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Question

Shadows www.webassign.net/web/Student/Assignme Response /submit? pos 28 dep 15843609 15. 013 points I Previous Answers SerPSE9 11.P034.Wi My Notes Ask Your Teacher A student sits on a freely rotating stool holding two dumbbells, each of mass 2.99 k (see figure below). When his arms are extended horizontally (Figure a), the dumbbells are 0.98 m g from the axis of rotation and the student rotates with an angular speed of 0.741 rads. The moment of inertia of the student plus stool is 2.67 kg m2 and is assumed to be constant The student pulls the dumbbells inward horizontally to a position 0.299 m from the rotation axis (Figure b (a) Find the new angular speed of the student. 2.66 The words freely rotating stool mean that there are no external torques applied. What is conserved in this case? rad/s (b) Find the kinetic energy of the rotating system before and after he pulls the dumbbells inward. 5.13 K before what is the moment of inertia of two 2.99 kg dumbbells, each a distance 0.98 from the axis of rotation? J 5.14 You cannot assume that kinetic energy is conserved here, so you must calculate the final kinetic energy from your previous results. J Need Help? Read it l watch 4 9 PM 4 3/28/2017 O Ask me anything

Explanation / Answer

here,
mass of dumbell, md = 2.99 kg
angular velocity of dumbells, w1 = 0.741 rad/s

moment of inertia student + stool, I = 2.67 kg.m^2

total inertia before, i1 = I + 2 * 2.99 * 0.98^2
total inertia before, i1 = 8.413 kg.m^2

Final inertia, I2 = I + 2 * 2.99 * 0.299^2
Final inertia, I2 = 3.205 kg.m^2


From conservation of angular momentum we have :
before extendeing the arms = after extensing the arms
w1 * I1 = w2 * I2

Final angular velocity, w2 = w1 * I1 / I2
Final angular velocity, w2 = 0.741 * 8.413 / 3.205
Final angular velocity, w2 = 1.945 rad/s

Intial Kinetic energy :
KEi = 0.5*I1*w1^2 = 0.5*8.413*0.741^2
KEi = 2.31 J

Final Kinetic energy :
KEf = 0.5*I2*w2^2 = 0.5*12.452*1.945^2
Kef = 23.553 J

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