The problem is, "A metal can containing condensed mushroom soup has mass 220 g, height 11.0 cm and diameter 6.38 cm. It is placed at rest on its side at the top of a 3.00-m-long incline that is at 30.0° to the horizontal and is then released to roll straight down. It reaches the bottom of the incline after 1.50 s.
(a) Assuming mechanical energy conservation, calculate the moment of inertia of the can.
(b) Which pieces of data, if any, are unnecessary for calculating the solution?
My attempt at solving (a):
I figured that I could use the equation $\Sigma W=\Delta K=1/2I\omega_f^2-1/2I\omega_i^2$ Since the force of gravity that is acting along the incline is applied constantly over a distance, $W_g=mg\cos(60^{\circ})(3.00~m)$; and since the can rolls down the incline in 1.50 s, $v=3.00/1.5 \implies 2~m/s$, which means that $\omega_f=2/0.0319 \implies 62.695925~rad/s$ With this, and knowing that $\omega_i=0$, $mg\cos(60^{\circ})(3.00)=1/2I(62.695925)^2 \implies I=\frac{(6.00)mg\cos(60^{\circ})}{(62.695925)^2}$ When I calculated this, I got $I=0.00165~kg\cdot m^2$; however, the true answer is $I=0.000187~kg\cdot m^2$ I've re-worked my solution several times, what am I doing incorrectly?
As for (b), the answer is that the height of the can is an irrelevant piece of information, why is that?