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  GR9277 #41
Problem
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\prob{41}
A cylinder with moment of inertia $4kgm^2$ 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 \omega by K = \frac{1}{2}I\omega^2.

The problem supplies I=4kgm^2 so one needs not calculate the moment of inertia. The angular velocity starts at 80 rad/s and ends at 40 rad/s.

Thus, the kinetic energy lost \Delta K = \frac{1}{2}I(\omega_f^2-\omega_0^2)=\frac{1}{2}(4)(40^2-80^2)=2(1600-6400)=-9600J, as in choice (D).

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Comments
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.NEC
aloha
2008-10-10 03:21:51
Maybe it's a silly question, but how can we know that I=4?
Thanks :)
Monk
2008-10-13 20:11:46
it tells you in the question...it is 4 kg m^2

kg m^2 is the units not a formula
tensordyne
2008-11-04 13:16:28
I am with Monk, from reading the problem it seems like they are saying I = 4 k g m^2 is a formula in k (proportionality factor maybe), g (9.8 m/s) and mass m instead of as units because there is no grouping of the units, such as in say I = 4 kg \cdot m^2, which would have made the use of units much more manifest.
NEC

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