Question

A clockwise torque of 6N-m is applied to the circular cylinder as shown in the figure. There is no friction between the cylinder and the block :

• A. the cylinder will be slipping but the system does not move forward
• B. the system cannot move forward for any torque applied to the cylinder
• C. the acceleration of the system will be $1m/s^2$ forward
• D. the angular acceleration of the cylinder is 10 rad $s^{-2}$

Answer 1

Answer: (C), (D)

The correct options are
C the angular acceleration of the cylinder is 10 rad $s^{-2}$
D the acceleration of the system will be $1m/s^2$ forward

$a=\frac{f_1-f_2}{m_1+m_2}=\frac{0.4\times 6g-0.5\times 3g}{6+3}$

$\Rightarrow a=1m{s}^{-2}$

$\alpha =\frac{{\tau }_{applied}-{\tau }_{friction}}{I}=\frac{6-f_{1}R}{\frac{1}{2}6(R{)}^2}$

$=\frac{6-0.4\times 6g\left(0.2\right)}{\frac{1}{2}\times 6\times (0.{)}^2}=10rad{s}^{-2}$