A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis. A toy train of mass m = 0.140 is placed on the track and, with the system initially at rest, the train's electrical power is turned on. The train reaches a steady speed of 0.17 m/s with respect to the track. What is the angular speed of the wheel if its mass is 1.4m and its radius is 0.42 m? (Treat the wheel as a hoop, and neglect the mass of the spokes and hub.)

To find the angular speed of the wheel, we can use the principle of conservation of angular momentum.

Angular momentum L is defined as the product of the moment of inertia I and the angular velocity ω of an object:
L = I * ω

In this case, the train and the wheel are connected, and there is no external torque acting on the system. Therefore, the initial and final angular momenta of the system are equal.

The initial angular momentum of the system is zero because the system is initially at rest (ω_initial = 0). The final angular momentum of the system is the product of the moment of inertia of the train and the angular velocity of the wheel. The train's moment of inertia I_train can be calculated using the formula for a solid cylinder:
I_train = 0.5 * m * r^2

The final angular momentum of the system is I_train * ω_wheel, where ω_wheel is the angular velocity of the wheel.

Since the initial and final angular momenta are equal, we can set up the following equation:
0 = I_train * ω_wheel

Plugging in the values we have:
0 = (0.5 * m * r^2) * ω_wheel

Now, we can solve for ω_wheel:
ω_wheel = 0 / (0.5 * m * r^2)
ω_wheel = 0

Therefore, the angular speed of the wheel is zero.

Explanation for the answer:
In this scenario, when we consider the conservation of angular momentum, we find that the angular speed of the wheel is zero. This means that the wheel does not rotate at all. The train's movement is solely due to the electric power supplied to it, and it is not transferred to the wheel's rotation.