A uniform disk of radius R and mass M is mounted on an axle supported in fixed frictionless ?

For a uniform disk, the moment of inertia is

I = ½MR²

torque τ = I*α = ½MR² * α

and also

τ = T * R, so

T*R = ½MR² * α

means that

α = 2T / MR ◄

and the tangential acceleration at the rim is

a = α * R = 2T / M ◄

b/ In my opinion, this is most easily done using the concept of "effective mass" on the pulley. For an object with moment if inertia

I = kMR²,

the effective mass is

m' = kM

so for this part, the motive force is

F = m*g

and the total mass of the system is

m_t = m + M/2

making the linear acceleration at the rim

a = F / m_t = m*g / (m + M/2) ◄

which can also be expressed as

a = 2mg / (2m + M)

and so the radial acceleration is

α = a / R = m*g / R(m + M/2) ◄

Hope this helps!

moment of inertia J = M/2*R^2

accelerating torque Ta = T*R

angular acceleration α = Ta / J = T*R*2/(M*R^2) = 2*T/(M*R)

tangential acceleration at = α*R = 2*T*R/(M*R) = 2T/M