By means of a rope whose mass is negligible, two blocks of mass 11.0 kg and 44.0

Question

By means of a rope whose mass is negligible, two blocks of mass 11.0 kg and 44.0 kg are suspended over a pulley. The pulley can be treated as a uniform solid cylindrical disk. The downward acceleration of the 44.0 kg block is observed to be exxactly one-half the acceleration due to gravity. Noting that the tension in the rope is not the same on each side of the pulley, find the mass of the pulley.

Explanation / Answer

Given : Masses of two blocks are m1 = 11.0 kg and m2 = 44.0 kg Here m2 > m1. So the block of mass m2 moving downwards and the block of mass m1 moving upwards. ---------------------------------------------------------------------------------------------------- The sum of forces on each box, andthe sum of torques for the pulley: First Block: T1 - m1 g = m1 a T1 = m1 a + m1 g = m1 (a + g) You are told that a is one half g so a= 4.9 m/s^2 Then T1 = (11.0kg) (4.9 m/s^2+ 9.8 m/s^2 ) =161.7 N --------------------------------------------------------------------------------------------- For the second block: m2 g - T2 = m2 a T2 = m2 (g - a ) T2 = (44kg)* (9.8 m/s^2 - 4.9 m/s^2) = 215.6 N -------------------------------------------------------------------------------------------- Pulley: T2 R - T1 R = I a Now in the pulley equation, we replace awith a / r and T2 R - T1 R = I (a /R) (R^2)( T2 - T1 ) = Ia now replace T2 and T1 with the above values, we get (R^2)( 215.6 N - 161.7 N) = (I)(4.9 m/s^2) I = (11)(R^2) the pulley is a disk, so I = (1/2) (M R^2)so (1/2) (M R^2) = (11)(R^2) M = 22 kg mass of thepulley

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