Question

An Atwood Machine with a massive pulley is set into motion. The geometry of the pulley is a solid disk with an axis of rotation through its center. The pulley’s mass is <0.275 + A/100> kg and has a radius of <0.30 + B/10> m. A mass m₁ =< 0.325 + C/1000> kg is hung on one side and m₂ = 0.425 kg on the other. They are connected by very strong, massless string!(pulley is frictionless)

What is the linear acceleration of the system in m/s²?

A=3 B=1 C=12

Answer 1

given
m = 0.275 + 3/100 = 0.278 kg

R = 0.3 + 1/10 = 0.40 m

m1 = 0.325 + 12/1000 = 0.337 kg

m2 = 0.425 kg

let T1 is the tension in the string connected to m1.

let T2 is the tension in the string connected to m2.

let a is the acceleration of the blocks and alfa is the angular acceleration of the pulley.

net force acting on m1, Fnet1 = T1 - m1*g

m1*a = T1 - m1*g

T1 = m1*g + m1*a -------(1)

Net force acting on m2, Fnet2 = m2*g - T2

m2*a = m2*g - T2

T2 = m2*g - m2*a -----(2)

Net torque acting on the pulley, Tnet = I*alfa

(T2 - T1)*R = (1/2)*m*R^2*alfa

T2 - T1 = (1/2)*m*R*alfa

T2 - T1 = m*a/2 (since a = R*alfa)

m2*g - m2*a - (m1*g + m1*a) = m*a/2

m2*g - m1*g = a*(m/2 + m1 + m2)

a = (m2 - m1)*g/(m/2 + m1 + m2)

= ( 0.425 - 0.337 )*9.8/( 0.278 + 0.425 + 0.337)

= 0.829 m/s^2 <<<<<<<<-------------Answer