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 W1 = 40 N and W2 = 50 N are suspended from points A1 and A2 by ropes which are wound round the two pulleys, determine acceleration of weight W2. Suppose g = 10 m/sec2.
Solution
Clockwise moment of W1 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, W2 shall have moved a distance s2 with acceleration a2 and shall have velocity of v2 = r2 ω , while W1 will move
Distance s1 = 2 s2 = 2 r2 θ .
with final velocity V1 = r1 ω = 2 V2
with acceleration .a1 = r1 α = 2 a2
Since, rk = 0.2 m.
I (m) = 28 × (0.2)2 = 1.12 kg m2
ω2 = 2 × α θ
By Using Principle of Conservation of Energy, and letting datum level as (A - A),
40 (s1 ) + 50 (so) = 50 (s1/2) + 50 (so) + (40/10) × (V1)2 /2+ 50 × (V2)2 /2+( Im ω2 )/2
40 (s ) - 25 s = ω2 [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 .