Question

The rope shown in figure wound around a cylinder of mass $4\text{kg}$ and moment of inertia $0.02{\text{kg m}}^2$ about the cylinder axis. If the cylinder rolls without slipping on a frictionless floor, then the linear acceleration of its centre of mass is -

• A. $7.6{\text{m/sec}}^2$
• B. $10{\text{m/sec}}^2$
• C. $6.7{\text{m/sec}}^2$
• D. $4.7{\text{m/sec}}^2$

Answer 1

The correct option is C $6.7{\text{m/sec}}^2$


Newton's 2nd law: $F=ma$
$(20+F)=ma$

Newton's 2nd law for rotational motion:
$\tau =I\alpha$
$(20-F)r=\frac{m{r}^2}{2}\times \frac{a}{r}$
$20-F=\frac{ma}{2}=\frac{4a}{2}$

From equation (1) and (2),
$20+F=4a$
& $20-F=2a$

Adding, $40=6a$
$\therefore a=\frac{40}{6}=\frac{20}{3}=6.67{\text{m/sec}}^2$