A string is wrapped tightly around a fixed pulley that has a moment of inertia of 0.0352 kg ∙ m² and a radius of 12.5 cm.A mass of 423 g is attached to the free end of the string.With the string vertical and taut,the mass is gently released so it can descend under the influence of gravity.As the mass descends,the string unwinds and causes the pulley to rotate,but does not slip on the pulley.What is the speed of the mass after it has fallen through 1.25 m?

Question

Quizplus · Accepted Answer

To find the speed of the mass after it has fallen 1.25 m, we can use the principle of conservation of mechanical energy. The initial mechanical energy (potential energy) is converted into translational kinetic energy of the mass and rotational kinetic energy of the pulley as it descends. The potential energy lost by the mass is equal to the sum of its kinetic energy and the pulley's kinetic energy.The potential energy lost by the mass is given by $PE = mgh$, where $m = 0.423 \, \text{kg}$ (mass of the object), $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity), and $h = 1.25 \, \text{m}$ (height fallen).The translational kinetic energy of the mass is given by $KE_{trans} = \frac{1}{2}mv^2$, where $v$ is the final velocity of the mass.The rotational kinetic energy of the pulley is given by $KE_{rot} = \frac{1}{2}I\omega^2$, where $I = 0.0352 \, \text{kg} \cdot \text{m}^2$ is the moment of inertia of the pulley, and $\omega$ is the angular velocity of the pulley. Since the string does not slip on the pulley, the linear speed of the mass $v$ is related to the angular velocity of the pulley by $v = r\omega$, where $r = 0.125 \, \text{m}$ is the radius of the pulley.Setting the initial potential energy equal to the sum of the translational and rotational kinetic energies, we get:$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v}{r}\right)^2$$Plugging in the values:$$0.423 \cdot 9.8 \cdot 1.25 = \frac{1}{2} \cdot 0.423 \cdot v^2 + \frac{1}{2} \cdot 0.0352 \cdot \left(\frac{v}{0.125}\right)^2$$Solving this equation for $v$ gives the speed of the mass after it has fallen 1.25 m. The correct answer after solving the equation is approximately $v = 1.97 \, \text{m/s}$, which corresponds to choice C.

A String Is Wrapped Tightly Around a Fixed Pulley That