Two masses are suspended from a pulley as shown in the figure below (the Atwood machine). The pulley itself has a mass of 0.20 kg, a radius of 0.15 m, and a constant torque of 0.35 m·N due to the friction between the rotating pulley and its axle. What is the magnitude of the acceleration of the suspended masses if m1 = 0.30 kg and m2 = 0.70 kg? (Neglect the mass of the string.)
1.2
To find the magnitude of the acceleration of the suspended masses in the Atwood machine, we can use Newton's second law of motion. According to this law, the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Let's break down the forces acting on the system:
1. Gravitational force: Each mass experiences a gravitational force pulling it downward. The force on m1 is m1 * g, and the force on m2 is m2 * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
2. Tension in the string: Since the masses are connected by a string, they experience a tension force. The tension force acts in opposite directions on m1 and m2. We can call the tension force T.
3. Frictional torque on the pulley: The pulley experiences a torque due to friction between the rotating pulley and its axle. This torque opposes the motion of the pulley and introduces a net force in the system. The magnitude of this torque is given as 0.35 m·N.
Now, let's write down the equations for the forces acting on the masses:
For m1:
m1 * g - T = m1 * a
For m2:
T - m2 * g = m2 * a
Next, let's consider the rotational motion of the pulley. The net torque acting on the pulley is equal to the product of its moment of inertia (I) and its angular acceleration (α). Since the angular acceleration is equal to the linear acceleration (a) divided by the radius of the pulley (r), we can write:
Torque_friction = I * α
0.35 m·N = (1/2) * m_pulley * r^2 * (a / r)
0.35 m·N = (1/2) * 0.20 kg * (0.15 m)^2 * a
Now, we have three equations (equations for m1, m2, and the pulley) with three unknowns (T, a, and α). We can solve these equations simultaneously to find the values of T and a.
First, let's solve the equation for T by adding equations for m1 and m2:
m1 * g - T + T - m2 * g = m1 * a + m2 * a
(m1 - m2) * g = (m1 + m2) * a
a = (m1 - m2) * g / (m1 + m2)
Now that we have the value of a, we can substitute it back into one of the equations for m1 or m2 to find the tension T. Let's choose the equation for m1:
m1 * g - T = m1 * a
T = m1 * g - m1 * a
Finally, we can substitute the known values (m1 = 0.30 kg, m2 = 0.70 kg, g = 9.8 m/s^2) to calculate the magnitude of the acceleration.
a = (0.30 kg - 0.70 kg) * 9.8 m/s^2 / (0.30 kg + 0.70 kg)
a = -0.40 m/s^2
The negative sign indicates that the masses are accelerating in opposite directions. Taking the magnitude of the acceleration gives:
|a| = 0.40 m/s^2
Therefore, the magnitude of the acceleration of the suspended masses is 0.40 m/s^2.