Question

The figure shows a uniform smooth solid cylinder A of radius 4 m rolling without slipping on the 8 kg plank which in turn is supported by a fixed smooth surface. Block B, is known to accelerate down with $6m/s^2$.
(Take $g=10m/s^2$)

What is the angular acceleration of the cylinder ?

• A. $\frac{4}{5}rad{s}^{-2}$
• B. $\frac{6}{5}rad{s}^{-2}$
• C. $2rad{s}^{-2}$
• D. $1rad{s}^{-2}$

Answer 1

The correct option is C $2rad{s}^{-2}$

Acceleration of the bottom of the cylinder = acceleration of plank = $2m{s}^{-2}$ towards right
Acceleration of the top of the cylinder = acceleration of $B=6m{s}^{-2}$ towards left
$\therefore a+R\alpha =6$ and $R\alpha -a=2$
$\Rightarrow \alpha =1rad{s}^{-2},a=2m{s}^{-2}$.