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GR9277 #41
Problem
 GREPhysics.NET Official Solution Alternate Solutions
\prob{41}
A cylinder with moment of inertia about a fixed axis initially rotates at 80 radians per second about this axis. A constant torque is applied to slow it down to 40 radians per second.

The kinetic energy lost by the cylinder is

1. 80 J
2. 800 J
3. 4000 J
4. 9600 J
5. 19,200 J

Mechanics}Energy

The kinetic energy is related to the inertia I and angular velocity by .

The problem supplies so one needs not calculate the moment of inertia. The angular velocity starts at and ends at .

Thus, the kinetic energy lost , as in choice (D).  Alternate Solutions
 Ryry0132019-10-06 09:10:04 A slightly faster way to do this problem is to know that if omega halves, then kinetic energy will decrease by 1/4. Then energy LOST is just (0.5)(4)(80^2) * (3/4) = (0.5)(6400)(3/4) = 9600JReply to this comment Ryry013
2019-10-06 09:10:04
A slightly faster way to do this problem is to know that if omega halves, then kinetic energy will decrease by 1/4. Then energy LOST is just (0.5)(4)(80^2) * (3/4) = (0.5)(6400)(3/4) = 9600J mnky9800n
2013-09-24 13:05:47
I also like to think in terms of the potential energy of a 100kg person who is about to fall 10 meters. That is mgh~(100 kg)(10 m/s^2)(10 m) = 10000 J. Thus A and B are probably too small and E is probably too big. aloha
2008-10-10 03:21:51
Maybe it's a silly question, but how can we know that I=4?
Thanks :)
 Monk2008-10-13 20:11:46 it tells you in the question...it is 4 kg m^2 kg m^2 is the units not a formula
 tensordyne2008-11-04 13:16:28 I am with Monk, from reading the problem it seems like they are saying is a formula in k (proportionality factor maybe), g () and mass m instead of as units because there is no grouping of the units, such as in say , which would have made the use of units much more manifest.      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$