PLEASE PROVIDE EXPLANATION

(a) 9.52

(b) 3.81

Explanation:

Draw a free body diagram of block m.  There are two forces:

Weight mg pulling down, and tension T₁ pulling up.

Draw a free body diagram of block M.  There are two forces:

Weight Mg pulling down, and tension T₂ pulling up.

Draw a free body diagram of the double pulley.  There are two torques:

Tension T₂ pulling counterclockwise at r, and tension T₁ pulling clockwise at R.

Let's assume that M moves down, m moves up, and the pulley rotates counterclockwise.

Sum of forces on block m in the +y direction:

∑F = ma

T₁ − mg = ma₁

T₁ = mg + ma₁

Sum of forces on block M in the -y direction:

∑F = ma

Mg − T₂ = Ma₂

T₂ = Mg − Ma₂

Sum of the torques about the pulley's center in the counterclockwise direction:

∑τ = Iα

T₂ r − T₁ R = (½ m₀ r² + ½ m₀ R²) α

T₂ r − T₁ R = ½ m₀ (r² + R²) α

Substitute:

(Mg − Ma₂) r − (mg + ma₁) R = ½ m₀ (r² + R²) α

The strings don't slip on the pulley, so a₁ = αR and a₂ = αr.

(Mg − Mαr) r − (mg + mαR) R = ½ m₀ (r² + R²) α

Mgr − Mαr² − mgR − mαR² = ½ m₀ (r² + R²) α

g (Mr − mR) − α (Mr² + mR²) = ½ m₀ (r² + R²) α

g (Mr − mR) = [½ m₀ (r² + R²) + Mr² + mR²] α

α = g (Mr − mR) / [½ m₀ (r² + R²) + Mr² + mR²]

α = (10) (9×0.2 − 1×0.5) / (½ (0.5) (0.2² + 0.5²) + 9(0.2)² + 1(0.5)²]

α = 19.0 rad/s²

Therefore:

a₁ = αR = (19.0 rad/s²) (0.5 m) = 9.52 m/s²

a₂ = αr = (19.0 rad/s²) (0.2 m) = 3.81 m/s²

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