Calculating Kinetic Energy of a Rotating Grinding Wheel

AI Thread Summary
The discussion focuses on calculating the kinetic energy of a rotating grinding wheel with a mass of 5 kg and a diameter of 0.20 m, rotating at 100 rad/s. The kinetic energy is determined to be 250 joules using the formula for kinetic energy, which involves the wheel's linear velocity derived from its radius and angular speed. Participants then discuss how to find the height from which the wheel must be dropped to achieve the same kinetic energy, using the gravitational potential energy formula. The correct height is calculated to be 5 meters. The conversation emphasizes understanding energy transformations rather than simply providing homework 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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