Question

A mass (M1=5.0kg) is connected by a light cord to a mass (M2=4.0kg) which slides on a smooth frctionless surface. The pulley( a solid disk r=0.20m) rotates about a frictionless axle. The accleration of M2 is 3.5 m/s^2. What is the mass of the pulley?

Answer 1

We know that
torque (T) = I*  
where I is moment of inertia of pulley and is angular acceleration of the pulley.
Torque (T) = (T₁ - T₂)r
where r is radius of pulley , hence
(T₁ - T₂) = I*( /r)
we know that = a/r
where a is linear acceleration , therfore
(T₁ - T₂) = I*(a/r²) -------------------(1)
Now for the mass M₂
T₂ = M₂a = 4*3.5 = 14 N
Now for the mass M₁
T₁ = M₁g - M₁a = M₁(g-a) = 5*(9.81 -3.5) =31.55 N
NOw putting the value in equation 1
(31.55 -14) = I*(3.5/0.2²)
I = 0.201 kg-m²
Now we know that moment of inertia of pulley (I) = mr²/2
0.201 = m(0.2²/2)
m = 10.03 kg