Question

the moment of inertia of the pulley system as shown in the figure is 3 kg -m2 cube the radii of bigger and smaller pulley are 2 and 1 m respectively. As the system is released from rest find the angular acceleration of the pulley system

Answer 1

M = mass of the heavier block = 6 kg

m = mass of the lighter block = 3 kg

r₁ = radius of larger pulley = 2 m

r₂ = radius of smaller pulley = 1 m

T₁ = tension force in the rope connected with the heavier block

T₂ = tension force in the rope connected with the lighter block

a = linear acceleration of the blocks

α = angular acceleration of the pulley

I = moment of inertia of the pulley = 3 kgm²

for the heavier block, force equation is given as

Mg - T₁ = Ma

T₁ = Mg - Ma                                eq-1

for the lighter block, force equation is given as

T₂ - mg = ma

T₂ = mg + ma                             eq-2

for the pulley , torque equation is given as

T₁ r₁ - T₂ r₂ = I α

using eq-1 and eq-2

(Mg - Ma) r₁ - (mg + ma)  r₂ =  I α

we know that , a = r α, hence

(Mg - M(r₁ α)) r₁ - (mg + m(r₂ α))  r₂ =  I α

inserting the values

((6 x 9.8) - 6 (2 α)) 2 - ((3 x 9.8) + 3(1 α))  1 =  3 α

α = 2.94 rad/s²

Answer 2

View answer in the attachment....and please pardon my handwriting.....

Hope it helps