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.)
To find the acceleration of the suspended masses, we need first to determine the net torque acting on the system. This can be obtained by considering the torques due to the gravitational forces acting on the masses and the frictional torque.
Let tension on string T1 be acting on m1 and tension T2 acting on m2, and let the angular acceleration of the pulley be α.
For mass m1:
m1 * g - T1 = m1 * a (1)
For mass m2:
T2 - m2 * g = m2 * a (2)
Next, we apply Newton's second law in rotational form, to the pulley.
The torque caused by T1 and T2 is given by:
τ_net = T2 * r - T1 * r = I * α (3)
Here, τ_net is the net torque acting due to tensions T1 and T2 and r is the radius of the pulley, whereas I is the moment of inertia of the pulley.
But the α is also equal to a / r (i.e., acceleration divided by the radius of the pulley).
Substitution α with a/r and the fact that the moment of inertia of a solid cylindrical pulley is given by I = (1/2) * M_pulley * r^2, we get:
τ_net = T2 * r - T1 * r = (1/2) * M_pulley * r * a (4)
Using equations (1) and (2), we can express T1 and T2 in terms of m1 * a and m2 * a, thus:
m1 * g - m1 * a = T1
T2 = m2 * a + m2 * g
Now substitute these expressions for T1 and T2 into equation 4 and obtain an equation in terms of a only:
τ_net = (m2 * a + m2 * g - m1 * g + m1 * a) * r = (1/2) * M_pulley * r * a
Substitute the given values to calculate the acceleration (remember to also include the frictional torque in your calculation):
0.35 N = ((0.7 kg * a + 0.7 kg * 9.8 m/s² - 0.3 kg * 9.8 m/s² + 0.3 kg * a) * 0.145 m) = (1/2) * 0.2 kg * 0.15 m * a
0.35 N = (0.7 * a + 6.86 - 2.94 + 0.3 * a) * 0.15
0.35 N = (1 * a + 3.92) * 0.15
0.35 N = 0.15 * a + 0.588
a = (0.35 - 0.588) / 0.15
a = -1.59 m/s²
Since we are asked for the magnitude of the acceleration, we take the absolute value:
a = 1.59 m/s²
To find the magnitude of the acceleration of the suspended masses, we can use Newton's second law of motion.
First, let's consider the forces on the system. The tension in the string will be the same on both sides of the pulley.
For mass m1:
The force acting on m1 is the tension in the string, T, and the force of gravity, which is its weight, mg. The weight of m1 is given by mg1 = m1 * g, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
For mass m2:
The force acting on m2 is the tension in the string, T, and the force of gravity, which is its weight, mg. The weight of m2 is given by mg2 = m2 * g.
Next, let's consider the torques acting on the pulley.
The net torque on the pulley is given by the torque due to the frictional force and the torque due to the tension in the string.
The torque due to friction is calculated as torque_friction = friction_force * radius, where the friction_force is the constant torque of 0.35 m·N given in the question and the radius is the radius of the pulley (0.15 m).
The torque due to the tension in the string is T * radius, where T is the tension in the string and radius is the radius of the pulley.
Since the pulley is in rotational equilibrium (not rotating), the net torque acting on the pulley will be zero.
Now we can write down the equations of motion for the system:
For mass m1:
T - m1 * g = m1 * a
For mass m2:
m2 * g - T = m2 * a
For the pulley:
torque_friction + T * radius = 0
Now we can solve these equations simultaneously to find the magnitude of the acceleration of the suspended masses.
To find the magnitude of the acceleration of the suspended masses in an Atwood machine, we can use the concept of torque and Newton's laws of motion.
First, let's analyze the forces acting on the system. We have two masses, m1 and m2, connected by a string that passes over a pulley. The gravity is acting downwards on both masses with a force of m1*g and m2*g, respectively, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Next, let's consider the torque acting on the pulley due to the friction between the rotating pulley and its axle. Torque is given by the formula torque = force * lever arm. In this case, the force is the constant torque (0.35 m·N) and the lever arm is the radius of the pulley (0.15 m). Therefore, the torque is 0.35 m·N * 0.15 m = 0.0525 N·m.
According to Newton's second law of motion, the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the net force is the difference between the forces due to gravity on m1 and m2, and the force due to the torque. Hence, we have the equation:
(m2 * g) - (m1 * g) - (torque / radius) = (m2 + m1) * acceleration
Substituting the given values, we have:
(0.70 kg * 9.8 m/s^2) - (0.30 kg * 9.8 m/s^2) - (0.0525 N·m / 0.15 m) = (0.30 kg + 0.70 kg) * acceleration
Simplifying the equation:
6.86 N - 2.94 N - 0.35 N = 1 kg * acceleration
3.57 N = 1 kg * acceleration
Now we can solve for the acceleration:
acceleration = 3.57 N / 1 kg
acceleration = 3.57 m/s^2
Therefore, the magnitude of the acceleration of the suspended masses is 3.57 m/s^2.