The Atwood machine consists of two masses hanging from the ends of a rope that passes over a pulley. The pulley can be approximated by a uniform disk with mass mp=4.93 kg and radius rp=0.250 m. The hanging masses are mL=19.1 kg and mR=11.5 kg. Calculate the magnitude of the masses' acceleration a, and the tension in the left and right ends of the rope, TL, and TR, respectively. The image depicts a basic physics setup known as an Atwood machine. It consists of two masses, \( m_L \) and \( m_R \), connected by a rope that passes over a pulley. **Description:**- **Pulley:** The pulley is shown as a gray circular object at the top of the image. It allows the rope to move freely around it, enabling the masses to move vertically.- **Mass \( m_L \):** This is the larger mass on the left side of the pulley, depicted as a blue cylinder labeled \( m_L \).- **Mass \( m_R \):** This is a smaller mass on the right side, also depicted as a blue cylinder labeled \( m_R \).The system is used to study the principles of mechanics, particularly Newton's second law, by analyzing the acceleration of the masses and the tension in the rope.

Name: The Atwood machine consists of two masses hanging from the ends of a r
Uploaded: 2024-03-20T06:58:04.000Z
Channel: Bartleby
Description: The Atwood machine consists of two masses hanging from the ends of a rope that passes over a pulley. The pulley can be approximated by a uniform disk with mass mp=4.93 kg and radius rp=0.250 m. The hanging masses are mL=19.1 kg and mR=11.5 kg. Calcul

Given,mass of pully =4.93 kg radius of pully =0.250 m mass of left block = 19.1 kg mass of right block = 11.5 kgFor left mass block TL=mLg-mLaTL=19.1*9.81-19.1a TL=187.371-19.1a ........ (1) and for right block TR=mRg+mRaTR=11.5*9.81+11.5a TR=112.815+11.5a ............. (2) and for pulley τ=Iαand moment of inertia I=12mpR2thenTL-TRR=12mpR2*aRTL-TR=12*4.93aTL-TR=2.465a from equation (1) and (2) we have 187.371-19.1a-112.815-11.5a=2.465a33.065a=74.556a=2.25 m/s2 TL=187.371-19.1*2.25TL=144.4 Nand TR=112.815+11.5*2.25TR=138.7 NAnswer (a) acceleration = 2.25 m/s2 (b) TL = 144.4 N (c) Tc =138.7 N