|
Flywheel from Witten, Germany |
A Flywheel is a large symmetrical wheel that is used to store kinetic energy. It is also used to "even out" the rate of rotation, making it less susceptible to variations in the driving force. Potter's wheels and Drop Spindles are millenia-old examples of the latter case – the large angular momentum of the spinning disc making it less likely that small interruptions or changes in the driving force will have a large effect on the angular velocity . In the 19th century large flywheels were used to store the large amounts of kinetic energy of water-driven machinery in factories, as in the photo above.
Assume that a flywheel consists of two joined discs of differing diameter, and that the force is applied tangentially to the smaller of these. What is the torque, and what are the angular velocity and the angle as a function of time?
Solution
System:
Interactions:
Model:
Approach:
Diagrammatic Representation
It is important to sketch the situation and to define linear and rotational coordinate axes.
Mathematical Representation
There are effectively no important external forces (gravity can be neglected since the astronaut and tool are in freefall), so linear momentum conservation will give us the astronaut's linear speed toward the ship. The relationship is straightforward:
\begin
[ m_
v_{a,{\rm f}} - m_
v_{t,{\rm f}} = 0 ]\end
This gives:
\begin
[ v_{a,{\rm f}} = \frac{m_
v_{t,
}}{m_{a}} ]\end
Since this velocity is assumed to be constant, we can use the (sole) Law of Change from One-Dimensional Motion with Constant Velocity to find that the time required to return to the ship is:
\begin
[ t = \frac{m_
x_{s}}{m_
v_{t,
}} ] \end
Similarly, angular momentum is conserved since there are no external torques. We can choose any non accelerating axis. For simplicity, we compute the angular momentum about the initial location of the astronaut's center of mass.
Since we have chosen an axis along the astronaut's center of mass' line of motion, the translation of the astronaut's center of mass will not contribute to the angular momentum. Further, since we are treating the tool as a point particle, it has no moment of inertia about its center of mass and so its rotations will not contribute to the angular momentum.
\begin
[ \frac
m_
h^
\omega_
- m_
v_{t,{\rm f}}h/2 = 0 ] \end
giving:
\begin
[ \omega_
= \frac{6 m_
v_{t,
}}{m_
h} ]\end
Then, using the time found above and the Law of Change for angular kinematics with constant angular velocity, we can find the total angle the astronaut rotates through before reaching the ship.
\begin
[ \theta = \omega_
t = \frac{6 x_{s}}
= \mbox
= \mbox
]\end