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+// SAMPLE PROBLEM 6/9
+clc;funcprot(0);
+// Given data
+F=100;// N
+m=40;// kg
+k=0.150;// m
+theta=15;// degree
+r_i=0.100;// m
+r_o=0.200;// m
+l=3;// The distance in m
+g=9.81;// The acceleration due to gravity in m/s^2
+
+// Calculation
+W=m*g;// N
+l=(r_o+r_i)/r_i;// m
+U_12=(F*((r_o+r_i)/r_i)*l)-((W*sind(theta)*l));// J
+T_1=0;// J
+// T_2=((1/2)*m*vbar^2)+((1/2)*Ibar*omega^2);
+// The work-energy equation gives
+omega=sqrt((T_1+U_12)/(((1/2)*m*(r_i)^2)+((1/2)*m*k^2)));// rad/s
+// Alternatively, the kinetic energy of the wheel may be written
+// T=(1/2)*I_C*omega^2
+P_100=F*(r_o+r_i)*omega;// W
+printf("The power input,P=%3.0f W",P_100);