A cylinder with moment of inertia I1 rotates with angular speed ω0 about a frictionless vertical axle. A second cylinder, with moment of inertia I , initially not rotating, drops onto the first cylinder. Since the surfaces are rough, the two eventually reach the same angular speed ωf.
a. Calculate ωf. (Use I1 for I1, I2 for I2, and omega_i for ωi, as necessary.)
b. Show that the kinetic energy of the system decreases in this interaction by calculating the ratio of the final to initial rotational energy. Express your answer in terms of ωi.
Answers: 3
radians per second. 
, then the kinetic energy of the system decreases due to the interaction. 
(1)
- Moment of inertia of the moving cylinder, in kilogram-square meters.
- Moment of inertia of the resting cylinder, in kilogram-square meters.
- Initial angular speed of the moving cylinder, in radians per second.
- Final angular speed of the resulting system, in radians per second.
![r = \left[\frac{(I_{1}+I_{2})}{I_{1}}\right]\cdot \left(\frac{I_{1}}{I_{1}+I_{2}} \right)^{2}](/tpl/images/0544/6712/0dab8.png)
,showing the the energy of th system will decrease.
' be the moment of inertia of an object and '
' be its angular velocity then the angular momentum '
' of the object can be written as
' and '
' be the initial angular velocity of the cylinders and '
' and '
' be their respective final angular velocity, then from conservation of angular momentum, 
. From the above expression
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A cylinder with moment of inertia I1 rotates with angular speed ω0 about a frictionless vertical axl...
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