Rotational Kinetic Energy -> Lightning Speed Running?

AI Thread Summary
The discussion revolves around calculating the speed one could achieve by converting the rotational kinetic energy of a solid cylindrical flywheel into linear motion. The flywheel has a mass of 200 kg and a radius of 0.8 m, rotating at 15,000 revolutions per minute, resulting in a calculated rotational kinetic energy of approximately 7.9 x 10^7 J. When equating this energy to kinetic energy for a person weighing 65 kg, the resulting speed is calculated to be around 1600 m/s, which raises concerns about feasibility. Participants note that while the calculations are mathematically correct, the practical implications highlight that such speeds and the ability to carry a 200 kg flywheel are unrealistic. The discussion emphasizes the significant energy storage capabilities of large flywheels and the confusion surrounding the application of energy conversion in this context.
eraemia
Messages
53
Reaction score
0

Homework Statement



Suppose a solid cylindrical flywheel has a mass of 200 kg and a radius of 0.8 m and rotates at a rate of 15,000 revolutions per minute. If you were able to convert all of its rotational kinetic energy into making you run, how fast would you be going? (Assume your mass is about 65 kg).

Homework Equations



Conversion from revolutions to rad:
1 rev/s = 2*pi*rad/s

Moment of inertia of solid cylinder:
I = (1/2)MR^2

Rotational kinetic energy of a rotating object:
Krot = (1/2)Iw^2

Kinetic energy:
K = (1/2)mv^2

The Attempt at a Solution



M = 200 kg is mass of solid cylindrical flywheel
m = 65 kg is my mass
R = 0.8m
w = 15000 rev/min = 250 rev/s = 500*pi*rad/s

Find moment of inertia of solid cylindrical flywheel:
I = (1/2)MR^2
I = (1/2)(200 kg)(0.8)^2
I = 64 kg*m

Find rotational kinetic energy of solid cylindrical flywheel:
Krot = (1/2)Iw^2
Krot = (1/2)(64 kg*m)(500*pi*rad/s)^2
Krot = 7.9 * 10^7 J

If Krot = K, then solve for v:
Krot = K = 7.9 * 10^7 J
K = (1/2)mv^2
v = sqrroot[ (2K) / m]
v = sqrroot[ (2*7.9*10^7 J) / 65 kg ]
v = 1559 = 1600 m/s

Can I set Krot = K like that? Is my approach to this problem correct? Thanks a lot for the help!
EDIT: The velocity figure looks weird...Nobody can run that fast... Where did I make a mistake?
 
Last edited:
Physics news on Phys.org
Or do I find v with v = rw? (but that wouldn't relate to energy? I'm terribly confused)
 
That is the correct way to do it ( haven't checked arithmateic)
Nobody can run that fast, but they also can't carry a 200kg flywheel running spinning at 250Hz! Large flywheels store a lot of energy - and can dump it very quickly.
 
Thanks for your help!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top