A 30kg bowling ball with a radius of 11 cm starts from rest at the top of an incline 1.5m in height. Find the translational speed of the bowling ball after it has rolled to the bottom of the incline. (assume that the ball is a uniform solid sphere) acceleration of gravity is 9.81m/s^2. Answer in units of m/s. (I=2/5mr^2) so I was thinking of treating this as a simple conservation of energy question, just substituting I and w for kinetic energy. mgh=1/2Iw^2 mgh=1/2(2/5mr^2)(v/r)^2 so: mgh=2/10mrv^2 I think that this works, but to be honest, I don't understand what translational speed is.