An Atwood machine is constructed using two wheels ( with themasses concentrated

Question

An Atwood machine is constructed using two wheels ( with themasses concentrated at the rims.) The left wheel has a mass of 2.3kg and radius 20.66 cm. The right wheel has a mass of 2.8 kg andradius 30.96 cm. The hanging mass on the left is 1.98 kg and on theright 1.31 kg. What is the acceleration of the system? An Atwood machine is constructed using two wheels ( with themasses concentrated at the rims.) The left wheel has a mass of 2.3kg and radius 20.66 cm. The right wheel has a mass of 2.8 kg andradius 30.96 cm. The hanging mass on the left is 1.98 kg and on theright 1.31 kg. What is the acceleration of the system?

Explanation / Answer

Let the tensions in left, middle and right sections of stringbe T1,T2,T3 respectively. So, if acceleration of m3 be a downwards, that of m4 wouldbe a up since the string is assumed ideal. So, m3g - T1 = m3a T3 - m4g = m4a Torque on m1 is (T1-T2)*R1 Torque on m2 = (T2-T3)*R2 R1 being radius of m1 and R2 being that of m2. Now their angular accelerations would be a/R1 and a/R2respectively in absence of slipping. So, the moments of inertia : I1 for m1 = m1*R1*R1,   I2 = m2*R2*R2 assuming the masses are at rims (ring like) So, (T1-T2)*R1 = m1*R1*R1*(a/R1) and (T2-T3)*R2 = m2*R2*R2*(a/R2) So, T1-T3 = (m1+m2)*a Also, for m3 and m4, (m3-m4)*g - (T1-T3) = (m3+m4)*a So, (m1+m2+m3+m4) * a = (m3-m4)*g => a = (m3-m4)*g / (m1+m2+m3+m4) Now putting the values, a = 0.7826 m/s2

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