A pulley (in the form of a uniform disk) with mass 93 kg and a radius 27 cm is a
ID: 2186618 • Letter: A
Question
A pulley (in the form of a uniform disk) with mass 93 kg and a radius 27 cm is attached to the ceiling in a uniform gravitational field and rotates with no friction about its pivot. The masses are connected by a massless inextensible cord and T1, T2, and T3 are magnitudes of the tensions. What is the magnitude of the tension T1? The acceleration due to gravity is 9.8 m/s^2. Assume up is positive. Answer in units of N.
A pulley (in the form of a uniform disk) with mass 93 kg and a radius 27 cm is attached to the ceiling in a uniform gravitational field and rotates with no friction about its pivot. The masses are connected by a massless inextensible cord and T1, T2, and T3 are magnitudes of the tensions. What is the magnitude of the tension T1? The acceleration due to gravity is 9.8 m/s^2. Assume up is positive. Answer in units of N.Explanation / Answer
1) Let the acceleration of the 20 kg mass = a The 35 kg mass moves down with the acceleration of the same magnitude as the magnitude of the acceleration with which the 20 kg mass moves up. Therefore the acceleration of the 35 kg mass = -a For the 20 kg mass, 20 * a = T1 - 20 * 9.8 20 * a = T1 - 196 T1 = 196 + 20 a------------(1) For the 35 kg mass, 35 * (-a) = T1 - 35 * 9.8 35 * (-a) = T2 - 343 T2 = 343 - 35a-------------(2) The 20 kg mass moves up and the 35 kg mass moves down. Therefore the pulley will rotate in clockwise direction. Let M = mass of the pulley, R = radius of the puelly In the clockwise direction:- Net torque on the pulley = (T2 R - T1 R) = (T2 - T1)R Angular acceleration of the puelly alpha = a/R The pulley's moment of inertia I = (1/2)MR^2 (T2 - T1)R = I * alpha (T2 - T1)R = (1/2)MR^2 * a/R (T2 - T1)R = (1/2)MaR Dividing by R, T2 - T1 = (1/2)Ma Substituting M = 50 kg, T2 - T1 = (1/2)50 a T2 - T1 = 25 a------------(3) Subtracting (1) from (2) T2 - T1 = (343 - 35a) - (196 + 20a) T2 - T1 = 147 - 15 a--------------(4) From (3) and (4) 25 a = 147 - 15 a 40 a = 147 a = 147/40 = 3.675 m/s^2 Ans: 3.675 m/s^2 2) Acceleration of 50 kg pulley = 0 Acceleration of 20 kg mass = 3.675 m/s^2 Acceleration of 35 kg mass = -a = -3.765 m/s^2 Therefore the acceleration of the center of mass of the system = [50 * 0 + 20 * 3.675 + 35 * (-3.675)](50 + 20 + 25) = - 0.58 m/s^2 Ans: -0.58 m/s^2
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