Question

The figure shows a spool with thread wound on it placed on a smooth plane inclined at angle from horizontal. The spool has mass $m$, edge radius $R$ and is wound up to a radius $r$, its moment of inertia about its own axis is $I$. The free end of the thread is attached as shown in figure. So that the thread is parallel to the inclined plane. $T$ is the tension in the thread. Which of the following is correct?

• A. The linear acceleration of the spool axis down the slope is $\frac{mg\sin \theta -T}{m}$
• B. Angular accelerations is $\frac{Tr}{2I}$
• C. The linear acceleration of the spool axis down the plane is $\frac{T{r}^2}{I}$
• D. The acceleration of the spool axis down the slope is $\frac{g\sin \theta }{1+\frac{1}{m{r}^2}}$

Answer 1

Answer: (A), (C), (D)

The correct options are
A The linear acceleration of the spool axis down the slope is $\frac{mg\sin \theta -T}{m}$
C The linear acceleration of the spool axis down the plane is $\frac{T{r}^2}{I}$
D The acceleration of the spool axis down the slope is $\frac{g\sin \theta }{1+\frac{1}{m{r}^2}}$

$a=r\alpha$ (no slipping)

$ma=mgsin\theta -T.......\left(1\right)⟹a=\frac{mgsin\theta -T}{m}$

Also, ${\tau }_o=I\alpha$

$Tr=\frac{Ia}{r}⟹a=\frac{T{r}^2}{I}$ and $\alpha =\frac{Tr}{I}$

Now, $ma=mgsin\theta -\frac{Ia}{r^2}$ [from 1]

$⟹a=\frac{gsin\theta }{1+\frac{I}{m{r}^2}}$