What is the magnitude of the linear acceleration a of the hanging masses?

Question

An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2.3 kg mass of the pulley is concentrated on its rim, which is a distance 24.5 cm from the axle. The mass on the right is 1.18 kg and on the left is 1.88 kg.

The acceleration of gravity is 9.8 m/s^2.

What is the magnitude of the linear acceleration a of the hanging masses? Answer in units of m/s^2.

Answer 1

Look at FBDs of each object with down for the mass on the left as positive and up for the mass on the right as positive, and counter-clockwise for the pulley being positive

Since the masses are connected by a pulley, the accelerations of the masses are equal

The mass on the left
1.88*9.8-T1=1.88*a

The mass on the right
T2-1.18*9.8=1.18*a

For the pulley, consider torques

(T1-T2)*R=I*anga

For a hoop, the moment of inertia is

I=m*R^2
I=2.3*.245^2
=0.138

The angular acceleration is related to the linear acceleration as
anga=a/r

Here are the equations
1.88*9.8-T1=1.88*a
T2-1.18*9.8=1.18*a

(T1-T2)*R=0.138*a/.245

Solving for a

1.88*(9.8-a)=T1
T2=1.18(9.8+a)


T1-T2=1.88*(9.8-a)-1.18(9.8+a)

T1-T2=1.88*(9.8-a)-1.18(9.8+a)

=(1.88-1.18)*9.8-(1.88+1.18)*a

=6.86-3.06*a
plugging in
(6.86-3.06*a)=2.3*a

a=6.86/(3.06+2.3)

=1.28 m/s^2
For the left this down, for the right this is up

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