A bowling ball has a mass of 3.00 kg, a moment of inertia of 1.60 10-2 kgm2, and a radius of 0.100 m. If it rolls down the lane without slipping at a linear speed of 6.00 m/s, what is its total energy?
I wanted to assume that because it's rolling and moving it would have both rotational kinetic and translational kinetic energy so the formula would be:
KE = 1/2mv^2 + 1/2Iw^2
(1/2)(3)(6^2) + 1/2((1.60x10^-2)^2)(w^2)
and for w^2, I thought that I could use the tangential velocity of 6.00 m/s and set it equal to r/w. Which would give me a value of:
w= .1/6 = .0166666....
Plugging in all the numbers gives me KE = 54.0. I tried both 54 and 54000, since it wants it in J. Where did I go wrong?
2006-11-07 06:03:51 · 1 answers · asked by flossie116 4 in Science & Mathematics ➔ Physics
the math is ok. what seems to be the problem?
I think one ans is in CGS and another is in SI unit.
2006-11-07 06:09:35 · answer #1 · answered by The Potter Boy 3 · 0⤊ 0⤋