Determine acceleration of weight:
Two pulleys of radii 200 mm and 400 mm are mounted co-axially and rigidly on a common-shaft.
These pulleys have a overall mass of 28 kg and radius of gyration of 200 mm of the combination. If weights W_{1} = 40 N and W_{2} = 50 N are suspended from points A_{1} and A_{2} by ropes which are wound round the two pulleys, determine acceleration of weight W_{2}. Suppose g = 10 m/sec^{2}.
Solution
Clockwise moment of W_{1} about centre of pulley
= 40 × 0.4 = 16 N.m.
Anticlockwise moment of W2 about centre of pulley = 10 N.m.
Therefore the pulley shall rotate clockwise with angular acceleration α about the centre. After some time, W_{2} shall have moved a distance s_{2} with acceleration a_{2} and shall have velocity of v_{2} = r_{2} ω , while W_{1} will move
Distance s_{1} = 2 s_{2} = 2 r_{2} θ .
with final velocity V_{1} = r_{1} ω = 2 V_{2}
with acceleration .a_{1} = r_{1} α = 2 a_{2}
Since, r_{k } = 0.2 m.
I _{(m) }= 28 × (0.2)^{2} = 1.12 kg m^{2}
ω^{2} = 2 × α θ
By Using Principle of Conservation of Energy, and letting datum level as (A - A),
40 (s_{1} ) + 50 (s_{o}) = 50 (s_{1}/2) + 50 (s_{o}) + (40/10) × (V_{1})^{2} /2+ 50 × (V_{2})^{2 }/2+( I_{m} ω^{2 })/2
40 (s ) - 25 s = ω^{2} [2 × (0.4)^{2} + 2.5 × (0.2 )^{2} +(1.12/2)]
15 × (0.4 θ) = 2 α θ [0.32 + 0.1 + 0.56]
∴ α = 3.06 rad / sec^{ 2}
α ≈ 3 rad / sec ^{2} .