Torque applied pully with mass

[sungju1203]

An Atwood's machine consists of blocks of masses m1 = 8.4 kg and m2 = 20.7 kg attached by a cord running over a pulley as in the figure below. The pulley is a solid cylinder with mass M = 7.40 kg and radius r = 0.200 m. The block of mass m2 is allowed to drop, and the cord turns the pulley without slipping.

(a) Why must the tension T2 be greater than the tension T1?

(b) What is the acceleration of the system, assuming the pulley axis is frictionless?

(c) Find the tensions T1 and T2.

Does anyone have idea how pully w/ mass works?

 

[Dr.D]

Treat the pulley as a third body and draw three FBDs. Since the cord does not slip on the pulley, you have the tensions acting on the two sides of the pulley acting to give it an angular acceleration.

 

[nlightner]

I can explain the concept of torque and how it applies to an Atwood's machine. Torque is the measure of the force that causes an object to rotate around a fixed point or axis. In this case, the pulley acts as the fixed point and the force from the masses causes it to rotate.

(a) In order for the pulley to rotate, there must be a net torque acting on it. This means that the torque from the mass m2 must be greater than the torque from the mass m1. Since the tension T1 is pulling in the opposite direction of the rotation, it will contribute a negative torque, while the tension T2 will contribute a positive torque in the same direction as the rotation. Therefore, T2 must be greater than T1 to produce a net positive torque and allow the pulley to rotate.

(b) The acceleration of the system can be calculated using the equation for torque: net torque = moment of inertia * angular acceleration. In this case, the moment of inertia is the mass of the pulley multiplied by the square of its radius (I = Mr^2). Since the pulley is assumed to be frictionless, the net torque is equal to the difference between the torque from the two masses (T2r - T1r). Thus, we can solve for the angular acceleration (alpha) and then use the equation a = r * alpha to find the linear acceleration of the masses.

(c) To find the tensions T1 and T2, we can use the equation for the net force on each mass in the y-direction. Since the masses are connected by a cord, the acceleration of both masses will be the same. Therefore, we can set the net forces equal to each other and solve for the tensions. This will give us T1 = m1g - m1a and T2 = m2g + m2a, where g is the acceleration due to gravity. Substituting in the value for acceleration found in part (b), we can solve for the tensions.

In summary, an Atwood's machine uses the concept of torque to explain the relationship between the tensions in the cord and the masses involved. By understanding the forces and accelerations involved, we can calculate the tensions and acceleration of the system.