Problem 5. The Figure below show two views of a double pulley system (two pulley

Question

Problem 5. The Figure below show two views of a double pulley system (two pulley wheels attached and sharing an axis of rotation) with radii r= 10 cm and R-30 cm. The pulleys are solid cylinders with mass 0.5 kg and 3.5 kg. A string is wrapped around the large pulley and attached to a mass m of 10 kg. A string is wrapped around the small pulley and attached to a mass M of 20 kg. The system is released from rest. What is the angular acceleration (both magnitude and direction) of the double pulley?

Explanation / Answer

The moment of inertia of the double pulley will be:

I = 1/2 (m r^2 + M R^2)

for the masses we can write

m1g - T1 = m1 a1

T1 = M g - M a

since, a1 = a2 = a

T2 - mg = m a2

T2 = m a + m g

Considering the pulley, the torque due to tension will be:

Tau = T1 r - T2 R

Also, Tau = I alpha

T1r - T2R = I alpha (1)

we know that, a = alpha R, So

T1 = M (g - alpha r)

T2 = m (g + alpha R)

putting I , T1 and T2 in (1) and solving for alpha we get

alpha = g(Mr - mR)/(R^2(m + m2/2) + R1^2 (M + m1/2))

alpha = 9.81 (10 x 0.1 - 20 x 0.3)/[0.1^2(20 + 0.5/2) + 0.3^2(10 + 3.5/2)] = 43 rad/s^2

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