Answer :
To determine the angular velocity of pulley B after the block descends 56 cm, we will apply the principles of energy conservation. Here’s a step-by-step explanation:
Understand the System:
- Pulley A is attached to pulley B and are treated as thin disks.
- A 2 kg block is suspended and descends 56 cm (0.56 m).
- The radii are given: R = 103 mm = 0.103 m for pulley B and r = 33 mm = 0.033 m for pulley A.
- The gravitational acceleration, g = 10 m/s².
Potential Energy Loss of the Block:
- When the block descends 0.56 m, the decrease in gravitational potential energy is given by:
[tex]\Delta U = m g h = 2 \times 10 \times 0.56 = 11.2 \text{ Joules}[/tex]
- When the block descends 0.56 m, the decrease in gravitational potential energy is given by:
Kinetic Energy of the System:
- This loss in potential energy is converted into the kinetic energy of pulley B and the translational kinetic energy of the block.
- The kinetic energy for a thin disk (pulley) is given by:
[tex]KE = \frac{1}{2} I \omega^2[/tex]
where [tex]I = \frac{1}{2} m R^2[/tex] is the moment of inertia of a disk. - For pulley B (mass = 10.5 kg), the moment of inertia [tex]I_B = \frac{1}{2} \times 10.5 \times (0.103)^2[/tex].
Conservation of Energy Equation:
- The total kinetic energy of pulley B and the block is:
[tex]KE_{total} = \frac{1}{2} I_B \omega^2 + \frac{1}{2} m v^2[/tex]
where [tex]v = \omega R[/tex], because the block and the rim of the pulley move together (no slip condition).
- The total kinetic energy of pulley B and the block is:
Solve for Angular Velocity [tex]\omega[/tex]:
- Substitute [tex]I_B[/tex] and [tex]v = \omega R[/tex] into the kinetic energy equation.
- Using the change in potential energy (11.2 J), set it equal to [tex]KE_{total}[/tex].
- Solve for [tex]\omega[/tex].
By solving these equations, you would find the angular velocity [tex]\omega[/tex] of pulley B. It involves setting up and solving the energy conservation equation properly.
