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+//to find maximum twist,apeed of flywheels when twist is maximum and when springs regains its shape
+clc
+//given
+m1=500//lb ft^2
+m2=1500//lb ft^2
+k=150//lb ft^2
+w1=150//rpm
+//angular momentum will be conserved as net external force is zero
+//let final angular velocity be N then (m1+m2)N=w1*m1
+N=(w1*m1)/(m1+m2)
+printf("Angular velocity at the instant when speeds of the flywheels are equalised is given by %.2f r.p.m\n",N)
+//kinetic energy at this instance
+ke1=(1/2)*((m1+m2)/32.2)*((%pi*N)/30)^2
+printf("The kinetic energy of the system at this instance is %.2f ft lb\n",ke1)
+printf("which is almost equal to 480 ft lb \n")
+//initial kinetic energy
+ke0=(1/2)*((m1)/32.2)*((%pi*w1)/30)^2
+printf("The initial kinetic energy of the system is %.2f ft lb\n",ke0)
+printf("which is almost equal to 1915 ft lb \n")
+//strain energy = s
+s=ke0-ke1
+printf("strain energy stored in the spring is %.2f ft lb which is approximately 1435 ft lb\n",s)
+//if x is the maximum anglular displacement of wheel and the mean torque applied by spring is i/2*k*x then work done or strain energy is given by 1/2 *k*x^2
+x=((1435*2)/150)^.5
+printf("Maximum angular displacement is %.2f in radians which is equal to 250 degrees\n",x)
+//na1 and na are initial and final speeds of the flywheel 1 and same nb1 and nb for flywheel 2
+na=2*N-w1//w1=na1
+nb=2*N-0//nb1=0
+printf ("Speed of flywheel a and b when spring regains its unstrained position are %.2f rpm and %.2f rpm respectively\n",na,nb)