An Atwood machine has a mass of 3.50 kg connected by a light string to a mass of 6.00 kg over a 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 masses after they have moved through 1.25 m? A) 2.00 m/s B) 2.28 m/s C) 4.00 m/s D) 4.95 m/s E) 6.00 m/s

Question

Quizplus · Accepted Answer

First, we need to find the net force acting on the system. Using Newton's second law, 
$F_{net} = (m_2 - m_1)g$, where $m_1$ = 3.50 kg and $m_2$ = 6.00 kg.
$F_{net} = (6.00 	ext{ kg} - 3.50 	ext{ kg})\cdot 9.81 	ext{ m/s}^2 = 23.4 	ext{ N}$

Next, we can use the torque equation, $	au = I\alpha$, to find the angular acceleration of the pulley. Since the pulley is a disk, $I = \frac{1}{2}MR^2$.
$	au = Fr = (m_2 - m_1)g\cdot r = 23.4	ext{ N}\cdot 0.125	ext{ m} = 2.93 	ext{ N}\cdot	ext{m}$
$I\alpha = 	au \Rightarrow \alpha = \frac{	au}{I} = \frac{2\cdot 2.93	ext{ N}\cdot	ext{m}}{0.0352	ext{ kg}\cdot	ext{m}^2} = 166.2 	ext{ rad/s}^2$

Now we can apply kinematic equations to find the speed of the masses after they have moved a distance of 1.25 m. Since the pulley does not slip, the distance travelled by the masses is equal to the distance travelled by the edge of the pulley, which is $1.25m\cdot\frac{2}{\pi} \approx 0.795m$.
Using the kinematic equation $v^2 = v_0^2 + 2a\Delta x$, where $v_0 = 0$ and $\Delta x = 0.795m$ (for the edge of the pulley):
$v = \sqrt{2a\Delta x} = \sqrt{2\cdot 166.2 	ext{ rad/s}^2\cdot 0.795	ext{ m}} \approx 2.28 	ext{ m/s}$
Therefore, the answer is (B) 2.28 m/s.

An Atwood Machine Has a Mass of 3