Vansf2700

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You can also use mechanical energy conservation Delta E(mec) = E (f)-E (i) = 0 -> E (f)= E (i) ---> K (f) + U(f) = K (i) +U (i) Since it is released from rest , intial linear speed Vi= 0, and angular speed w(i) =0, and when it touches the floor h= 0, and...

Your part b can be shortened by just replacing R1 with R2 in a1 = mg/ (m + I/( R1)^2) -> a2= mg/(m+ I/(R2)^2) Instead of repeating the same steps again