An green hoop with mass mh = 2.8 kg and radius Rh = 0.12 m hangs from a string t

Question

An green hoop with mass mh = 2.8 kg and radius Rh = 0.12 m hangs from a string that goes over a blue solid disk pulley with mass md = 2.0 kg and radius Rd = 0.09 m. The other end of the string is attached to a massless axel through the center of an orange sphere on a flat horizontal surface that rolls without slipping and has mass ms = 4.1 kg and radius Rs = 0.24 m. The system is released from rest. 1) What is magnitude of the linear acceleration of the hoop? m/s2 2) What is magnitude of the linear acceleration of the sphere? m/s2 3) What is the magnitude of the angular acceleration of the disk pulley? rad/s2 4) What is the magnitude of the angular acceleration of the sphere? rad/s2 5) What is the tension in the string between the sphere and disk pulley? N 6) What is the tension in the string between the hoop and disk pulley? N 7) The green hoop falls a distance d = 1.52 m. (After being released from rest.) How much time does the hoop take to fall 1.52 m? s 8) What is the magnitude of the velocity of the green hoop after it has dropped 1.52 m? m/s 9) What is the magnitude of the final angular speed of the orange sphere (after the green hoop has fallen the 1.52 m)?

Explanation / Answer

the only force on the system is the weight of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The inert masses that must be accelerated are the ring, pulley, and the rolling sphere. The mass equivalent of M the pulley is found by torque t = F*R = I*a = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m The mass equivalent of the rolling sphere is found by: the sphere rotates around the contact point with the table. So using the theorem of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2 ---------------- 1)linear acceleration of hoop= 2.922 m/s^2 2) lin. accel. of sphere = 2.922 m/s 3) angular acceleration disk pulley: a = a/R = 2.922/0.09 = 32.46 rad/s^2 4) ang. acceleration sphere = a/R = 2.922/0.2 = 14.61 rad/s^2 5) tension between pulley and sphere = M*a = 7/5*4*2.922 = 16.3632 N 6) tension between hoop and pulley = m(hoop) (g - a) = 2.8(9.81 - 2.922) = 19.29 N 7) s = 1/2*at^2 --> t = v(2s/a) = v(2*1,56/2.922) --> t = 1.068 seconds 8) v = v(2as) = v(2*2.922*1.59) = 3.05 m/s 9) ? = v/R = 3.05/0.2 = 15.24 rad/s

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