A toy train track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis (see the figure). A toy train of mass m = 2.5 kg is placed on the track and, with the system initially at rest, the electrical power is turned on. The train reaches a steady speed v = 0.4 m/s with respect to the track. What is the angular velocity ω of the wheel, if its mass is M = 4.50 kg and its radius R = 0.6 m? (Treat the wheel as a hoop, and neglect the mass of the spokes and hub.)
To solve this problem, we will use the principle of conservation of angular momentum.
The angular momentum of a system remains constant unless acted upon by an external torque. In this case, the only torque acting on the system is the friction between the train and the track.
First, let's find the initial angular momentum of the system. Since the system is initially at rest, the initial angular momentum is zero.
Next, let's find the final angular momentum of the system. The final angular momentum can be calculated by summing the angular momentum of the train and the angular momentum of the wheel.
The angular momentum of the train is given by L_train = m * v * R, where m is the mass of the train, v is the velocity of the train with respect to the track, and R is the radius of the wheel. Substituting the given values, we have L_train = 2.5 kg * 0.4 m/s * 0.6 m = 0.6 kg·m²/s.
The angular momentum of the wheel is given by L_wheel = I * ω, where I is the moment of inertia of the wheel and ω is the angular velocity of the wheel. The moment of inertia of a hoop about its axis of rotation is I = m * R², where m is the mass of the hoop and R is its radius. Substituting the given values, we have I = 4.5 kg * (0.6 m)² = 1.62 kg·m².
Since the wheel and the train are connected and the angular momentum is conserved, the final angular momentum of the system is L_wheel + L_train.
So, final angular momentum = L_wheel + L_train = I * ω + m * v * R.
Since the initial angular momentum is zero and the final angular momentum is L_wheel + L_train, we have:
0 = I * ω + m * v * R.
Solving for ω, we get:
ω = - (m * v * R) / I.
Substituting the given values, we have ω = - (2.5 kg * 0.4 m/s * 0.6 m) / 1.62 kg·m².
Evaluating this expression, we find:
ω = -0.3704 rad/s.
Therefore, the angular velocity of the wheel is approximately -0.3704 rad/s.