moment of momentum theorem

Joε=(m 1-m2)gR？ ( 1)

Jo=J+(m 1+m2)R^2

Angular acceleration of pulley? ε = (m1-m2) gr/(j+(m1+m2) R2 direction as shown in the figure.

The acceleration of m 1? a=εr=(m 1-m2)gr^2/(j+(m 1+m2)r^2)？ (a) The direction is shown in the figure.

Go m 1:

m 1 * a = m 1 * g-t 1(2)

Rope tension t1t1= m1(g-a)

Take m2:

m2*a=T2-m2*g？ (3)

Rope tension T2? T2=m2(g+a)

When T 1=T2? m 1(g-a)=m2(g+a)

a=( m 1-m2)g/(m2+m 1)？ (4)

(4)=(a)？ J=0

That is to say, regardless of the quality of the pulley, the rope tension on both sides is equal.