Two masses, mA = 40.0 kg and mB = 55.0 kg, are connected by a massless cord that passes over a pulley that is free to rotate about a fixed axis. This device is known as an Atwood’s Machine. The pulley is a solid cylinder of radius R = 0.360 m and mass of 7.50 kg. Note: The tensions on the two masses are not equal. There is a figure showing a pulley in the center and the two masses hanging freely on either side vertically. mA is going up, mB is going down.
A) Determine the linear acceleration of the two masses.
B) Determine the linear acceleration of the two masses if the pulley mass were ignored so that I = 0 for the pulley
To determine the linear acceleration of the two masses in the given Atwood's Machine setup, we need to analyze the forces and use Newton's second law.
A) To determine the linear acceleration of the two masses:
1. Identify the forces acting on the masses:
- Tension force (T1) acting on mass mA (upward)
- Tension force (T2) acting on mass mB (downward)
- Weight of mass mA (mg1) acting downward
- Weight of mass mB (mg2) acting downward
2. Write down the equation of motion for both masses using Newton's second law (F = ma), where a is the linear acceleration:
- For mass mA: T1 - mg1 = ma1
- For mass mB: mg2 - T2 = ma2
3. Recognize that the tensions T1 and T2 are related because they are connected by the same cord passing over the pulley. The relation is given by:
- T2 = T1 - (2/3)ma1a
4. Apply the rotational equilibrium condition for the pulley:
- The sum of the torques about the axis of rotation is zero. In this case, the only torque comes from the tension forces on the pulley. The torque is given by:
- T2R - T1R = (1/2)Iα
5. Recognize that the tension forces on the pulley can be related to the linear acceleration using the radius of the pulley, R:
- T1 = (1/2)mAa
- T2 = (1/2)mBa
6. Use the equations obtained in steps 1-5 to eliminate the tensions and simplify the equations of motion. The final equations should be:
- 2mAa - 2/3mAa = (3/2)mAa = mBa
- 3/2mAa + mBa = (mA + mB)g
7. Solve the obtained equations simultaneously to find the linear acceleration, a.
B) To determine the linear acceleration of the two masses if the pulley mass were ignored (I = 0 for the pulley):
1. Follow steps 1-4 as mentioned in part A.
2. Recognize that since the pulley mass is ignored, the tensions T1 and T2 are equal: T1 = T2 = T.
3. Apply the rotational equilibrium condition for the pulley (without considering pulley mass):
- The sum of the torques about the axis of rotation is zero. In this case, the only torque comes from the tension forces on the pulley. The torque is given by:
- TR - TR = 0
4. Recognize that the tension forces on the pulley can be related to the linear acceleration using the radius of the pulley, R:
- T1 = mAa
- T2 = mBa
5. Use the equations obtained in steps 1-4 to eliminate the tensions and simplify the equations of motion. The final equations should be:
- mAa = mBa
- mAa + mBa = (mA + mB)g
6. Solve the obtained equations simultaneously to find the linear acceleration, a.