Question

A force $F=12\mu mg$ acts tangentially at the given point of the disc which is at a distance of $\frac{2}{3}R$ from centre of disc as shown in figure. Disc is kept on a rough horizontal plane of coefficient of friction $\mu$ . The angular acceleration of the disc is (mass of the disc is $m$ and its radius is $R$ )

• A. $\frac{40\mu g}{3R}$
• B. $\frac{18\mu g}{R}$
• C. $\frac{20\mu g}{3R}$
• D. $\frac{14\mu g}{R}$

Answer 1

Answer: (D)

$r=\frac{I_{cm}}{mR}=\frac{m{R}^2}{mR}=\frac{R}{2}$
But force acts above to it so friction on the disc acts along the applied force. Torque about centre of mass $\frac{m{R}^2\alpha }{2}=12\mu mg\left(\frac{2}{3}R\right)-\mu mgR=7\mu mgR$
$\alpha =\frac{14\mu g}{R}$