Answer
The speed at the lowest end of the board is $~2.90~m/s$
Work Step by Step
We can use conservation of energy to find the speed at the lowest end of the board:
$U_g = KE_{tr}+KE_{rot}$
$Mgh = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$
$Mgh = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{5}MR^2)~(\frac{v}{R})^2$
$Mgh = \frac{1}{2}Mv^2+\frac{1}{5}Mv^2$
$Mgh = \frac{7}{10}Mv^2$
$v^2 = \frac{10gh}{7}$
$v = \sqrt{\frac{10gh}{7}}$
$v = \sqrt{\frac{(10)(9.80~m/s^2)(0.60~m)}{7}}$
$v = 2.90~m/s$
The speed at the lowest end of the board is $~2.90~m/s$
