A cylinder weighing $450 \mathrm{N}$ with a radius of $30 \mathrm{cm}$ is held fixed on an incline that is rotating at $0.5 \mathrm{rad} s^{-1}$. The cylinder is released when the incline is at position $\mathrm{\theta }$ equal to $30°$. If the cylinder is $6 \mathrm{m}$ from the bottom $\mathrm{A}$ at the instant of release, what is the initial acceleration of the centre of the cylinder relative to the incline, if there is no slipping? $\left(\mathrm{g} = 10 \mathrm{m} {\mathrm{s}}^{-2} \right)$

• Correct answer
  A

  $2.33 \mathrm{m} {\mathrm{s}}^{-2}$

   
• B

  $4.66 \mathrm{m} {\mathrm{s}}^{-2}$

   
• C

  $1.33 \mathrm{m} {\mathrm{s}}^{-2}$

   
• D

  $3.33 \mathrm{m} {\mathrm{s}}^{-2}$

   

Solution

Free body diagram of the cylinder w.r.t. the inclined plane. The torque about the point of contact O.

$\left(\text{mg}\text{sin}\theta -\text{mr}{\omega }^2\right)R=I\text{.}\alpha =\frac{3}{2}{\text{mR}}^2\times \frac{a}{R}$    ($r$= radius of circle, $R=6 \mathrm{m}$)

$\therefore \text{mgsin}\theta -\text{mr}{\omega }^2=\frac{3}{2}\mathrm{ma}$

$\therefore a=\frac{2}{3}\left(\text{g}\text{sin}\theta -r{\omega }^2\right)=\frac{2}{3}\left[10\times \frac{1}{2}-6\times \frac{1}{4}\right]=2\text{.}33{\text{m s}}^{-2}$

Hence, the correct answer is option: 'A'.