# Rotational Motion FInding the total ENergy

by Schu
Tags: energy, motion, rotational
 P: 12 Particulars: ball has a radius of 2.5 cm a mass of .125 and is rolling across a table with a speed of .547 m/s, this table is 1.04 m off the ground. It rolls to the edge and down a ramp How fast will it be rolling across the floor? First I found the Gravitational Potential Energy: Ep=mgh Initial of 1.2753 FInal = 0 THen the Linear Kinetic ENergy : 1/2 mv^2 Initial .0187005625 FInal .0625v^2 Elastic Potential Energy: .5k(delta)x^2 0 0 Rotational Kinetic Energy: 1/5mv^2 initial .007480225 FInal .025v^2 Now I need to bring them all togther and solve the final velocity. Is the Sum of the inital energy's = to the SUM of the final energy's? If that's true then 1.30148075 = .0875v^2 so v = 3.85 m/s Is that at all right??
 P: 12 Is anyone out there???? I would appreciate the help
 Sci Advisor HW Helper P: 2,002 Looks ok to me. (I got 3.86 m/s, by rounding off)
Mentor
P: 41,304
Rotational Motion FInding the total ENergy

The rotational KE is ${KE}_{rot} = 1/2 I \omega^2$.
You will also need the "rolling condition": $V = \omega R$.