Answer
The speed of the solid cylinder is $~1.84~m/s$
Work Step by Step
We can find the total kinetic energy of a solid cylinder moving at a speed $v$:
$KE = KE_{tr}+KE_{rot}$
$KE = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$
$KE = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{1}{2}MR^2)~(\frac{v}{R})^2$
$KE = \frac{1}{2}Mv^2+\frac{1}{4}Mv^2$
$KE = \frac{3}{4}Mv^2$
We can use conservation of energy to find the speed of the solid cylinder:
$KE_f = U_0$
$\frac{3}{4}Mv^2 = Mgh$
$\frac{3}{4}Mv^2 = Mgd~sin~\theta$
$v^2 = \frac{4gd~sin~\theta}{3}$
$v = \sqrt{\frac{4gd~sin~\theta}{3}}$
$v = \sqrt{\frac{(4)(9.80~m/s^2)(0.30~m)~sin~60.0^{\circ}}{3}}$
$v = 1.84~m/s$
The speed of the solid cylinder is $~1.84~m/s$
