A hollow cylinder of mass m1, filled with water of mass m2, is pure rolling on a sufficiently rough surface with a speed v. What is the total kinetic energy possessed by the cylinder+water system?
Options:
(a) 0.5*(m1+m2)*v2
(b) 0.5*(2m1+m2)*v2
(c) 0.5*(2m1+1.5m2)*v2
(d) 0.5*(m1+1.5m2)*v2
62oops sorry was it not for me.. but for other users..
I will delete my reply..
33answer is clearly (b)....
given that between fluid and cylinder there is no friction..;)
water having energy by translation only = 0.5(m2)v2
energy of cylinder=0.5(2m1R2)ω2.. but ω=v/R
therefore energy of cylinder=0.5(2m1)v2
total energy=0.5(m2)v2+0.5(2m1)v2
=0.5(2m1+m2)v2
:)
1@priyam
even if the cylinder's inner surface was rough, the answer would not change. Quoting from H C Verma, "a liquid cannot bear shearing stress",i.e.,liquids cannot bear frictional forces.
