The system shown in the figure below consists of a m1 = 5.52-kg block resting on a frictionless horizontal ledge. This block is attached to a string that passes over a pulley, and the other end of the string is attached to a hanging m2 = 2.76-kg block.The pulley is a uniform disk of radius 7.86 cm and mass 0.592 kg. Calculate the speed of the m2 = 2.76-kg block after it is released from rest and falls a distance of 2.08 m. Calculate the angular speed of the pulley at this instant.

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To solve this problem, we can use the principles of conservation of energy and the relationships between linear motion and rotational motion. Here's how you can proceed:

1. Start by calculating the potential energy of the m2 block when it is at the initial height of 2.08 m. The potential energy of an object at height h is given by PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height.

Potential energy of m2 block: PE = (2.76 kg) * (9.8 m/s^2) * (2.08 m)

2. Next, convert the potential energy into kinetic energy when m2 falls:

Kinetic energy of m2 block: KE = PE

3. The total kinetic energy of the system will be equal to the sum of the kinetic energy of m2 and the rotational kinetic energy of the pulley. The kinetic energy of an object in linear motion is given by KE = (1/2) * m * v^2, where m is the mass of the object and v is the velocity. The rotational kinetic energy of a uniform disk is given by KE = (1/2) * I * ω^2, where I is the moment of inertia of the disk and ω is the angular velocity.

Total kinetic energy of the system: KE_total = KE_m2 + KE_pulley

4. Since m2 will move downward, the velocity of m2 will be positive. We can assume the rotational direction of the pulley as positive (counterclockwise) as well. Use the following equations:

KE_m2 = (1/2) * m2 * v^2
KE_pulley = (1/2) * I_pulley * ω^2

We need to find v (velocity of m2) and ω (angular velocity of the pulley).

5. Rewrite the above equations by substituting the appropriate values:

(1/2) * m2 * v^2 + (1/2) * I_pulley * ω^2 = PE

(1/2) * (2.76 kg) * v^2 + (1/2) * (0.5 * m_pulley * r_pulley^2) * ω^2 = PE

Here, m_pulley is the mass of the pulley and r_pulley is the radius of the pulley.

6. Rearrange the equation to solve for ω:

(1/2) * (2.76 kg) * v^2 + (1/2) * (0.5 * (0.592 kg) * (0.0786 m)^2) * ω^2 = PE

(1.38 kg) * v^2 + (0.01843 kg * m^2) * ω^2 = PE

7. Substitute the previously calculated potential energy into the equation:

(1.38 kg) * v^2 + (0.01843 kg * m^2) * ω^2 = (2.76 kg) * (9.8 m/s^2) * (2.08 m)

8. Simplify the equation and solve for v. This will give you the speed of the m2 block:

v^2 + (0.13 m^2) * ω^2 = (40.73 m^2/s^2) / (1.38 kg)

v^2 + 0.0942 m^2/s^2 * ω^2 = 29.58 m^2/s^2

9. To find ω (angular velocity), we need an additional equation. Since the m1 block is at rest, the tension in the string between m1 and m2 must be equal to the weight of m1 (Tension = m1 * g):

Tension = m1 * g

Tension = (5.52 kg) * (9.8 m/s^2)

Tension = 54.096 N

10. The torque developed by the tension in the string causes the angular acceleration of the pulley. The moment of inertia of a uniform disk is given by I = (1/2) * m * r^2, where m is the mass of the disk and r is its radius. The torque acting on the pulley is given by Torque = Tension * r, where r is the radius of the pulley.

Torque = Tension * r

Torque = (54.096 N) * (0.0786 m)

11. The torque developed by the tension is equal to the moment of inertia multiplied by the angular acceleration:

Torque = I_pulley * α

(54.096 N) * (0.0786 m) = (0.5 * (0.592 kg) * (0.0786 m)^2) * α

12. Solve the equation for α (angular acceleration) and substitute it into the equation for ω (angular velocity):

α = (54.096 N * 0.0786 m) / (0.5 * 0.592 kg * 0.0786 m^2)

ω = α * t

Here, t is the time it takes for m2 to fall the given distance of 2.08 m. Use the equation t = √(2h / g) to find t.

13. Calculate t using the given distance:

t = √(2 * 2.08 m / 9.8 m/s^2)

14. Substitute the calculated α and t into the equation for ω:

ω = α * t

15. Finally, substitute the obtained ω value into the equation for v from step 8 and solve for v.

This will give you the speed of the m2 block after it is released and the angular speed of the pulley at that instant.