Answer:
(a) 9.52
(b) 3.81
Step-by-step 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