An Atwood machine consists of a mass of 3.5 kg connected by a light string to a mass of 6.0 kg over a frictionless pulley with a moment of inertia of 0.0352 $\mathrm { kg } \cdot \mathrm { m } ^ { 2 }$ and a radius of 12.5 cm. If the system is released from rest, what is the speed of the masses after they have moved through 1.25 m if the string does not slip on the pulley?
A) 4.0 m/s
B) 2.0 m/s
C) 2.3 m/s
D) 6.0 m/s
E) 5.0 m/s

Question

An Atwood machine consists of a mass of  3.5 kg connected by a light string to a mass of  6.0 kg over a frictionless pulley with a moment of inertia of  0.0352 $\mathrm { kg } \cdot \mathrm { m } ^ { 2 }$ and a radius of  12.5 cm. If the system is released from rest, what is the speed of the masses after they have moved through  1.25 m if the string does not slip on the pulley?
A)  4.0 m/s
B)  2.0 m/s
C)  2.3 m/s
D)  6.0 m/s
E)  5.0 m/s

Quizplus · Accepted Answer

To find the speed of the masses after they have moved through 1.25 m, we can use the principle of conservation of mechanical energy. Initially, the system has potential energy due to the height difference between the masses and no kinetic energy since it starts from rest. As the system is released, 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 $m_2 > m_1$, meaning $m_2$ will move downward and $m_1$ will move upward. The moment of inertia of the pulley is $I = 0.0352 \, \text{kg} \cdot \text{m}^2$ and the radius of the pulley is $r = 0.125 \, \text{m}$. The distance moved by the masses is $d = 1.25 \, \text{m}$.The change in gravitational potential energy ($\Delta U$) of the system as $m_2$ moves down and $m_1$ moves up by the same distance $d$ is given by $\Delta U = m_2gd - m_1gd = (m_2 - m_1)gd$, 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 $\omega = v/r$ 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 C.

An Atwood Machine Consists of a Mass of 3 kg⋅m2\mathrm { kg } \cdot \mathrm { m } ^ { 2 }kg⋅m2

An Atwood Machine Consists of a Mass of 3 $\mathrm { kg } \cdot \mathrm { m } ^ { 2 }$