original poster presents a scenario involving a grinding wheel's mass, diameter, and angular speed, seeking to determine its kinetic energy and the height from which it would need to be dropped to acquire that energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of kinetic energy using the wheel's mass and angular speed. Questions arise regarding the formula for gravitational potential energy and its application to find the height corresponding to the calculated kinetic energy.

Discussion Status

There is an ongoing exploration of the relationship between kinetic energy and gravitational potential energy. Some participants provide guidance on relevant formulas, while others seek clarification on the nature of the assistance being offered. The discussion reflects a mix of confirmations and inquiries without reaching a definitive conclusion.

Contextual Notes

Participants note the importance of energy transformations and the potential ambiguity in the original problem statement, which may not have been posted verbatim. The conversation emphasizes the learning process rather than providing direct answers.

physaru86
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1. A grinding wheel of mass 5 kg and diameter 0.20 m is rotating with an angular speed of 100 rad/s. Calculate its kinetic energy. Through what distance would it have to be dropped in free fall to acquire this kinetic energy? m = 5 kg, r = 0.20/2 m = 0.10 m, g = 10 m/s2, ω = 100 rad/s 2. v = r*ω, kinetic energy = (1/2)*m*v2

3. v = (0.10 m)*(100 rad/s) = 10 m/s, kinetic energy = (1/2)*(5 kg)*(10 m/s)2 = 250 kgm2/s2..... is this right?

 
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Yes it is. Just calculate the second part, it should be easy for you now.
 
PWiz said:
Yes it is. Just calculate the second part, it should be easy for you now.
what formula to use?
 
Do you know the formula for gravitational potential energy?
 
PWiz said:
Do you know the formula for gravitational potential energy?
U = m*g*h...?
 
Correct. Now for what value of ##h## does this expression equal 250 joules?
 
PWiz said:
Correct. Now for what value of ##h## does this expression equal 250 joules?
um... is it a question or are you helping me with the solution..?
 
physaru86 said:
um... is it a question or are you helping me with the solution..?
We don't provide direct answers to homework questions here. And I wouldn't be posting in this thread trying to help you if I didn't know the concept myself :)
 
PWiz said:
We don't provide direct answers to homework questions here. And I wouldn't be posting in this thread trying to help you if I didn't know the concept myself :)
Is it related to
Energy conservation during free-fall
 
  • #10
Not exactly. It is more to do with energy transformations. The question is simply asking you from which height you'd have to drop this mass for it to have the same kinetic energy when it's about to strike the surface as the value you have calculated (that's what I think the question is asking, since it appears that the question has not been posted verbatim).
 
  • #11
PWiz said:
Correct. Now for what value of ##h## does this expression equal 250 joules?
m = 5 kg, g = 10 m/s2, m*g*h = 250 joules, h = 250/5*10 m, h = 5 m... ?
 
  • #12
physaru86 said:
m = 5 kg, g = 10 m/s2, m*g*h = 250 joules, h = 250/5*10 m, h = 5 m... ?
That's right.
 
  • #13
PWiz said:
That's right.
So the problem is solved, right...? thanks
 
  • #14
Your welcome :wink:
 

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