Two blocks with masses m1 = kg and m2 = kg are connected by a string that hangs over a pulley of mass M = kg and radius R = m as shown above. The string does not slip. Assuming the system starts from rest, use energy principle find the speed of m2 after it has fallen by m. Treat the pulley as a disk.

To find the speed of m2 after it has fallen by m using energy principles, we can utilize the conservation of energy. The potential energy lost by m2 as it falls is converted into kinetic energy. We can set up the energy equations as follows:

1. Potential energy at the start: m2 * g * h1, where h1 is the initial height of m2.
2. Kinetic energy at the end: (1/2) * m2 * v^2, where v is the final velocity of m2.

The potential energy is converted to kinetic energy as the mass m2 falls. However, we need to consider two additional factors: the rotational kinetic energy of the pulley and the tension in the string.

1. Rotational kinetic energy of the pulley:
The moment of inertia of a disk is given by: I = (1/2) * M * R^2
The rotational kinetic energy is given by: (1/2) * I * ω^2, where ω is the angular velocity.

2. Tension in the string:
The tension in the string is given by: T = m1 * g - m2 * g

Now, let's set up the energy equation:

Potential energy at the start (m2 * g * h1) = Kinetic energy at the end ((1/2) * m2 * v^2) + Rotational kinetic energy of the pulley + Tension in the string.

Since the system starts from rest, the initial velocity is zero (v0 = 0), and we have h1 = m.

Now, let's substitute the expressions and solve for v:

m2 * g * m = (1/2) * m2 * v^2 + (1/2) * (1/2) * M * R^2 * ω^2 + (m1 * g - m2 * g) * m

Simplifying the equation, we get:

m2 * g * m = (1/2) * m2 * v^2 + (1/4) * M * R^2 * ω^2 + (m1 * g - m2 * g) * m

Now, we need to relate the linear velocity of m2 to the angular velocity of the pulley. The relationship is given by:

v = R * ω (since the string does not slip)

Substituting v = R * ω into the equation, we get:

m2 * g * m = (1/2) * m2 * (R * ω)^2 + (1/4) * M * R^2 * ω^2 + (m1 * g - m2 * g) * m

Simplifying further:

m2 * g * m = (1/2) * m2 * R^2 * ω^2 + (1/4) * M * R^2 * ω^2 + (m1 * g - m2 * g) * m

Now, we can solve for ω by rearranging the equation:

ω^2 = (2 * g * (m1 * m + m2 * m)) / (m2 * R^2 + M * R^2 / 2)

Taking the square root of both sides, we get:

ω = sqrt((2 * g * (m1 * m + m2 * m)) / (m2 * R^2 + M * R^2 / 2))

Finally, we can substitute the relationship v = R * ω to find the velocity of m2:

v = R * sqrt((2 * g * (m1 * m + m2 * m)) / (m2 * R^2 + M * R^2 / 2))

This equation gives us the speed of m2 after it has fallen by m, using the energy principle.