A 1.2 kg mass is held at rest on top of a frictionless and horizontal table. A l

Question

A 1.2 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 1.2 kg. The light string then supports a mass of 1.2 kg which is hanging in air a.The mass on the table is released and the suspended mass falls. What is the acceleration of the falling mass. b.What is the tension in the string which is attached to the sliding mass on the table? c.What is the tension of the string which supports the hanging mass?

Explanation / Answer

Assuming g = 10m/s^2 and considering that block m1 accelerates towards pulley and block m3 accelerates down and the 2 blocks have same magnitude of acc. as the string does not slip on the pulley. For m1, Forces on m1 are friction and tension in the horizontal and normal and gravity in the vertical. Since normal is a self adjusting force, therefore it can balance gravity. Therefore, by Newton's second law, we can write, T1 - µN = (m1)a & N = (m1)g = 10 => T1 - 6.4 = a .... (1) Similarly for m3, we have (m3)g - T3 = (m3)a => 30 - T3 = 3a ... (2) Now for the pulley m2, By Newton's Second law for rotation, I = (m2)(0.15)^2 = 0.01125 kgm^2 Angular acc. = a/0.15 rad/s^2 Torque = (T3)(0.15) - (T1)(0.15) Therefore, Torque = I(angular Acc.) => T3 - T1 = 0.5a ... (3) Solving (1), (2) and (3); a = 47.2/9 = 5.244 m/s^2 T1 = a + 6.4 = 11.644 N T3 = 0.5a + T1 = 14.267 N

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