## College Physics (4th Edition)

The speed at the lowest end of the board is $~2.90~m/s$
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 $~2.90~m/s$