A 1.5 kg mass is held at rest on top of a frictionless and horizontal table. A l

Question

A 1.5 kg mass is held at rest on top of a frictionless and horizontal table. A light string loops over a pulley which is in the shape of a 10 cm radius solid disk which has a mass of 1.5 kg. The light string then supports a mass of 1.5 kg which is hanging in air. The mass on the table is released and the suspended mass falls. What is the acceleration of the falling mass. What is the tension in the string which is attached to the sliding mass on the table? What is the tension of the string which supports the hanging mass?

Explanation / Answer

mass on table: T1 = M a hanging mass M g - T2 = Ma pulley torque = I alpha r T2 - r T1 = 1/2 M r^2 (a/r) T2 - T1 = 1/2 M a add the three equations together Mg = 5/2 M a a = 2/5 g = 3.924 m/s^2 b) T1 = 1.5*3.924 = 5.886 N c) 1.5*9.81 - T2 = 1.5*3.924 T2 = 1.5*(9.81-3.924)=8.829 N

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