Answer
(a) $I = 0.00225~kg~m^2$
(b) $v = 3.40~m/s$
(c) $v = 4.95~m/s$
The final speed is greater when the string is farther away from the axis of rotation since it is easier for the tension in the string to rotate the disks. Because the tension in the string decreases, the final speed of the block increase
Work Step by Step
(a) $I = \frac{1}{2}M_1R_1^2+\frac{1}{2}M_2R_2^2$
$I = \frac{1}{2}(0.80~kg)(0.0250~m)^2+\frac{1}{2}(1.60~kg)(0.0500~m)^2$
$I = 0.00225~kg~m^2$
(b) The kinetic energy of the block and the disks will be equal to the magnitude of the change in potential energy.
$KE = PE$
$\frac{1}{2}mv^2+\frac{1}{2}I\omega^2 = mgh$
$mv^2+I(\frac{v}{R_1})^2 = 2mgh$
$v^2 = \frac{2mgh}{m+\frac{I}{R_1^2}}$
$v = \sqrt{\frac{2mgh}{m+\frac{I}{R_1^2}}}$
$v = \sqrt{\frac{(2)(1.50~kg)(9.80~m/s^2)(2.00~m)}{1.50~kg+\frac{0.00225~kg~m^2}{(0.0250~m)^2}}}$
$v = 3.40~m/s$
(c) The kinetic energy of the block and the disks will be equal to the magnitude of the change in potential energy.
$KE = PE$
$\frac{1}{2}mv^2+\frac{1}{2}I\omega^2 = mgh$
$mv^2+I(\frac{v}{R_2})^2 = 2mgh$
$v^2 = \frac{2mgh}{m+\frac{I}{R_2^2}}$
$v = \sqrt{\frac{2mgh}{m+\frac{I}{R_2^2}}}$
$v = \sqrt{\frac{(2)(1.50~kg)(9.80~m/s^2)(2.00~m)}{1.50~kg+\frac{0.00225~kg~m^2}{(0.0500~m)^2}}}$
$v = 4.95~m/s$
The final speed is greater when the string is farther away from the axis of rotation since it is easier for the tension in the string to rotate the disks. Because the tension in the string decreases, the final speed of the block increases.
