An Atwood machine consists of an object of mass of 3.5 kg connected by a light string to an object of mass of 6.0 kg over a frictionless pulley with a moment of inertia of 0.0352 kg ∙ m² and a radius of 12.5 cm.If the system is released from rest,what is the speed of the objects after they have moved through 1.25 m if the string does not slip on the pulley? A)2.0 m/s B)2.3 m/s C)4.0 m/s D)5.0 m/s E)6.0 m/s

Question

Quizplus · Accepted Answer

To solve this problem, we can use the principle of conservation of mechanical energy. Since the system is released from rest, the initial kinetic energy is zero. The only initial energy in the system is the gravitational potential energy of the two masses. As the system moves, some of this potential energy is converted into kinetic energy of the masses and rotational kinetic energy of the pulley.Let $m_1 = 3.5 \, \text{kg}$ and $m_2 = 6.0 \, \text{kg}$, with $m_2$ being the heavier mass that moves down. The distance moved by the masses is $d = 1.25 \, \text{m}$, and the radius of the pulley is $r = 0.125 \, \text{m}$. The moment of inertia of the pulley is $I = 0.0352 \, \text{kg} \cdot \text{m}^2$.The change in gravitational potential energy ($\Delta U$) of the system as $m_2$ moves down and $m_1$ moves up by distance $d$ is given by $\Delta U = (m_2 - m_1)gd$, where $g = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity.This change in potential energy is equal to the sum of the kinetic energy of the two masses and the rotational kinetic energy of the pulley. The kinetic energy ($K$) of the masses is given by $K = \frac{1}{2}(m_1 + m_2)v^2$, where $v$ is the final speed of the masses. The rotational kinetic energy ($K_{rot}$) of the pulley is given by $K_{rot} = \frac{1}{2}I\omega^2$, where $\omega = \frac{v}{r}$ is the angular velocity of the pulley.Setting the change in potential energy equal to the total kinetic energy gives:$$(m_2 - m_1)gd = \frac{1}{2}(m_1 + m_2)v^2 + \frac{1}{2}I\left(\frac{v}{r}\right)^2$$Plugging in the given values:$$(6.0 \, \text{kg} - 3.5 \, \text{kg})(9.8 \, \text{m/s}^2)(1.25 \, \text{m}) = \frac{1}{2}(3.5 \, \text{kg} + 6.0 \, \text{kg})v^2 + \frac{1}{2}(0.0352 \, \text{kg} \cdot \text{m}^2)\left(\frac{v}{0.125 \, \text{m}}\right)^2$$Solving this equation for $v$ gives a value of approximately $2.3 \, \text{m/s}$, which corresponds to choice B.

An Atwood Machine Consists of an Object of Mass of 3.5