Question

a circular hoop and two solid disks are placed on a horizontal surface where they can move with negligible friction. the hoop and disks are set into motion such that each moves forward and rotates about its center of mass, as shown in the top view figure. the hoop and disks have the same mass and angular speeds. the radii, rotational inertia, and translational speeds of the hoop and disks are listed in the table. which of the following correctly compares the total kinetic energies ( k_{\text{hoop}} ), ( k_{\text{disk 1}} ), and ( k_{\text{disk 2}} ) of the hoop, disk 1, and disk 2, respectively?
a ( k_{\text{disk 2}} > k_{\text{disk 1}} = k_{\text{hoop}} )
b ( k_{\text{disk 2}} > k_{\text{hoop}} > k_{\text{disk 1}} )
c ( k_{\text{hoop}} > k_{\text{disk 1}} > k_{\text{disk 2}} )

Explanation:

Step1: Recall total kinetic energy formula

Total kinetic energy $K = K_{trans} + K_{rot} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

Step2: Calculate $K_{Hoop}$

Substitute $I=I_0$, $v=v_0$, $\omega=\omega_0$, and $I_0 = mr_0^2$, $v_0 = r_0\omega_0$:
$K_{Hoop} = \frac{1}{2}mv_0^2 + \frac{1}{2}I_0\omega_0^2 = \frac{1}{2}mv_0^2 + \frac{1}{2}(mr_0^2)(\frac{v_0}{r_0})^2 = \frac{1}{2}mv_0^2 + \frac{1}{2}mv_0^2 = mv_0^2$

Step3: Calculate $K_{Disk 1}$

Substitute $I=\frac{1}{2}I_0$, $v=v_0$, $\omega=\omega_0$:
$K_{Disk 1} = \frac{1}{2}mv_0^2 + \frac{1}{2}(\frac{1}{2}I_0)\omega_0^2 = \frac{1}{2}mv_0^2 + \frac{1}{2}(\frac{1}{2}mr_0^2)(\frac{v_0}{r_0})^2 = \frac{1}{2}mv_0^2 + \frac{1}{4}mv_0^2 = \frac{3}{4}mv_0^2$

Step4: Calculate $K_{Disk 2}$

Substitute $I=\frac{1}{8}I_0$, $v=2v_0$, $\omega=\omega_0$, and $I_0 = mr_0^2$, $v_0 = r_0\omega_0$ (so $2v_0 = 2r_0\omega_0$, and $\omega_0 = \frac{2v_0}{2r_0} = \frac{2v_0}{r_0'}$ where $r_0'=\frac{1}{2}r_0$):
$K_{Disk 2} = \frac{1}{2}m(2v_0)^2 + \frac{1}{2}(\frac{1}{8}I_0)\omega_0^2 = \frac{1}{2}m(4v_0^2) + \frac{1}{2}(\frac{1}{8}mr_0^2)(\frac{2v_0}{r_0})^2 = 2mv_0^2 + \frac{1}{2}(\frac{1}{8}mr_0^2)(\frac{4v_0^2}{r_0^2}) = 2mv_0^2 + \frac{1}{4}mv_0^2 = \frac{9}{4}mv_0^2$

Step5: Compare the values

$\frac{9}{4}mv_0^2 > mv_0^2 > \frac{3}{4}mv_0^2$, so $K_{Disk 2} > K_{Hoop} > K_{Disk 1}$

Answer:

B. $K_{\text{Disk 2}} > K_{\text{hoop}} > K_{\text{Disk 1}}$