11.3 Conservation of angular momentum  (Page 2/7)

Coupled flywheels

A flywheel rotates without friction at an angular velocity ${\omega }_0=600\text{rev}\text{/}\text{min}$ on a frictionless, vertical shaft of negligible rotational inertia. A second flywheel, which is at rest and has a moment of inertia three times that of the rotating flywheel, is dropped onto it ( [link] ). Because friction exists between the surfaces, the flywheels very quickly reach the same rotational velocity, after which they spin together. (a) Use the law of conservation of angular momentum to determine the angular velocity $\omega$ of the combination. (b) What fraction of the initial kinetic energy is lost in the coupling of the flywheels?

Strategy

Part (a) is straightforward to solve for the angular velocity of the coupled system. We use the result of (a) to compare the initial and final kinetic energies of the system in part (b).

Solution

a. No external torques act on the system. The force due to friction produces an internal torque, which does not affect the angular momentum of the system. Therefore conservation of angular momentum gives

b. Before contact, only one flywheel is rotating. The rotational kinetic energy of this flywheel is the initial rotational kinetic energy of the system, $\frac{1}{2}{I}_0{\omega }_0^2$ . The final kinetic energy is $\frac{1}{2}(4I_0){\omega }^2=\frac{1}{2}(4I_0){\left(\frac{{\omega }_0}{4}\right)}^2=\frac{1}{8}{I}_0{\omega }_0^2.$

Therefore, the ratio of the final kinetic energy to the initial kinetic energy is

Thus, 3/4 of the initial kinetic energy is lost to the coupling of the two flywheels.

Significance

Since the rotational inertia of the system increased, the angular velocity decreased, as expected from the law of conservation of angular momentum. In this example, we see that the final kinetic energy of the system has decreased, as energy is lost to the coupling of the flywheels. Compare this to the example of the skater in [link] doing work to bring her arms inward and adding rotational kinetic energy.

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Dismount from a high bar

An 80.0-kg gymnast dismounts from a high bar. He starts the dismount at full extension, then tucks to complete a number of revolutions before landing. His moment of inertia when fully extended can be approximated as a rod of length 1.8 m and when in the tuck a rod of half that length. If his rotation rate at full extension is 1.0 rev/s and he enters the tuck when his center of mass is at 3.0 m height moving horizontally to the floor, how many revolutions can he execute if he comes out of the tuck at 1.8 m height? See [link] .

Strategy

Using conservation of angular momentum, we can find his rotation rate when in the tuck. Using the equations of kinematics, we can find the time interval from a height of 3.0 m to 1.8 m. Since he is moving horizontally with respect to the ground, the equations of free fall simplify. This will allow the number of revolutions that can be executed to be calculated. Since we are using a ratio, we can keep the units as rev/s and don’t need to convert to radians/s.

Solution

The moment of inertia at full extension is $I_0=\frac{1}{12}m{L}^2=\frac{1}{12}80.0\text{kg}(1{.8\text{m})}^2=21.6\text{kg}·{\text{m}}^2$ .

The moment of inertia in the tuck is $I_{\text{f}}=\frac{1}{12}m{L}_{\text{f}}^{\text{2}}=\frac{1}{12}80.0\text{kg}(0{.9\text{m})}^2=5.4\text{kg}·{\text{m}}^2$ .

Conservation of angular momentum: $I_{\text{f}}{\omega }_{\text{f}}=I_0{\omega }_0\Rightarrow {\omega }_{\text{f}}=\frac{I_0{\omega }_0}{I_{\text{f}}}=\frac{21.6\text{kg}·{\text{m}}^2(1.0\text{rev}\text{/}\text{s})}{5.4\text{kg}·{\text{m}}^2}=4.0\text{rev}\text{/}\text{s}$ .

Time interval in the tuck: $t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(3.0-1.8)\text{m}}{9.8\text{m}\text{/}\text{s}}}=0.5\text{s}$ .

In 0.5 s, he will be able to execute two revolutions at 4.0 rev/s.

Significance

Note that the number of revolutions he can complete will depend on how long he is in the air. In the problem, he is exiting the high bar horizontally to the ground. He could also exit at an angle with respect to the ground, giving him more or less time in the air depending on the angle, positive or negative, with respect to the ground. Gymnasts must take this into account when they are executing their dismounts.

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