How Does a Bowling Ball's Speed Change Down an Incline?

[BoldKnight399]

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.

 

[tiny-tim]

Hi BoldKnight399!

(have an omega: ω and try using the X² tag just above the Reply box )

… 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

No, KE of a rotating object is rotational KE plus ordinary KE (ie as if all the mass was moving with the velocity of the centre of mass) …

1/2 Iω² + 1/2 mv² ​

(And translational speed or velocity is simply the speed or velocity of the centre of mass)