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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$\Rightarrow$}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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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 $I = 4 k g m^2$ 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.

LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces $\alpha^2_0$. type this... to get...$\int_0^\infty$$\int_0^\infty$$\partial$$\partial$$\Rightarrow$$\Rightarrow$$\ddot{x},\dot{x}$$\ddot{x},\dot{x}$$\sqrt{z}$$\sqrt{z}$$\langle my \rangle$$\langle my \rangle$$\left( abacadabra \right)_{me}$$\left( abacadabra \right)_{me}$$\vec{E}$$\vec{E}$$\frac{a}{b}\$ $\frac{a}{b}$