A pulley with a radius of 3.0 cm and a rotational inertia of 4.5 * 10^-3 kg∙m² is suspended from the ceiling.A rope passes over it with a 2.0-kg block attached to one end and a 4.0-kg block attached to the other.The rope does not slip on the pulley.When the velocity of the heavier block is 2.0 m/s the total kinetic energy of the pulley and blocks is: A)2.0 J B)12 J C)14 J D)22 J E)28 J

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Quizplus · Accepted Answer

The total kinetic energy (KE) of the system includes the translational kinetic energy of the two blocks and the rotational kinetic energy of the pulley. The translational kinetic energy (KE_trans) for each block is given by $KE_{trans} = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the velocity. The rotational kinetic energy (KE_rot) of the pulley is given by $KE_{rot} = \frac{1}{2}I\omega^2$, where $I$ is the moment of inertia (rotational inertia) and $\omega$ is the angular velocity.Given:- Mass of block 1 (m1) = 2.0 kg- Mass of block 2 (m2) = 4.0 kg- Velocity of blocks (v) = 2.0 m/s- Radius of pulley (r) = 0.03 m (3.0 cm)- Rotational inertia of pulley (I) = $4.5 \times 10^{-3}$ kg·m²First, calculate the translational kinetic energy for both blocks:$$KE_{trans1} = \frac{1}{2}m_1v^2 = \frac{1}{2} \times 2.0 \times (2.0)^2 = 4.0 \, \text{J}$$$$KE_{trans2} = \frac{1}{2}m_2v^2 = \frac{1}{2} \times 4.0 \times (2.0)^2 = 8.0 \, \text{J}$$To find the rotational kinetic energy of the pulley, we first need to find the angular velocity ($\omega$). The linear velocity (v) is related to angular velocity ($\omega$) by $v = r\omega$, so:$$\omega = \frac{v}{r} = \frac{2.0}{0.03} = \frac{200}{3} \, \text{rad/s}$$Now, calculate the rotational kinetic energy of the pulley:$$KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 4.5 \times 10^{-3} \times \left(\frac{200}{3}\right)^2$$$$KE_{rot} = \frac{1}{2} \times 4.5 \times 10^{-3} \times \frac{40000}{9} = 4.5 \times 10^{-3} \times \frac{40000}{18}$$$$KE_{rot} = 10 \, \text{J}$$Finally, the total kinetic energy of the system is the sum of the translational kinetic energies of both blocks and the rotational kinetic energy of the pulley:$$KE_{total} = KE_{trans1} + KE_{trans2} + KE_{rot} = 4.0 + 8.0 + 10.0 = 22.0 \, \text{J}$$

A Pulley with a Radius of 3