A textbook of mass 2.05 rests on a frictionless, horizontal surface. A cord atta

Question

A textbook of mass 2.05 rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.190 , to a hanging book with mass 2.94 . The system is released from rest, and the books are observed to move a distance 1.29 over a time interval of 0.760 . the tension in the part of the cord attached to the textbook is 9.16 N. The tension in the part of the cord attached to the book is 15.7 N. Then!!!! What is the moment of inertia of the pulley about its rotation axis? Take the free fall acceleration to be = 9.80 .

Explanation / Answer

we find the linear acceleration by knowing the masses moved 1.29 m in 0.760s distance = 1/2 at^2 => a=2d/t^2 a=2*1.29m/0.760s^2=4.66m/s/s now apply newton's second law to the 2.05kg book: T1= 2a =>T1=2.05kg*4.66m/s/s=9.553N apply newton's second law to T2: T2-mg = -ma T2=2.94g-2.94a=2.94(9.8m/s/s-4.66m/s/s)=15.11N now, consider the pulley T1 exerts a force of 9.533 N in one direction, and T2 exerts a force of 15.11N in the opposite direction, the net force on the pulley of 5.577N generates a torque the amount of torque =(T2-T1)R since this force acts a distance R from the rotation axis of the pulley this torque produces an angular acceleration equal to torque = I alpha where I is the moment of inertia and alpha is the angular acceleration alpha is related to linear acceleration according to a=R alpha or alpha =a/R, so we combine all these and get (T2-T1)R=I(a/R) I=(T2-T1)R^2/a =5.577N*(0.095m)^2/4.66m/s/s I=10.8x10^(-3)kgm^2

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