Question

A m₁ = 16.6 kg mass and a m₂ = 10.1 kg mass are suspended by a pulley that has a radius of R = 12.4 cm and a mass of M = 2.60 kg, as seen in the figure below. The cord has a negligible mass and causes the pulley to rotate without slipping. The pulley rotates without friction. The masses start from rest d = 3.30 m apart. Treating the pulley as a uniform disk, determine the speeds of the two masses as they pass each other.

The answer is not 2.42 m/s. Thanks!

Answer 1

Since m1 > m2, it will accelerate downwards

m1a = m1g - T1------1

m2a = T2 - m2g--------2

For the Pulley, net torque

Tnet = T - T' = T1r - T2r = r(T1 - T2) = I.alpha = (1/2)MR^2.(a/R)

=> r(T1 - T2) = (1/2)MR^2.(a/R)

=> T1 - T2 = (1/2)Ma---------3

From 1, 2 and 3,

m1g - m1a - m2g - m2a = (1/2)Ma

16.6*9.81 - 16.6*a - 10.1*9.81 - 10.1*a = (1/2)(2.6)a

a = 2.28 m/s^2

Keep in mind that both boxes are accelerating, so their closing acceleration will be twice the value of a or 4.56 m/s^2

vf² = vi² + 2ad

vf² = (0 m/s)² + 2(4.56 m/s²)(3.30 m)

vf = 5.48 m/s