Chapter 9, Problem 11MCQ

Two disks are cut from the same uniform board. The radius of disk B is twice the radius of the disk A. The disks can rotate around axes with negligible friction. Two very light battery-powered fans are attached to the disks, as shown in Figure Q9.11. When switched on, the fans exert equal forces on the disks. Which of the following correctly compares and explains the rotational accelerations of the disks after the fans are switched on?

${\alpha }_{\text{A}}=4{\alpha }_{\text{B}}\text{because}{F}_{\text{B}}=F_{\text{A}}\text{and}{m}_{\text{B}}=4m_{\text{A}}$.

${\alpha }_{\text{A}}={\alpha }_{\text{B}}\text{because}{\tau }_{\text{B}}=2{\tau }_{\text{A}}\text{and}{I}_{\text{B}}=2I_{\text{A}}$.

${\alpha }_{\text{A}}=2{\alpha }_{\text{B}}\text{because}{\tau }_{\text{B}}=2{\tau }_{\text{A}}\text{and}{I}_{\text{B}}=4I_{\text{A}}$.

${\alpha }_{\text{A}}=4{\alpha }_{\text{B}}\text{because}{\tau }_{\text{B}}=2{\tau }_{\text{A}}\text{and}{I}_{\text{B}}=8I_{\text{A}}$

${\alpha }_{\text{A}}=8{\alpha }_{\text{B}}\text{because}{\tau }_{\text{B}}=2{\tau }_{\text{A}}\text{and}{I}_{\text{B}}=16I_{\text{A}}$.